10 - Rotation of a Rigid Object About a Fixed Axis

10 - Rotation of a Rigid Object About a Fixed Axis -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PUZZLER Did you know that the CD inside this player spins at different speeds, depending on which song is playing? Why would such a strange characteristic be incorporated into the design of every CD player? (George Semple) chapter Rotation of a Rigid Object About a Fixed Axis Chapter Outline 10.1 Angular Displacement, Velocity, and Acceleration 10.2 Rotational Kinematics: Rotational Motion with Constant Angular Acceleration 10.3 Angular and Linear Quantities 10.4 Rotational Energy 292 10.5 Calculation of Moments of Inertia 10.6 Torque 10.7 Relationship Between Torque and Angular Acceleration 10.8 Work, Power, and Energy in Rotational Motion 293 10.1 Angular Displacement, Velocity, and Acceleration W hen an extended object, such as a wheel, rotates about its axis, the motion cannot be analyzed by treating the object as a particle because at any given time different parts of the object have different linear velocities and linear accelerations. For this reason, it is convenient to consider an extended object as a large number of particles, each of which has its own linear velocity and linear acceleration. In dealing with a rotating object, analysis is greatly simplified by assuming that the object is rigid. A rigid object is one that is nondeformable — that is, it is an object in which the separations between all pairs of particles remain constant. All real bodies are deformable to some extent; however, our rigid-object model is useful in many situations in which deformation is negligible. In this chapter, we treat the rotation of a rigid object about a fixed axis, which is commonly referred to as pure rotational motion. 10.1 Rigid object ANGULAR DISPLACEMENT, VELOCITY, AND ACCELERATION Figure 10.1 illustrates a planar (flat), rigid object of arbitrary shape confined to the xy plane and rotating about a fixed axis through O. The axis is perpendicular to the plane of the figure, and O is the origin of an xy coordinate system. Let us look at the motion of only one of the millions of “particles” making up this object. A particle at P is at a fixed distance r from the origin and rotates about it in a circle of radius r. (In fact, every particle on the object undergoes circular motion about O.) It is convenient to represent the position of P with its polar coordinates (r, ), where r is the distance from the origin to P and is measured counterclockwise from some preferred direction — in this case, the positive x axis. In this representation, the only coordinate that changes in time is the angle ; r remains constant. (In cartesian coordinates, both x and y vary in time.) As the particle moves along the circle from the positive x axis ( 0) to P, it moves through an arc of length s, which is related to the angular position through the relationship r (10.1a) s r s (10.1b) y P r θ O Because the circumference of a circle is 2 r, it follows from Equation 10.1b that 360° corresponds to an angle of 2 r/r rad 2 rad (one revolution). Hence, 1 rad 360°/2 57.3°. To convert an angle in degrees to an angle in radians, we use the fact that 2 rad 360°: (rad) 180° (deg) For example, 60° equals /3 rad, and 45° equals /4 rad. x Figure 10.1 A rigid object rotating about a fixed axis through O perpendicular to the plane of the figure. (In other words, the axis of rotation is the z axis.) A particle at P rotates in a circle of radius r centered at O. It is important to note the units of in Equation 10.1b. Because is the ratio of an arc length and the radius of the circle, it is a pure number. However, we commonly give the artificial unit radian (rad), where one radian is the angle subtended by an arc length equal to the radius of the arc. s Radian 294 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis y Q ,t f r P, ti θf In a short track event, such as a 200-m or 400-m sprint, the runners begin from staggered positions on the track. Why don’t they all begin from the same line? θi x O Figure 10.2 A particle on a rotating rigid object moves from P to Q along the arc of a circle. In the time interval t t f t i , the radius vector sweeps out an angle f i. As the particle in question on our rigid object travels from position P to position Q in a time t as shown in Figure 10.2, the radius vector sweeps out an angle f is defined as the angular displacement of the particle: i . This quantity f We define the average angular speed placement to the time interval t : (omega) as the ratio of this angular dis- f i tf Average angular speed (10.2) i ti t In analogy to linear speed, the instantaneous angular speed the limit of the ratio / t as t approaches zero: lim Instantaneous angular speed t t:0 d dt (10.3) is defined as (10.4) Angular speed has units of radians per second (rad/s), or rather second 1 (s because radians are not dimensional. We take to be positive when is increasing (counterclockwise motion) and negative when is decreasing (clockwise motion). If the instantaneous angular speed of an object changes from i to f in the time interval t, the object has an angular acceleration. The average angular acceleration (alpha) of a rotating object is defined as the ratio of the change in the angular speed to the time interval t : 1) Average angular acceleration f tf i ti t (10.5) 295 10.1 Angular Displacement, Velocity, and Acceleration ω Figure 10.3 The right-hand rule for determining the direction of the angular velocity vector. ω In analogy to linear acceleration, the instantaneous angular acceleration is defined as the limit of the ratio / t as t approaches zero: lim t:0 t d dt (10.6) Angular acceleration has units of radians per second squared (rad/s2 ), or just second 2 (s 2 ). Note that is positive when the rate of counterclockwise rotation is increasing or when the rate of clockwise rotation is decreasing. When rotating about a fixed axis, every particle on a rigid object rotates through the same angle and has the same angular speed and the same angular acceleration. That is, the quantities , , and characterize the rotational motion of the entire rigid object. Using these quantities, we can greatly simplify the analysis of rigid-body rotation. Angular position ( ), angular speed ( ), and angular acceleration ( ) are analogous to linear position (x), linear speed (v), and linear acceleration (a). The variables , , and differ dimensionally from the variables x, v, and a only by a factor having the unit of length. We have not specified any direction for and . Strictly speaking, these variables are the magnitudes of the angular velocity and the angular acceleration vectors and , respectively, and they should always be positive. Because we are considering rotation about a fixed axis, however, we can indicate the directions of the vectors by assigning a positive or negative sign to and , as discussed earlier with regard to Equations 10.4 and 10.6. For rotation about a fixed axis, the only direction that uniquely specifies the rotational motion is the direction along the axis of rotation. Therefore, the directions of and are along this axis. If an object rotates in the xy plane as in Figure 10.1, the direction of is out of the plane of the diagram when the rotation is counterclockwise and into the plane of the diagram when the rotation is clockwise. To illustrate this convention, it is convenient to use the right-hand rule demonstrated in Figure 10.3. When the four fingers of the right hand are wrapped in the direction of rotation, the extended right thumb points in the direction of . The direction of follows from its definition d /dt. It is the same as the direction of if the angular speed is increasing in time, and it is antiparallel to if the angular speed is decreasing in time. Instantaneous angular acceleration 296 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis Quick Quiz 10.1 Describe a situation in which 10.2 7.2 0 and and are antiparallel. ROTATIONAL KINEMATICS: ROTATIONAL MOTION WITH CONSTANT ANGULAR ACCELERATION In our study of linear motion, we found that the simplest form of accelerated motion to analyze is motion under constant linear acceleration. Likewise, for rotational motion about a fixed axis, the simplest accelerated motion to analyze is motion under constant angular acceleration. Therefore, we next develop kinematic relationships for this type of motion. If we write Equation 10.6 in the form d dt, and let ti 0 and tf t, we can integrate this expression directly: f (for constant ) t i (10.7) Substituting Equation 10.7 into Equation 10.4 and integrating once more we obtain Rotational kinematic equations f 1 2 it i t2 (for constant ) (10.8) If we eliminate t from Equations 10.7 and 10.8, we obtain f 2 i 2 2( i) f (for constant ) (10.9) Notice that these kinematic expressions for rotational motion under constant angular acceleration are of the same form as those for linear motion under constant linear acceleration with the substitutions x : , v : , and a : . Table 10.1 compares the kinematic equations for rotational and linear motion. EXAMPLE 10.1 Rotating Wheel A wheel rotates with a constant angular acceleration of 3.50 rad/s2. If the angular speed of the wheel is 2.00 rad/s at ti 0, (a) through what angle does the wheel rotate in 2.00 s? Solution We can use Figure 10.2 to represent the wheel, and so we do not need a new drawing. This is a straightforward application of an equation from Table 10.1: f i 1 2 it 1 2 t2 (2.00 rad/s)(2.00 s) (3.50 rad/s2)(2.00 s)2 11.0 rad 630° 360°/rev (11.0 rad)(57.3°/rad) 630° Solution Because the angular acceleration and the angular speed are both positive, we can be sure our answer must be greater than 2.00 rad/s. f i t We could also obtain this result using Equation 10.9 and the results of part (a). Try it! You also may want to see if you can formulate the linear motion analog to this problem. Exercise Find the angle through which the wheel rotates 2.00 s and t 3.00 s. 1.75 rev Answer 2.00 s? (3.50 rad/s2)(2.00 s) 9.00 rad/s between t (b) What is the angular speed at t 2.00 rad/s 10.8 rad. 297 10.3 Angular and Linear Quantities TABLE 10.1 Kinematic Equations for Rotational and Linear Motion Under Constant Acceleration Rotational Motion About a Fixed Axis f f 10.3 f 2 t it i i i 2 2( 1 2 f Linear Motion vf xf t2 i) vf 2 vi xi vi at vit 2 2a (xf 1 2 at 2 xi ) y ANGULAR AND LINEAR QUANTITIES In this section we derive some useful relationships between the angular speed and acceleration of a rotating rigid object and the linear speed and acceleration of an arbitrary point in the object. To do so, we must keep in mind that when a rigid object rotates about a fixed axis, as in Figure 10.4, every particle of the object moves in a circle whose center is the axis of rotation. We can relate the angular speed of the rotating object to the tangential speed of a point P on the object. Because point P moves in a circle, the linear velocity vector v is always tangent to the circular path and hence is called tangential velocity. The magnitude of the tangential velocity of the point P is by definition the tangential speed v ds/dt, where s is the distance traveled by this point measured along the circular path. Recalling that s r ( Eq. 10.1a) and noting that r is constant, we obtain v Because d /dt ds dt r d dt v P r θ O x Figure 10.4 As a rigid object rotates about the fixed axis through O, the point P has a linear velocity v that is always tangent to the circular path of radius r. ( see Eq. 10.4), we can say v (10.10) r That is, the tangential speed of a point on a rotating rigid object equals the perpendicular distance of that point from the axis of rotation multiplied by the angular speed. Therefore, although every point on the rigid object has the same angular speed, not every point has the same linear speed because r is not the same for all points on the object. Equation 10.10 shows that the linear speed of a point on the rotating object increases as one moves outward from the center of rotation, as we would intuitively expect. The outer end of a swinging baseball bat moves much faster than the handle. We can relate the angular acceleration of the rotating rigid object to the tangential acceleration of the point P by taking the time derivative of v : at dv dt at r r d dt (10.11) That is, the tangential component of the linear acceleration of a point on a rotating rigid object equals the point’s distance from the axis of rotation multiplied by the angular acceleration. Relationship between linear and angular speed QuickLab Spin a tennis ball or basketball and watch it gradually slow down and stop. Estimate and at as accurately as you can. Relationship between linear and angular acceleration 298 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis In Section 4.4 we found that a point rotating in a circular path undergoes a centripetal, or radial, acceleration ar of magnitude v 2/r directed toward the center of rotation (Fig. 10.5). Because v r for a point P on a rotating object, we can express the radial acceleration of that point as y at a ar x O v2 r ar P r 2 (10.12) The total linear acceleration vector of the point is a at ar . (at describes the change in how fast the point is moving, and ar represents the change in its direction of travel.) Because a is a vector having a radial and a tangential component, the magnitude of a for the point P on the rotating rigid object is a √a t 2 a r2 √r 2 2 r2 4 r √ 2 4 (10.13) Figure 10.5 As a rigid object rotates about a fixed axis through O, the point P experiences a tangential component of linear acceleration at and a radial component of linear acceleration ar . The total linear acceleration of this point is a at ar . EXAMPLE 10.2 Quick Quiz 10.2 When a wheel of radius R rotates about a fixed axis, do all points on the wheel have (a) the same angular speed and (b) the same linear speed? If the angular speed is constant and equal to , describe the linear speeds and linear accelerations of the points located at (c) r 0, (d) r R/2, and (e) r R, all measured from the center of the wheel. CD Player On a compact disc, audio information is stored in a series of pits and flat areas on the surface of the disc. The information is stored digitally, and the alternations between pits and flat areas on the surface represent binary ones and zeroes to be read by the compact disc player and converted back to sound waves. The pits and flat areas are detected by a system consisting of a laser and lenses. The length of a certain number of ones and zeroes is the same everywhere on the disc, whether the information is near the center of the disc or near its outer edge. In order that this length of ones and zeroes always passes by the laser – lens system in the same time period, the linear speed of the disc surface at the location of the lens must be constant. This requires, according to Equation 10.10, that the angular speed vary as the laser – lens system moves radially along the disc. In a typical compact disc player, the disc spins counterclockwise (Fig. 10.6), and the constant speed of the surface at the point of the laser – lens system is 1.3 m/s. (a) Find the angular speed of the disc in revolutions per minute when information is being read from the innermost first track (r 23 mm) and the outermost final track (r 58 mm). 1 (56.5 rad/s) 2 rev/rad (60 s/min) 5.4 10 2 rev/min 23 mm 58 mm Solution Using Equation 10.10, we can find the angular speed; this will give us the required linear speed at the position of the inner track, i v ri 1.3 m/s 2.3 10 2 m 56.5 rad/s Figure 10.6 A compact disc. 299 10.4 Rotational Energy (c) What total length of track moves past the objective lens during this time? For the outer track, v rf 1.3 m/s 5.8 10 2 m 2.1 f 10 2 rev/min 22.4 rad/s Solution Because we know the (constant) linear velocity and the time interval, this is a straightforward calculation: xf (1.3 m/s)(4 473 s) v it 5.8 10 3 m The player adjusts the angular speed of the disc within this range so that information moves past the objective lens at a constant rate. These angular velocity values are positive because the direction of rotation is counterclockwise. More than 3.6 miles of track spins past the objective lens! (b) The maximum playing time of a standard music CD is 74 minutes and 33 seconds. How many revolutions does the disc make during that time? Solution Solution We know that the angular speed is always decreasing, and we assume that it is decreasing steadily, with constant. The time interval t is (74 min)(60 s/min) 33 s 4 473 s. We are looking for the angular position f , where we set the initial angular position i 0. We can use Equation 10.3, replacing the average angular speed with its mathematical equivalent ( i f )/2: f i 0 1 2( i 1 2 (540 We have several choices for approaching this problem. Let us use the most direct approach by utilizing Equation 10.5, which is based on the definition of the term we are seeking. We should obtain a negative number for the angular acceleration because the disc spins more and more slowly in the positive direction as time goes on. Our answer should also be fairly small because it takes such a long time — more than an hour — for the change in angular speed to be accomplished: f f )t rev/min 22.4 rad/s 56.5 rad/s 4 473 s i t 210 rev/min) 7.6 (1 min/60 s)(4 473 s) 2.8 (d) What is the angular acceleration of the CD over the 4 473-s time interval? Assume that is constant. 10 3 rad/s2 The disc experiences a very gradual decrease in its rotation rate, as expected. 10 4 rev web 10.4 7.3 R OTATIONAL ENERGY Let us now look at the kinetic energy of a rotating rigid object, considering the object as a collection of particles and assuming it rotates about a fixed z axis with an angular speed ( Fig. 10.7). Each particle has kinetic energy determined by its mass and linear speed. If the mass of the i th particle is mi and its linear speed is vi , its kinetic energy is To proceed further, we must recall that although every particle in the rigid object has the same angular speed , the individual linear speeds depend on the distance ri from the axis of rotation according to the expression vi ri ( see Eq. 10.10). The total kinetic energy of the rotating rigid object is the sum of the kinetic energies of the individual particles: Ki i i 1 2 2m i v i 1 2 mi ri2 where we have factored 2 1 2 vi m i ri 2 2 mi ri θ O x 2 i We can write this expression in the form KR y 1 2 2 m iv i Ki KR If you want to learn more about the physics of CD players, visit the Special Interest Group on CD Applications and Technology at www.sigcat.org (10.14) i from the sum because it is common to every particle. Figure 10.7 A rigid object rotating about a z axis with angular speed . The kinetic energy of the particle of mass mi is 1 m iv i 2. 2 The total kinetic energy of the object is called its rotational kinetic energy. 300 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis We simplify this expression by defining the quantity in parentheses as the moment of inertia I : Moment of inertia mi ri 2 I (10.15) i From the definition of moment of inertia, we see that it has dimensions of ML2 (kg m2 in SI units).1 With this notation, Equation 10.14 becomes Rotational kinetic energy KR 12 2I (10.16) Although we commonly refer to the quantity 1I 2 as rotational kinetic energy, 2 it is not a new form of energy. It is ordinary kinetic energy because it is derived from a sum over individual kinetic energies of the particles contained in the rigid object. However, the mathematical form of the kinetic energy given by Equation 10.16 is a convenient one when we are dealing with rotational motion, provided we know how to calculate I. It is important that you recognize the analogy between kinetic energy associated with linear motion 1mv 2 and rotational kinetic energy 1 I 2. The quantities I 2 2 and in rotational motion are analogous to m and v in linear motion, respectively. (In fact, I takes the place of m every time we compare a linear-motion equation with its rotational counterpart.) The moment of inertia is a measure of the resistance of an object to changes in its rotational motion, just as mass is a measure of the tendency of an object to resist changes in its linear motion. Note, however, that mass is an intrinsic property of an object, whereas I depends on the physical arrangement of that mass. Can you think of a situation in which an object’s moment of inertia changes even though its mass does not? EXAMPLE 10.3 The Oxygen Molecule Consider an oxygen molecule (O2 ) rotating in the xy plane about the z axis. The axis passes through the center of the molecule, perpendicular to its length. The mass of each oxygen atom is 2.66 10 26 kg, and at room temperature the average separation between the two atoms is d 1.21 10 10 m (the atoms are treated as point masses). (a) Calculate the moment of inertia of the molecule about the z axis. This is a very small number, consistent with the minuscule masses and distances involved. Solution Solution This is a straightforward application of the definition of I. Because each atom is a distance d/2 from the z axis, the moment of inertia about the axis is mi ri 2 I m i 1 2 (2.66 10 d 2 26 2 m kg)(1.21 1 d 2 1 2 2 md 10 m)2 10 46 kg m2 (b) If the angular speed of the molecule about the z axis is 4.60 1012 rad/s, what is its rotational kinetic energy? We apply the result we just calculated for the moment of inertia in the formula for K R : KR 2 10 1.95 1 2 2I 1 2 (1.95 10 46 kg m2)(4.60 2.06 10 21 J 1012 rad/s)2 Civil engineers use moment of inertia to characterize the elastic properties (rigidity) of such structures as loaded beams. Hence, it is often useful even in a nonrotational context. 301 10.5 Calculation of Moments of Inertia EXAMPLE 10.4 Four Rotating Masses Four tiny spheres are fastened to the corners of a frame of negligible mass lying in the xy plane (Fig. 10.8). We shall assume that the spheres’ radii are small compared with the dimensions of the frame. (a) If the system rotates about the y axis with an angular speed , find the moment of inertia and the rotational kinetic energy about this axis. Solution First, note that the two spheres of mass m, which lie on the y axis, do not contribute to Iy (that is, ri 0 for these spheres about this axis). Applying Equation 10.15, we obtain m i ri 2 Iy Ma 2 Ma 2 Therefore, the rotational kinetic energy about the y axis is KR 1 2 Iy 2 1 2 2 (2Ma ) 2 Ma 2 2 The fact that the two spheres of mass m do not enter into this result makes sense because they have no motion about the axis of rotation; hence, they have no rotational kinetic energy. By similar logic, we expect the moment of inertia about the x axis to be Ix 2mb 2 with a rotational kinetic energy about that axis of K R mb 2 2. (b) Suppose the system rotates in the xy plane about an axis through O (the z axis). Calculate the moment of inertia and rotational kinetic energy about this axis. 2Ma 2 i y Solution Because ri in Equation 10.15 is the perpendicular distance to the axis of rotation, we obtain m mi ri 2 Iz Ma 2 Ma 2 mb 2 mb 2 2Ma 2 2mb 2 i b M a M a O KR m Figure 10.8 The four spheres are at a fixed separation as shown. The moment of inertia of the system depends on the axis about which it is evaluated. 7.5 2 1 2 2 (2Ma We can evaluate the moment of inertia of an extended rigid object by imagining the object divided into many small volume elements, each of which has mass m. r i 2 m i and take the limit of this sum as m : 0. In We use the definition I i this limit, the sum becomes an integral over the whole object: lim mi :0 i ri 2 mi 2 (Ma 2 mb 2) 2 Comparing the results for parts (a) and (b), we conclude that the moment of inertia and therefore the rotational kinetic energy associated with a given angular speed depend on the axis of rotation. In part (b), we expect the result to include all four spheres and distances because all four spheres are rotating in the xy plane. Furthermore, the fact that the rotational kinetic energy in part (a) is smaller than that in part (b) indicates that it would take less effort (work) to set the system into rotation about the y axis than about the z axis. CALCULATION OF MOMENTS OF INERTIA I 2mb 2) x b 10.5 1 2 Iz r 2 dm (10.17) It is usually easier to calculate moments of inertia in terms of the volume of the elements rather than their mass, and we can easily make that change by using m/V, where is the density of the object and V is its volume. We Equation 1.1, want this expression in its differential form dm/dV because the volumes we are dealing with are very small. Solving for dm dV and substituting the result 302 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis into Equation 10.17 gives r 2 dV I If the object is homogeneous, then is constant and the integral can be evaluated for a known geometry. If is not constant, then its variation with position must be known to complete the integration. The density given by m/V sometimes is referred to as volume density for the obvious reason that it relates to volume. Often we use other ways of expressing density. For instance, when dealing with a sheet of uniform thickness t, we can define a surface density t, which signifies mass per unit area. Finally, when mass is distributed along a uniform rod of cross-sectional area A, we sometimes use linear density M/L A, which is the mass per unit length. EXAMPLE 10.5 Uniform Hoop y Find the moment of inertia of a uniform hoop of mass M and radius R about an axis perpendicular to the plane of the hoop and passing through its center (Fig. 10.9). dm Solution All mass elements dm are the same distance r R from the axis, and so, applying Equation 10.17, we obtain for the moment of inertia about the z axis through O : Iz r 2 dm R2 dm O x R MR 2 Note that this moment of inertia is the same as that of a single particle of mass M located a distance R from the axis of rotation. Figure 10.9 The mass elements dm of a uniform hoop are all the same distance from O. Quick Quiz 10.3 (a) Based on what you have learned from Example 10.5, what do you expect to find for the moment of inertia of two particles, each of mass M/2, located anywhere on a circle of radius R around the axis of rotation? (b) How about the moment of inertia of four particles, each of mass M/4, again located a distance R from the rotation axis? EXAMPLE 10.6 Uniform Rigid Rod Calculate the moment of inertia of a uniform rigid rod of length L and mass M (Fig. 10.10) about an axis perpendicular to the rod (the y axis) and passing through its center of mass. Solution The shaded length element dx has a mass dm equal to the mass per unit length multiplied by dx : dm dx M dx L Substituting this expression for dm into Equation 10.17, with r x, we obtain L/2 Iy r 2 dm x2 L/2 M L x3 3 L/2 L/2 M dx L 1 2 12 ML M L L/2 x 2 dx L/2 303 10.5 Calculation of Moments of Inertia y′ y dx x O x Figure 10.10 A uniform rigid rod of length L . The moment of inertia about the y axis is less than that about the y axis. The latter axis is examined in Example 10.8. L EXAMPLE 10.7 Uniform Solid Cylinder A uniform solid cylinder has a radius R, mass M, and length L. Calculate its moment of inertia about its central axis (the z axis in Fig. 10.11). Solution It is convenient to divide the cylinder into many z cylindrical shells, each of which has radius r, thickness dr, and length L, as shown in Figure 10.11. The volume dV of each shell is its cross-sectional area multiplied by its length: dV dA L (2 r dr )L. If the mass per unit volume is , then the mass of this differential volume element is dm dV 2 r L dr. Substituting this expression for dm into Equation 10.17, we obtain R dr Iz r 2 dm 2 r 3 dr L 0 r 1 2 LR 4 Because the total volume of the cylinder is R 2L, we see that M/V M/ R 2L. Substituting this value for into the above result gives R L (1) Figure 10.11 Calculating I about the z axis for a uniform solid cylinder. Iz 1 2 2 MR Note that this result does not depend on L, the length of the cylinder. In other words, it applies equally well to a long cylinder and a flat disc. Also note that this is exactly half the value we would expect were all the mass concentrated at the outer edge of the cylinder or disc. (See Example 10.5.) Table 10.2 gives the moments of inertia for a number of bodies about specific axes. The moments of inertia of rigid bodies with simple geometry (high symmetry) are relatively easy to calculate provided the rotation axis coincides with an axis of symmetry. The calculation of moments of inertia about an arbitrary axis can be cumbersome, however, even for a highly symmetric object. Fortunately, use of an important theorem, called the parallel-axis theorem, often simplifies the calculation. Suppose the moment of inertia about an axis through the center of mass of an object is ICM . The parallel-axis theorem states that the moment of inertia about any axis parallel to and a distance D away from this axis is I ICM MD 2 (10.18) Parallel-axis theorem 304 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis TABLE 10.2 Moments of Inertia of Homogeneous Rigid Bodies with Different Geometries Hoop or cylindrical shell I CM = MR 2 Hollow cylinder I CM = 1 M(R 12 + R 22) 2 R Solid cylinder or disk I CM = 1 MR 2 2 R1 R2 Rectangular plate I CM = 1 M(a 2 + b 2) 12 R b a Long thin rod with rotation axis through center I CM = 1 ML 2 12 Long thin rod with rotation axis through end L I = 1 ML 2 3 Solid sphere I CM = 2 MR 2 5 L Thin spherical shell I CM = 2 MR 2 3 R R Proof of the Parallel-Axis Theorem (Optional). Suppose that an object rotates in the xy plane about the z axis, as shown in Figure 10.12, and that the coordinates of the center of mass are x CM , y CM . Let the mass element dm have coordinates x, y. Because this element is a distance r √x 2 y 2 from the z axis, the moment of inertia about the z axis is I r 2 dm (x 2 y 2) dm However, we can relate the coordinates x, y of the mass element dm to the coordinates of this same element located in a coordinate system having the object’s center of mass as its origin. If the coordinates of the center of mass are x CM , y CM in the original coordinate system centered on O, then from Figure 10.12a we see that the relationships between the unprimed and primed coordinates are x x x CM 305 10.5 Calculation of Moments of Inertia y dm x, y z y′ Axis through CM Rotation axis r y y CM xCM, yCM O yCM CM D x O x x′ xCM x (a) (b) Figure 10.12 (a) The parallel-axis theorem: If the moment of inertia about an axis perpendicular to the figure through the center of mass is ICM , then the moment of inertia about the z axis is Iz I CM MD 2. (b) Perspective drawing showing the z axis (the axis of rotation) and the parallel axis through the CM. and y I y [(x [(x )2 y CM . Therefore, x CM)2 (y (y )2] dm y CM)2] dm 2x CM x dm 2y CM y dm (x CM2 y CM 2) dm The first integral is, by definition, the moment of inertia about an axis that is parallel to the z axis and passes through the center of mass. The second two integrals are zero because, by definition of the center of mass, x dm y dm 0. The last integral is simply MD 2 because dm M and D 2 x CM2 y CM2. Therefore, we conclude that I EXAMPLE 10.8 ICM MD 2 Applying the Parallel-Axis Theorem Consider once again the uniform rigid rod of mass M and length L shown in Figure 10.10. Find the moment of inertia of the rod about an axis perpendicular to the rod through one end (the y axis in Fig. 10.10). Solution Intuitively, we expect the moment of inertia to 1 be greater than ICM 12ML 2 because it should be more difficult to change the rotational motion of a rod spinning about an axis at one end than one that is spinning about its center. Because the distance between the center-of-mass axis and the y axis is D L/2, the parallel-axis theorem gives I MD 2 ICM 1 12 ML 2 M L 2 2 1 3 ML 2 So, it is four times more difficult to change the rotation of a rod spinning about its end than it is to change the motion of one spinning about its center. Exercise Calculate the moment of inertia of the rod about a perpendicular axis through the point x L/4. Answer I 7 48 ML 2. 306 CHAPTER 10 10.6 7.6 Rotation of a Rigid Object About a Fixed Axis TORQUE Why are a door’s doorknob and hinges placed near opposite edges of the door? This question actually has an answer based on common sense ideas. The harder we push against the door and the farther we are from the hinges, the more likely we are to open or close the door. When a force is exerted on a rigid object pivoted about an axis, the object tends to rotate about that axis. The tendency of a force to rotate an object about some axis is measured by a vector quantity called torque (tau). Consider the wrench pivoted on the axis through O in Figure 10.13. The applied force F acts at an angle to the horizontal. We define the magnitude of the torque associated with the force F by the expression r F sin Definition of torque Moment arm (10.19) Fd where r is the distance between the pivot point and the point of application of F and d is the perpendicular distance from the pivot point to the line of action of F. (The line of action of a force is an imaginary line extending out both ends of the vector representing the force. The dashed line extending from the tail of F in Figure 10.13 is part of the line of action of F.) From the right triangle in Figure 10.13 that has the wrench as its hypotenuse, we see that d r sin . This quantity d is called the moment arm (or lever arm) of F. It is very important that you recognize that torque is defined only when a reference axis is specified. Torque is the product of a force and the moment arm of that force, and moment arm is defined only in terms of an axis of rotation. In Figure 10.13, the only component of F that tends to cause rotation is F sin , the component perpendicular to r. The horizontal component F cos , because it passes through O, has no tendency to produce rotation. From the definition of torque, we see that the rotating tendency increases as F increases and as d increases. This explains the observation that it is easier to close a door if we push at the doorknob rather than at a point close to the hinge. We also want to apply our push as close to perpendicular to the door as we can. Pushing sideways on the doorknob will not cause the door to rotate. If two or more forces are acting on a rigid object, as shown in Figure 10.14, each tends to produce rotation about the pivot at O. In this example, F2 tends to F1 F sin φ F r d1 φ φ O d F cos φ Line of action Figure 10.13 The force F has a greater rotating tendency about O as F increases and as the moment arm d increases. It is the component F sin that tends to rotate the wrench about O. O d2 F2 Figure 10.14 The force F1 tends to rotate the object counterclockwise about O, and F2 tends to rotate it clockwise. 307 10.7 Relationship Between Torque and Angular Acceleration rotate the object clockwise, and F1 tends to rotate it counterclockwise. We use the convention that the sign of the torque resulting from a force is positive if the turning tendency of the force is counterclockwise and is negative if the turning tendency is clockwise. For example, in Figure 10.14, the torque resulting from F1 , which has a moment arm d 1 , is positive and equal to F1 d 1 ; the torque from F2 is negative and equal to F2 d 2 . Hence, the net torque about O is 1 2 F 1d 1 F 2d 2 Torque should not be confused with force. Forces can cause a change in linear motion, as described by Newton’s second law. Forces can also cause a change in rotational motion, but the effectiveness of the forces in causing this change depends on both the forces and the moment arms of the forces, in the combination that we call torque. Torque has units of force times length — newton meters in SI units — and should be reported in these units. Do not confuse torque and work, which have the same units but are very different concepts. EXAMPLE 10.9 The Net Torque on a Cylinder A one-piece cylinder is shaped as shown in Figure 10.15, with a core section protruding from the larger drum. The cylinder is free to rotate around the central axis shown in the drawing. A rope wrapped around the drum, which has radius R 1 , exerts a force F1 to the right on the cylinder. A rope wrapped around the core, which has radius R 2 , exerts a force F2 downward on the cylinder. (a) What is the net torque acting on the cylinder about the rotation axis (which is the z axis in Figure 10.15)? y F1 R1 R2 O F2 A solid cylinder pivoted about the z axis through O. The moment arm of F1 is R 1 , and the moment arm of F2 is R 2 . 10.7 7.6 The torque due to F1 is R 1 F1 (the sign is negative because the torque tends to produce clockwise rotation). The torque due to F2 is R 2 F 2 (the sign is positive because the torque tends to produce counterclockwise rotation). Therefore, the net torque about the rotation axis is 1 2 R 1F 1 R 2F 2 We can make a quick check by noting that if the two forces are of equal magnitude, the net torque is negative because R 1 R 2 . Starting from rest with both forces acting on it, the cylinder would rotate clockwise because F 1 would be more effective at turning it than would F 2 . (b) Suppose F 1 5.0 N, R 1 1.0 m, F 2 15.0 N, and R 2 0.50 m. What is the net torque about the rotation axis, and which way does the cylinder rotate from rest? x z Figure 10.15 Solution (5.0 N)(1.0 m) (15.0 N)(0.50 m) 2.5 N m Because the net torque is positive, if the cylinder starts from rest, it will commence rotating counterclockwise with increasing angular velocity. (If the cylinder’s initial rotation is clockwise, it will slow to a stop and then rotate counterclockwise with increasing angular speed.) RELATIONSHIP BETWEEN TORQUE AND ANGULAR ACCELERATION In this section we show that the angular acceleration of a rigid object rotating about a fixed axis is proportional to the net torque acting about that axis. Before discussing the more complex case of rigid-body rotation, however, it is instructive 308 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis first to discuss the case of a particle rotating about some fixed point under the influence of an external force. Consider a particle of mass m rotating in a circle of radius r under the influence of a tangential force Ft and a radial force Fr , as shown in Figure 10.16. (As we learned in Chapter 6, the radial force must be present to keep the particle moving in its circular path.) The tangential force provides a tangential acceleration at , and Ft ma t The torque about the center of the circle due to Ft is A particle rotating in a circle under the influence of a tangential force Ft . A force Fr in the radial direction also must be present to maintain the circular motion. (ma t )r Ft r Figure 10.16 Because the tangential acceleration is related to the angular acceleration through the relationship at r (see Eq. 10.11), the torque can be expressed as (mr 2) (mr )r Recall from Equation 10.15 that mr 2 is the moment of inertia of the rotating particle about the z axis passing through the origin, so that Relationship between torque and angular acceleration y That is, the torque acting on the particle is proportional to its angular acceleration, and the proportionality constant is the moment of inertia. It is important to note that I is the rotational analog of Newton’s second law of motion, F ma. Now let us extend this discussion to a rigid object of arbitrary shape rotating about a fixed axis, as shown in Figure 10.17. The object can be regarded as an infinite number of mass elements dm of infinitesimal size. If we impose a cartesian coordinate system on the object, then each mass element rotates in a circle about the origin, and each has a tangential acceleration at produced by an external tangential force d Ft . For any given element, we know from Newton’s second law that d Ft dm dFt r O (10.20) I (dm)a t The torque d associated with the force d Ft acts about the origin and is given by x d Because at Figure 10.17 A rigid object rotating about an axis through O. Each mass element dm rotates about O with the same angular acceleration , and the net torque on the object is proportional to . r dFt (r dm)a t r , the expression for d becomes d (r 2 dm) (r dm)r It is important to recognize that although each mass element of the rigid object may have a different linear acceleration at , they all have the same angular acceleration . With this in mind, we can integrate the above expression to obtain the net torque about O due to the external forces: (r 2 dm) r 2 dm where can be taken outside the integral because it is common to all mass elements. From Equation 10.17, we know that r 2 dm is the moment of inertia of the object about the rotation axis through O, and so the expression for becomes Torque is proportional to angular acceleration I (10.21) Note that this is the same relationship we found for a particle rotating in a circle (see Eq. 10.20). So, again we see that the net torque about the rotation axis is pro- 309 10.7 Relationship Between Torque and Angular Acceleration portional to the angular acceleration of the object, with the proportionality factor being I, a quantity that depends upon the axis of rotation and upon the size and shape of the object. In view of the complex nature of the system, it is interesting to note that the relationship I is strikingly simple and in complete agreement with experimental observations. The simplicity of the result lies in the manner in which the motion is described. Although each point on a rigid object rotating about a fixed axis may not experience the same force, linear acceleration, or linear speed, each point experiences the same angular acceleration and angular speed at any instant. Therefore, at any instant the rotating rigid object as a whole is characterized by specific values for angular acceleration, net torque, and angular speed. Every point has the same Q uickLab Finally, note that the result I also applies when the forces acting on the mass elements have radial components as well as tangential components. This is because the line of action of all radial components must pass through the axis of rotation, and hence all radial components produce zero torque about that axis. EXAMPLE 10.10 Solution We cannot use our kinematic equations to find or a because the torque exerted on the rod varies with its position, and so neither acceleration is constant. We have enough information to find the torque, however, which we can then use in the torque – angular acceleration relationship (Eq. 10.21) to find and then a. The only force contributing to torque about an axis through the pivot is the gravitational force M g exerted on the rod. (The force exerted by the pivot on the rod has zero torque about the pivot because its moment arm is zero.) To compute the torque on the rod, we can assume that the gravitational force acts at the center of mass of the rod, as shown in Figure 10.18. The torque due to this force about an axis through the pivot is g 1 2 3 ML With I , and I Table 10.2), we obtain L/2 Mg The uniform rod is pivoted at the left end. L 2 for this axis of rotation (see g (L/2) I 13 L2 3g 2L All points on the rod have this angular acceleration. To find the linear acceleration of the right end of the rod, we use the relationship a t r (Eq. 10.11), with r L : at Figure 10.18 Tip over a child’s tall tower of blocks. Try this several times. Does the tower “break” at the same place each time? What affects where the tower comes apart as it tips? If the tower were made of toy bricks that snap together, what would happen? (Refer to Conceptual Example 10.11.) Rotating Rod A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical plane, as shown in Figure 10.18. The rod is released from rest in the horizontal position. What is the initial angular acceleration of the rod and the initial linear acceleration of its right end? Pivot and L 3 2g This result — that at g for the free end of the rod — is rather interesting. It means that if we place a coin at the tip of the rod, hold the rod in the horizontal position, and then release the rod, the tip of the rod falls faster than the coin does! Other points on the rod have a linear acceleration that is less than 3 g. For example, the middle of the rod has 2 an acceleration of 3 g. 4 310 CHAPTER 10 CONCEPTUAL EXAMPLE 10.11 Rotation of a Rigid Object About a Fixed Axis Falling Smokestacks and Tumbling Blocks When a tall smokestack falls over, it often breaks somewhere along its length before it hits the ground, as shown in Figure 10.19. The same thing happens with a tall tower of children’s toy blocks. Why does this happen? Solution As the smokestack rotates around its base, each higher portion of the smokestack falls with an increasing tangential acceleration. (The tangential acceleration of a given point on the smokestack is proportional to the distance of that portion from the base.) As the acceleration increases, higher portions of the smokestack experience an acceleration greater than that which could result from gravity alone; this is similar to the situation described in Example 10.10. This can happen only if these portions are being pulled downward by a force in addition to the gravitational force. The force that causes this to occur is the shear force from lower portions of the smokestack. Eventually the shear force that provides this acceleration is greater than the smokestack can withstand, and the smokestack breaks. EXAMPLE 10.12 Figure 10.19 A falling smokestack. Angular Acceleration of a Wheel A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axle, as shown in Figure 10.20. A light cord wrapped around the wheel supports an object of mass m. Calculate the angular acceleration of the wheel, the linear acceleration of the object, and the tension in the cord. M O Solution The torque acting on the wheel about its axis of rotation is TR, where T is the force exerted by the cord on the rim of the wheel. (The gravitational force exerted by the Earth on the wheel and the normal force exerted by the axle on the wheel both pass through the axis of rotation and thus produce no torque.) Because I, we obtain I TR R T T TR I (1) m Now let us apply Newton’s second law to the motion of the object, taking the downward direction to be positive: Fy (2) a mg T mg ma T m Equations (1) and (2) have three unknowns, , a, and T. Because the object and wheel are connected by a string that does not slip, the linear acceleration of the suspended object is equal to the linear acceleration of a point on the rim of the mg Figure 10.20 The tension in the cord produces a torque about the axle passing through O. 311 10.7 Relationship Between Torque and Angular Acceleration wheel. Therefore, the angular acceleration of the wheel and this linear acceleration are related by a R . Using this fact together with Equations (1) and (2), we obtain (3) (4) a TR 2 I R a R m R The wheel in Figure 10.20 is a solid disk of M 2.00 kg, R 30.0 cm, and I 0.090 0 kg m2. The suspended object has a mass of m 0.500 kg. Find the tension in the cord and the angular acceleration of the wheel. Substituting Equation (4) into Equation (2), and solving for a and , we find that EXAMPLE 10.13 g I/mR Exercise mR 2 I 1 g I/mR 2 1 mg mg T a 3.27 N; 10.9 rad/s2. Answer Atwood’s Machine Revisited Two blocks having masses m1 and m 2 are connected to each other by a light cord that passes over two identical, frictionless pulleys, each having a moment of inertia I and radius R, as shown in Figure 10.21a. Find the acceleration of each block and the tensions T1 , T2 , and T3 in the cord. (Assume no slipping between cord and pulleys.) Substituting Equation (6) into Equation (5), we have [(m 1 m 2)g (m 1 m 2)a]R 2I a/R, this expression can be simplified to Because (m 1 m 2)g (m 1 m 2)a a R2 2I Solution We shall define the downward direction as positive for m 1 and upward as the positive direction for m 2 . This allows us to represent the acceleration of both masses by a single variable a and also enables us to relate a positive a to a positive (counterclockwise) angular acceleration . Let us write Newton’s second law of motion for each block, using the free-body diagrams for the two blocks as shown in Figure 10.21b: (1) m 1g T1 m 1a (2) T3 m 2g (7) m1 T2)R (T2 T3)R I T1 T3 + T1 m1 T3 m1 m2 m1g m2 m2g + (a) (b) (5) (T1 T3)R (6) m 1g T1 T1 n2 T2 mpg mpg T3 (c) 2I Figure 10.21 Adding Equations (1) and (2) gives T1 T2 n1 We now have four equations with four unknowns: a, T1 , T2 , and T3 . These can be solved simultaneously. Adding Equations (3) and (4) gives T3 I R2 I (4) 2 m2 T2 Next, we must include the effect of the pulleys on the motion. Free-body diagrams for the pulleys are shown in Figure 10.21c. The net torque about the axle for the pulley on the left is ( T1 T2 )R, while the net torque for the pulley on the right is ( T2 T3 )R. Using the relation I for each pulley and noting that each pulley has the same angular acceleration , we obtain (T1 m 2)g This value of a can then be substituted into Equations (1) m 2a (3) (m 1 a m 2g (m 1 m 2)a T3 (m 1 m 2)g (m 1 m 2)a (a) Another look at Atwood’s machine. (b) Free-body diagrams for the blocks. (c) Free-body diagrams for the pulleys, where m p g represents the force of gravity acting on each pulley. 312 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis and (2) to give T1 and T3 . Finally, T2 can be found from Equation (3) or Equation (4). Note that if m 1 m 2 , the acceleration is positive; this means that the left block accelerates downward, the right block accelerates upward, and both 10.8 F φ ds dθ pulleys accelerate counterclockwise. If m 1 m 2 , then all the values are negative and the motions are reversed. If m 1 m 2 , then no acceleration occurs at all. You should compare these results with those found in Example 5.9 on page 129. WORK, POWER, AND ENERGY IN ROTATIONAL MOTION In this section, we consider the relationship between the torque acting on a rigid object and its resulting rotational motion in order to generate expressions for the power and a rotational analog to the work – kinetic energy theorem. Consider the rigid object pivoted at O in Figure 10.22. Suppose a single external force F is applied at P, where F lies in the plane of the page. The work done by F as the object rotates through an infinitesimal distance ds r d in a time dt is P r O Figure 10.22 A rigid object rotates about an axis through O under the action of an external force F applied at P. F ds dW (F sin )r d where F sin is the tangential component of F, or, in other words, the component of the force along the displacement. Note that the radial component of F does no work because it is perpendicular to the displacement. Because the magnitude of the torque due to F about O is defined as r F sin by Equation 10.19, we can write the work done for the infinitesimal rotation as dW (10.22) d The rate at which work is being done by F as the object rotates about the fixed axis is d dt dW dt Because dW/dt is the instantaneous power (see Section 7.5) delivered by the force, and because d /dt , this expression reduces to dW dt Power delivered to a rigid object (10.23) Fv in the case of linear motion, and the exThis expression is analogous to pression dW d is analogous to dW Fx dx. Work and Energy in Rotational Motion In studying linear motion, we found the energy concept — and, in particular, the work – kinetic energy theorem — extremely useful in describing the motion of a system. The energy concept can be equally useful in describing rotational motion. From what we learned of linear motion, we expect that when a symmetric object rotates about a fixed axis, the work done by external forces equals the change in the rotational energy. To show that this is in fact the case, let us begin with I . Using the chain rule from the calculus, we can express the resultant torque as I I d dt I d d d dt I d d 10.8 Work, Power, and Energy in Rotational Motion 313 TABLE 10.3 Useful Equations in Rotational and Linear Motion Rotational Motion About a Fixed Axis Linear Motion Angular speed d /dt Angular acceleration d /dt Resultant torque I Linear speed v dx/dt Linear acceleration a dv/dt Resultant force F ma If f constant f f t t i i i 2 i 2 1 2 2( If a t2 vi xi vf 2 i) f at vit vi2 1 2 at 2 2a(xf xi ) xf f Work W vf xf constant Work W d xi i Rotational kinetic energy K R Power Angular momentum L I Resultant torque dL/dt 1 2 2I Kinetic energy K 1mv 2 2 Power Fv Linear momentum p mv Resultant force F dp/dt Rearranging this expression and noting that d Fx dx dW, we obtain d dW Id Integrating this expression, we get for the total work done by the net external force acting on a rotating system f W f d i where the angular speed changes from from i to f . That is, 1 2 2I f Id 1 2 2I i (10.24) i i to f as the angular position changes the net work done by external forces in rotating a symmetric rigid object about a fixed axis equals the change in the object’s rotational energy. Table 10.3 lists the various equations we have discussed pertaining to rotational motion, together with the analogous expressions for linear motion. The last two equations in Table 10.3, involving angular momentum L , are discussed in Chapter 11 and are included here only for the sake of completeness. Quick Quiz 10.4 For a hoop lying in the xy plane, which of the following requires that more work be done by an external agent to accelerate the hoop from rest to an angular speed : (a) rotation about the z axis through the center of the hoop, or (b) rotation about an axis parallel to z passing through a point P on the hoop rim? Work – kinetic energy theorem for rotational motion 314 CHAPTER 10 EXAMPLE 10.14 Rotation of a Rigid Object About a Fixed Axis Rotating Rod Revisited A uniform rod of length L and mass M is free to rotate on a frictionless pin passing through one end (Fig 10.23). The rod is released from rest in the horizontal position. (a) What is its angular speed when it reaches its lowest position? 1 energy is entirely rotational energy, 2 I 2, where I is the moment of inertia about the pivot. Because I 1 ML 2 (see Table 3 10.2) and because mechanical energy is constant, we have Ei Ef or 1 2 MgL Solution The question can be answered by considering the mechanical energy of the system. When the rod is horizontal, it has no rotational energy. The potential energy relative to the lowest position of the center of mass of the rod (O ) is MgL/2. When the rod reaches its lowest position, the 1 2 2I 11 2 2 2 (3 ML ) √ 3g L (b) Determine the linear speed of the center of mass and the linear speed of the lowest point on the rod when it is in the vertical position. E i = U = MgL/2 O Solution These two values can be determined from the relationship between linear and angular speeds. We know from part (a), and so the linear speed of the center of mass is L/2 O′ v CM 1 2 √3gL Because r for the lowest point on the rod is twice what it is for the center of mass, the lowest point has a linear speed equal to 1 E f = K R = – Iω 2 2 Figure 10.23 A uniform rigid rod pivoted at O rotates in a vertical plane under the action of gravity. EXAMPLE 10.15 L 2 r √3gL 2v CM Connected Cylinders Consider two cylinders having masses m1 and m 2 , where m1 m2 , connected by a string passing over a pulley, as shown in Figure 10.24. The pulley has a radius R and moment of inertia I about its axis of rotation. The string does not slip on the pulley, and the system is released from rest. Find the linear speeds of the cylinders after cylinder 2 descends through a distance h, and the angular speed of the pulley at this time. Solution We are now able to account for the effect of a massive pulley. Because the string does not slip, the pulley rotates. We neglect friction in the axle about which the pulley rotates for the following reason: Because the axle’s radius is small relative to that of the pulley, the frictional torque is much smaller than the torque applied by the two cylinders, provided that their masses are quite different. Mechanical energy is constant; hence, the increase in the system’s kinetic energy (the system being the two cylinders, the pulley, and the Earth) equals the decrease in its potential energy. Because K i 0 (the system is initially at rest), we have R m2 h h m1 Figure 10.24 K Kf (1m 1v f 2 2 Ki 1 2 2 m 2v f 1 2 2I f ) where vf is the same for both blocks. Because vf expression becomes K 1 2 m1 m2 I v2 R2 f 0 R f , this 315 Summary From Figure 10.24, we see that the system loses potential energy as cylinder 2 descends and gains potential energy as cylinder 1 rises. That is, U 2 m 2 gh and U 1 m 1gh. Applying the principle of conservation of energy in the form K U1 U 2 0 gives 1 2 m1 m2 vf I v2 R2 f m 1gh 2(m 2 m1 m2 R f , the angular speed of the pulley at this invf f R 1 R 0 m 2gh m 1)gh Because v f stant is Exercise I R2 SUMMARY If a particle rotates in a circle of radius r through an angle (measured in radians), the arc length it moves through is s r . The angular displacement of a particle rotating in a circle or of a rigid object rotating about a fixed axis is (10.2) i The instantaneous angular speed of a particle rotating in a circle or of a rigid object rotating about a fixed axis is d dt (10.4) The instantaneous angular acceleration of a rotating object is d dt (10.6) When a rigid object rotates about a fixed axis, every part of the object has the same angular speed and the same angular acceleration. If a particle or object rotates about a fixed axis under constant angular acceleration, one can apply equations of kinematics that are analogous to those for linear motion under constant linear acceleration: f f f 2 (10.7) t i it i i 2 m1 m 1)gh m2 1/2 I R2 Repeat the calculation of vf , using I applied to the pulley and Newton’s second law applied to the two cylinders. Use the procedures presented in Examples 10.12 and 10.13. 1/2 f 2(m 2 1 2 2( f t2 (10.8) i) (10.9) A useful technique in solving problems dealing with rotation is to visualize a linear version of the same problem. When a rigid object rotates about a fixed axis, the angular position, angular speed, and angular acceleration are related to the linear position, linear speed, and linear acceleration through the relationships s ru (10.1a) v r (10.10) at r (10.11) 316 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis You must be able to easily alternate between the linear and rotational variables that describe a given situation. The moment of inertia of a system of particles is mi ri 2 I (10.15) i If a rigid object rotates about a fixed axis with angular speed , its rotational energy can be written KR 1 2 2I (10.16) where I is the moment of inertia about the axis of rotation. The moment of inertia of a rigid object is I r 2 dm (10.17) where r is the distance from the mass element dm to the axis of rotation. The magnitude of the torque associated with a force F acting on an object is (10.19) Fd where d is the moment arm of the force, which is the perpendicular distance from some origin to the line of action of the force. Torque is a measure of the tendency of the force to change the rotation of the object about some axis. If a rigid object free to rotate about a fixed axis has a net external torque acting on it, the object undergoes an angular acceleration , where (10.21) I The rate at which work is done by an external force in rotating a rigid object about a fixed axis, or the power delivered, is (10.23) The net work done by external forces in rotating a rigid object about a fixed axis equals the change in the rotational kinetic energy of the object: W 1 2 2I f 1 2 2I i (10.24) QUESTIONS 1. What is the angular speed of the second hand of a clock? What is the direction of as you view a clock hanging vertically? What is the magnitude of the angular acceleration vector of the second hand? 2. A wheel rotates counterclockwise in the xy plane. What is the direction of ? What is the direction of if the angular velocity is decreasing in time? 3. Are the kinematic expressions for , , and valid when the angular displacement is measured in degrees instead of in radians? 4. A turntable rotates at a constant rate of 45 rev/min. What is its angular speed in radians per second? What is the magnitude of its angular acceleration? 5. Suppose a b and M m for the system of particles described in Figure 10.8. About what axis (x, y, or z ) does the moment of inertia have the smallest value? the largest value? 6. Suppose the rod in Figure 10.10 has a nonuniform mass distribution. In general, would the moment of inertia about the y axis still equal ML2/12? If not, could the moment of inertia be calculated without knowledge of the manner in which the mass is distributed? 7. Suppose that only two external forces act on a rigid body, and the two forces are equal in magnitude but opposite in direction. Under what condition does the body rotate? 8. Explain how you might use the apparatus described in Example 10.12 to determine the moment of inertia of the wheel. (If the wheel does not have a uniform mass density, the moment of inertia is not necessarily equal to 1 2 2 MR .) 317 Problems 9. Using the results from Example 10.12, how would you calculate the angular speed of the wheel and the linear speed of the suspended mass at t 2 s, if the system is released from rest at t 0? Is the expression v R valid in this situation? 10. If a small sphere of mass M were placed at the end of the rod in Figure 10.23, would the result for be greater than, less than, or equal to the value obtained in Example 10.14? 11. Explain why changing the axis of rotation of an object changes its moment of inertia. 12. Is it possible to change the translational kinetic energy of an object without changing its rotational energy? 13. Two cylinders having the same dimensions are set into rotation about their long axes with the same angular speed. 14. 15. 16. 17. One is hollow, and the other is filled with water. Which cylinder will be easier to stop rotating? Explain your answer. Must an object be rotating to have a nonzero moment of inertia? If you see an object rotating, is there necessarily a net torque acting on it? Can a (momentarily) stationary object have a nonzero angular acceleration? The polar diameter of the Earth is slightly less than the equatorial diameter. How would the moment of inertia of the Earth change if some mass from near the equator were removed and transferred to the polar regions to make the Earth a perfect sphere? PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide WEB = solution posted at http://www.saunderscollege.com/physics/ = Computer useful in solving problem = Interactive Physics = paired numerical/symbolic problems Section 10.2 Rotational Kinematics: Rotational Motion with Constant Angular Acceleration WEB 1. A wheel starts from rest and rotates with constant angular acceleration and reaches an angular speed of 12.0 rad/s in 3.00 s. Find (a) the magnitude of the angular acceleration of the wheel and (b) the angle (in radians) through which it rotates in this time. 2. What is the angular speed in radians per second of (a) the Earth in its orbit about the Sun and (b) the Moon in its orbit about the Earth? 3. An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angular speed of 2 000 rad/s. The engine’s rotation slows with an angular acceleration of magnitude 80.0 rad/s2. (a) Determine the angular speed after 10.0 s. (b) How long does it take for the rotor to come to rest? 4. (a) The positions of the hour and minute hand on a clock face coincide at 12 o’clock. Determine all other times (up to the second) at which the positions of the hands coincide. (b) If the clock also has a second hand, determine all times at which the positions of all three hands coincide, given that they all coincide at 12 o’clock. 5. An electric motor rotating a grinding wheel at 100 rev/min is switched off. Assuming constant negative acceleration of magnitude 2.00 rad/s2, (a) how long does it take the wheel to stop? (b) Through how many radians does it turn during the time found in part (a)? 6. A centrifuge in a medical laboratory rotates at a rotational speed of 3 600 rev/min. When switched off, it rotates 50.0 times before coming to rest. Find the constant angular acceleration of the centrifuge. 7. The angular position of a swinging door is described by 5.00 10.0t 2.00t 2 rad. Determine the angular position, angular speed, and angular acceleration of the door (a) at t 0 and (b) at t 3.00 s. 8. The tub of a washer goes into its spin cycle, starting from rest and gaining angular speed steadily for 8.00 s, when it is turning at 5.00 rev/s. At this point the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub smoothly slows to rest in 12.0 s. Through how many revolutions does the tub turn while it is in motion? 9. A rotating wheel requires 3.00 s to complete 37.0 revolutions. Its angular speed at the end of the 3.00-s interval is 98.0 rad/s. What is the constant angular acceleration of the wheel? 10. (a) Find the angular speed of the Earth’s rotation on its axis. As the Earth turns toward the east, we see the sky turning toward the west at this same rate. (b) The rainy Pleiads wester And seek beyond the sea The head that I shall dream of That shall not dream of me. A. E. Housman (© Robert E. Symons) Cambridge, England, is at longitude 0°, and Saskatoon, Saskatchewan, is at longitude 107° west. How much time elapses after the Pleiades set in Cambridge until these stars fall below the western horizon in Saskatoon? Section 10.3 Angular and Linear Quantities 11. Make an order-of-magnitude estimate of the number of revolutions through which a typical automobile tire 318 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis sume the discus moves on the arc of a circle 1.00 m in radius. (a) Calculate the final angular speed of the discus. (b) Determine the magnitude of the angular acceleration of the discus, assuming it to be constant. (c) Calculate the acceleration time. 17. A car accelerates uniformly from rest and reaches a speed of 22.0 m/s in 9.00 s. If the diameter of a tire is 58.0 cm, find (a) the number of revolutions the tire makes during this motion, assuming that no slipping occurs. (b) What is the final rotational speed of a tire in revolutions per second? 18. A 6.00-kg block is released from A on the frictionless track shown in Figure P10.18. Determine the radial and tangential components of acceleration for the block at P. turns in 1 yr. State the quantities you measure or estimate and their values. 12. The diameters of the main rotor and tail rotor of a single-engine helicopter are 7.60 m and 1.02 m, respectively. The respective rotational speeds are 450 rev/min and 4 138 rev/min. Calculate the speeds of the tips of both rotors. Compare these speeds with the speed of sound, 343 m/s. A Figure P10.12 (Ross Harrrison Koty/Tony Stone Images) 13. A racing car travels on a circular track with a radius of 250 m. If the car moves with a constant linear speed of 45.0 m/s, find (a) its angular speed and (b) the magnitude and direction of its acceleration. 14. A car is traveling at 36.0 km/h on a straight road. The radius of its tires is 25.0 cm. Find the angular speed of one of the tires, with its axle taken as the axis of rotation. 15. A wheel 2.00 m in diameter lies in a vertical plane and rotates with a constant angular acceleration of 4.00 rad/s2. The wheel starts at rest at t 0, and the radius vector of point P on the rim makes an angle of 57.3° with the horizontal at this time. At t 2.00 s, find (a) the angular speed of the wheel, (b) the linear speed and acceleration of the point P, and (c) the angular position of the point P. 16. A discus thrower accelerates a discus from rest to a speed of 25.0 m/s by whirling it through 1.25 rev. As- Figure P10.16 (Bruce Ayers/Tony Stone Images) h = 5.00 m P R = 2.00 m Figure P10.18 WEB 19. A disc 8.00 cm in radius rotates at a constant rate of 1 200 rev/min about its central axis. Determine (a) its angular speed, (b) the linear speed at a point 3.00 cm from its center, (c) the radial acceleration of a point on the rim, and (d) the total distance a point on the rim moves in 2.00 s. 20. A car traveling on a flat (unbanked) circular track accelerates uniformly from rest with a tangential acceleration of 1.70 m/s2. The car makes it one quarter of the way around the circle before it skids off the track. Determine the coefficient of static friction between the car and track from these data. 21. A small object with mass 4.00 kg moves counterclockwise with constant speed 4.50 m/s in a circle of radius 3.00 m centered at the origin. (a) It started at the point with cartesian coordinates (3 m, 0). When its angular displacement is 9.00 rad, what is its position vector, in cartesian unit-vector notation? (b) In what quadrant is the particle located, and what angle does its position vector make with the positive x axis? (c) What is its velocity vector, in unit – vector notation? (d) In what direction is it moving? Make a sketch of the position and velocity vectors. (e) What is its acceleration, expressed in unit – vector notation? (f) What total force acts on the object? (Express your answer in unit vector notation.) 319 Problems 22. A standard cassette tape is placed in a standard cassette player. Each side plays for 30 min. The two tape wheels of the cassette fit onto two spindles in the player. Suppose that a motor drives one spindle at a constant angular speed of 1 rad/s and that the other spindle is free to rotate at any angular speed. Estimate the order of magnitude of the thickness of the tape. y(m) 3.00 kg 2.00 kg 6.00 m x(m) O Section 10.4 Rotational Energy 23. Three small particles are connected by rigid rods of negligible mass lying along the y axis (Fig. P10.23). If the system rotates about the x axis with an angular speed of 2.00 rad/s, find (a) the moment of inertia about the x axis and the total rotational kinetic energy evaluated from 1I 2 and (b) the linear speed of each 2 particle and the total kinetic energy evaluated from 1 2 2 m iv i . 4.00 m 2.00 kg 4.00 kg Figure P10.25 y 4.00 kg y = 3.00 m x O 2.00 kg 3.00 kg y = –2.00 m y = – 4.00 m Figure P10.26 Figure P10.23 WEB 24. The center of mass of a pitched baseball (3.80-cm radius) moves at 38.0 m/s. The ball spins about an axis through its center of mass with an angular speed of 125 rad/s. Calculate the ratio of the rotational energy to the translational kinetic energy. Treat the ball as a uniform sphere. 25. The four particles in Figure P10.25 are connected by rigid rods of negligible mass. The origin is at the center of the rectangle. If the system rotates in the xy plane about the z axis with an angular speed of 6.00 rad/s, calculate (a) the moment of inertia of the system about the z axis and (b) the rotational energy of the system. 26. The hour hand and the minute hand of Big Ben, the famous Parliament tower clock in London, are 2.70 m long and 4.50 m long and have masses of 60.0 kg and 100 kg, respectively. Calculate the total rotational kinetic energy of the two hands about the axis of rotation. (You may model the hands as long thin rods.) Problems 26 and 74. ( John Lawrence/Tony Stone Images) 27. Two masses M and m are connected by a rigid rod of length L and of negligible mass, as shown in Figure P10.27. For an axis perpendicular to the rod, show that the system has the minimum moment of inertia when the axis passes through the center of mass. Show that this moment of inertia is I L2, where mM/(m M ). L m M x L–x Figure P10.27 Section 10.5 Calculation of Moments of Inertia 28. Three identical thin rods, each of length L and mass m, are welded perpendicular to each other, as shown in Figure P10.28. The entire setup is rotated about an axis 320 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis 31. Attention! About face! Compute an order-of-magnitude estimate for the moment of inertia of your body as you stand tall and turn around a vertical axis passing through the top of your head and the point halfway between your ankles. In your solution state the quantities you measure or estimate and their values. Section 10.6 Torque 32. Find the mass m needed to balance the 1 500-kg truck on the incline shown in Figure P10.32. Assume all pulleys are frictionless and massless. 3r Axis of rotation r Figure P10.28 that passes through the end of one rod and is parallel to another. Determine the moment of inertia of this arrangement. 29. Figure P10.29 shows a side view of a car tire and its radial dimensions. The rubber tire has two sidewalls of uniform thickness 0.635 cm and a tread wall of uniform thickness 2.50 cm and width 20.0 cm. Suppose its density is uniform, with the value 1.10 103 kg/m3. Find its moment of inertia about an axis through its center perpendicular to the plane of the sidewalls. m 1500 kg θ = 45° Sidewall Figure P10.32 33.0 cm WEB 16.5 cm 33. Find the net torque on the wheel in Figure P10.33 about the axle through O if a 10.0 cm and b 25.0 cm. 10.0 N 30.5 cm 30.0° Tread a O 12.0 N Figure P10.29 30. Use the parallel-axis theorem and Table 10.2 to find the moments of inertia of (a) a solid cylinder about an axis parallel to the center-of-mass axis and passing through the edge of the cylinder and (b) a solid sphere about an axis tangent to its surface. b 9.00 N Figure P10.33 34. The fishing pole in Figure P10.34 makes an angle of 20.0° with the horizontal. What is the torque exerted by 321 Problems 2.00 m m1 20.0° 37.0° I, R 20.0° m2 100 N θ Figure P10.39 Figure P10.34 the fish about an axis perpendicular to the page and passing through the fisher’s hand? 35. The tires of a 1 500-kg car are 0.600 m in diameter, and the coefficients of friction with the road surface are 0.800 and k 0.600. Assuming that the weight is s evenly distributed on the four wheels, calculate the maximum torque that can be exerted by the engine on a driving wheel such that the wheel does not spin. If you wish, you may suppose that the car is at rest. 36. Suppose that the car in Problem 35 has a disk brake system. Each wheel is slowed by the frictional force between a single brake pad and the disk-shaped rotor. On this particular car, the brake pad comes into contact with the rotor at an average distance of 22.0 cm from the axis. The coefficients of friction between the brake pad and the disk are s 0.600 and k 0.500. Calculate the normal force that must be applied to the rotor such that the car slows as quickly as possible. Section 10.7 Relationship Between Torque and Angular Acceleration WEB 37. A model airplane having a mass of 0.750 kg is tethered by a wire so that it flies in a circle 30.0 m in radius. The airplane engine provides a net thrust of 0.800 N perpendicular to the tethering wire. (a) Find the torque the net thrust produces about the center of the circle. (b) Find the angular acceleration of the airplane when it is in level flight. (c) Find the linear acceleration of the airplane tangent to its flight path. 38. The combination of an applied force and a frictional force produces a constant total torque of 36.0 N m on a wheel rotating about a fixed axis. The applied force acts for 6.00 s; during this time the angular speed of the wheel increases from 0 to 10.0 rad/s. The applied force is then removed, and the wheel comes to rest in 60.0 s. Find (a) the moment of inertia of the wheel, (b) the magnitude of the frictional torque, and (c) the total number of revolutions of the wheel. 39. A block of mass m1 2.00 kg and a block of mass m 2 6.00 kg are connected by a massless string over a pulley in the shape of a disk having radius R 0.250 m and mass M 10.0 kg. These blocks are allowed to move on a fixed block – wedge of angle 30.0°, as shown in Figure P10.39. The coefficient of kinetic friction for both blocks is 0.360. Draw free-body diagrams of both blocks and of the pulley. Determine (a) the acceleration of the two blocks and (b) the tensions in the string on both sides of the pulley. 40. A potter’s wheel a thick stone disk with a radius of 0.500 m and a mass of 100 kg — is freely rotating at 50.0 rev/min. The potter can stop the wheel in 6.00 s by pressing a wet rag against the rim and exerting a radially inward force of 70.0 N. Find the effective coefficient of kinetic friction between the wheel and the rag. 41. A bicycle wheel has a diameter of 64.0 cm and a mass of 1.80 kg. Assume that the wheel is a hoop with all of its mass concentrated on the outside radius. The bicycle is placed on a stationary stand on rollers, and a resistive force of 120 N is applied tangent to the rim of the tire. (a) What force must be applied by a chain passing over a 9.00-cm-diameter sprocket if the wheel is to attain an acceleration of 4.50 rad/s2? (b) What force is required if the chain shifts to a 5.60-cm-diameter sprocket? Section 10.8 Work , Power, and Energy in Rotational Motion 42. A cylindrical rod 24.0 cm long with a mass of 1.20 kg and a radius of 1.50 cm has a ball with a diameter of 8.00 cm and a mass of 2.00 kg attached to one end. The arrangement is originally vertical and stationary, with the ball at the top. The apparatus is free to pivot about the bottom end of the rod. (a) After it falls through 90°, what is its rotational kinetic energy? (b) What is the angular speed of the rod and ball? (c) What is the linear speed of the ball? (d) How does this compare with the speed if the ball had fallen freely through the same distance of 28 cm? 43. A 15.0-kg mass and a 10.0-kg mass are suspended by a pulley that has a radius of 10.0 cm and a mass of 3.00 kg (Fig. P10.43). The cord has a negligible mass and causes the pulley to rotate without slipping. The pulley 322 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis rotates without friction. The masses start from rest 3.00 m apart. Treating the pulley as a uniform disk, determine the speeds of the two masses as they pass each other. 44. A mass m 1 and a mass m 2 are suspended by a pulley that has a radius R and a mass M (see Fig. P10.43). The cord has a negligible mass and causes the pulley to rotate without slipping. The pulley rotates without friction. The masses start from rest a distance d apart. Treating the pulley as a uniform disk, determine the speeds of the two masses as they pass each other. m Figure P10.47 M R M = 3.00 kg R = 10.0 cm m1 = 15.0 kg m2 = 10.0 kg m1 3.00 m m2 Figure P10.43 v. Show that the moment of inertia I of the equipment (including the turntable) is mr 2(2gh/v 2 1). 48. A bus is designed to draw its power from a rotating flywheel that is brought up to its maximum rate of rotation (3 000 rev/min) by an electric motor. The flywheel is a solid cylinder with a mass of 1 000 kg and a diameter of 1.00 m. If the bus requires an average power of 10.0 kW, how long does the flywheel rotate? 49. (a) A uniform, solid disk of radius R and mass M is free to rotate on a frictionless pivot through a point on its rim (Fig. P10.49). If the disk is released from rest in the position shown by the blue circle, what is the speed of its center of mass when the disk reaches the position indicated by the dashed circle? (b) What is the speed of the lowest point on the disk in the dashed position? (c) Repeat part (a), using a uniform hoop. Problems 43 and 44. 45. A weight of 50.0 N is attached to the free end of a light string wrapped around a reel with a radius of 0.250 m and a mass of 3.00 kg. The reel is a solid disk, free to rotate in a vertical plane about the horizontal axis passing through its center. The weight is released 6.00 m above the floor. (a) Determine the tension in the string, the acceleration of the mass, and the speed with which the weight hits the floor. (b) Find the speed calculated in part (a), using the principle of conservation of energy. 46. A constant torque of 25.0 N m is applied to a grindstone whose moment of inertia is 0.130 kg m2. Using energy principles, find the angular speed after the grindstone has made 15.0 revolutions. (Neglect friction.) 47. This problem describes one experimental method of determining the moment of inertia of an irregularly shaped object such as the payload for a satellite. Figure P10.47 shows a mass m suspended by a cord wound around a spool of radius r, forming part of a turntable supporting the object. When the mass is released from rest, it descends through a distance h, acquiring a speed Pivot R g Figure P10.49 50. A horizontal 800-N merry-go-round is a solid disk of radius 1.50 m and is started from rest by a constant horizontal force of 50.0 N applied tangentially to the cylinder. Find the kinetic energy of the solid cylinder after 3.00 s. ADDITIONAL PROBLEMS 51. Toppling chimneys often break apart in mid-fall (Fig. P10.51) because the mortar between the bricks cannot 323 Problems withstand much shear stress. As the chimney begins to fall, shear forces must act on the topmost sections to accelerate them tangentially so that they can keep up with the rotation of the lower part of the stack. For simplicity, let us model the chimney as a uniform rod of length pivoted at the lower end. The rod starts at rest in a vertical position (with the frictionless pivot at the bottom) and falls over under the influence of gravity. What fraction of the length of the rod has a tangential acceleration greater than g sin , where is the angle the chimney makes with the vertical? exerts on the wheel. (a) How long does the wheel take to reach its final rotational speed of 1 200 rev/min? (b) Through how many revolutions does it turn while accelerating? 54. The density of the Earth, at any distance r from its center, is approximately [14.2 11.6 r/R] where R is the radius of the Earth. Show that this density leads to a moment of inertia I 0.330MR 2 about an axis through the center, where M is the mass of the Earth. 55. A 4.00-m length of light nylon cord is wound around a uniform cylindrical spool of radius 0.500 m and mass 1.00 kg. The spool is mounted on a frictionless axle and is initially at rest. The cord is pulled from the spool with a constant acceleration of magnitude 2.50 m/s2. (a) How much work has been done on the spool when it reaches an angular speed of 8.00 rad/s? (b) Assuming that there is enough cord on the spool, how long does it take the spool to reach this angular speed? (c) Is there enough cord on the spool? 56. A flywheel in the form of a heavy circular disk of diameter 0.600 m and mass 200 kg is mounted on a frictionless bearing. A motor connected to the flywheel accelerates it from rest to 1 000 rev/min. (a) What is the moment of inertia of the flywheel? (b) How much work is done on it during this acceleration? (c) When the angular speed reaches 1 000 rev/min, the motor is disengaged. A friction brake is used to slow the rotational rate to 500 rev/min. How much energy is dissipated as internal energy in the friction brake? 57. A shaft is turning at 65.0 rad/s at time zero. Thereafter, its angular acceleration is given by 10 rad/s2 Figure P10.51 A building demolition site in Baltimore, MD. At the left is a chimney, mostly concealed by the building, that has broken apart on its way down. Compare with Figure 10.19. ( Jerry Wachter/Photo Researchers, Inc.) 52. Review Problem. A mixing beater consists of three thin rods: Each is 10.0 cm long, diverges from a central hub, and is separated from the others by 120°. All turn in the same plane. A ball is attached to the end of each rod. Each ball has a cross-sectional area of 4.00 cm2 and is shaped so that it has a drag coefficient of 0.600. Calculate the power input required to spin the beater at 1 000 rev/min (a) in air and (b) in water. 53. A grinding wheel is in the form of a uniform solid disk having a radius of 7.00 cm and a mass of 2.00 kg. It starts from rest and accelerates uniformly under the action of the constant torque of 0.600 N m that the motor 10 3 kg/m3 5t rad/s3 where t is the elapsed time. (a) Find its angular speed at t 3.00 s. (b) How far does it turn in these 3 s? 58. For any given rotational axis, the radius of gyration K of a rigid body is defined by the expression K 2 I/M, where M is the total mass of the body and I is its moment of inertia. Thus, the radius of gyration is equal to the distance between an imaginary point mass M and the axis of rotation such that I for the point mass about that axis is the same as that for the rigid body. Find the radius of gyration of (a) a solid disk of radius R, (b) a uniform rod of length L, and (c) a solid sphere of radius R, all three of which are rotating about a central axis. 59. A long, uniform rod of length L and mass M is pivoted about a horizontal, frictionless pin passing through one end. The rod is released from rest in a vertical position, as shown in Figure P10.59. At the instant the rod is horizontal, find (a) its angular speed, (b) the magnitude of its angular acceleration, (c) the x and y components of the acceleration of its center of mass, and (d) the components of the reaction force at the pivot. 324 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis y A drop that breaks loose on the next turn rises a distance h 2 h1 above the tangent point. The height to which the drops rise decreases because the angular speed of the wheel decreases. From this information, determine the magnitude of the average angular acceleration of the wheel. L Pivot x Figure P10.59 62. The top shown in Figure P10.62 has a moment of inertia of 4.00 10 4 kg m2 and is initially at rest. It is free to rotate about the stationary axis AA . A string, wrapped around a peg along the axis of the top, is pulled in such a manner that a constant tension of 5.57 N is maintained. If the string does not slip while it is unwound from the peg, what is the angular speed of the top after 80.0 cm of string has been pulled off the peg? A′ h F A Figure P10.62 Figure P10.60 Problems 60 and 61. 60. A bicycle is turned upside down while its owner repairs a flat tire. A friend spins the other wheel, of radius 0.381 m, and observes that drops of water fly off tangentially. She measures the height reached by drops moving vertically (Fig. P10.60). A drop that breaks loose from the tire on one turn rises h 54.0 cm above the tangent point. A drop that breaks loose on the next turn rises 51.0 cm above the tangent point. The height to which the drops rise decreases because the angular speed of the wheel decreases. From this information, determine the magnitude of the average angular acceleration of the wheel. 61. A bicycle is turned upside down while its owner repairs a flat tire. A friend spins the other wheel of radius R and observes that drops of water fly off tangentially. She measures the height reached by drops moving vertically (see Fig. P10.60). A drop that breaks loose from the tire on one turn rises a distance h1 above the tangent point. 63. A cord is wrapped around a pulley of mass m and of radius r. The free end of the cord is connected to a block of mass M. The block starts from rest and then slides down an incline that makes an angle with the horizontal. The coefficient of kinetic friction between block and incline is . (a) Use energy methods to show that the block’s speed as a function of displacement d down the incline is v [4gdM(m 2M ) 1(sin cos )]1/2 (b) Find the magnitude of the acceleration of the block in terms of , m, M, g, and . 64. (a) What is the rotational energy of the Earth about its spin axis? The radius of the Earth is 6 370 km, and its mass is 5.98 1024 kg. Treat the Earth as a sphere of moment of inertia 2MR 2. (b) The rotational energy of 5 the Earth is decreasing steadily because of tidal friction. Estimate the change in one day, given that the rotational period increases by about 10 s each year. 65. The speed of a moving bullet can be determined by allowing the bullet to pass through two rotating paper disks mounted a distance d apart on the same axle (Fig. P10.65). From the angular displacement of the two 325 Problems ∆ θ = 31° v ω d Figure P10.65 66. 67. 68. 69. bullet holes in the disks and the rotational speed of the disks, we can determine the speed v of the bullet. Find the bullet speed for the following data: d 80 cm, 900 rev/min, and 31.0. A wheel is formed from a hoop and n equally spaced spokes extending from the center of the hoop to its rim. The mass of the hoop is M, and the radius of the hoop (and hence the length of each spoke) is R. The mass of each spoke is m. Determine (a) the moment of inertia of the wheel about an axis through its center and perpendicular to the plane of the wheel and (b) the moment of inertia of the wheel about an axis through its rim and perpendicular to the plane of the wheel. A uniform, thin, solid door has a height of 2.20 m, a width of 0.870 m, and a mass of 23.0 kg. Find its moment of inertia for rotation on its hinges. Are any of the data unnecessary? A uniform, hollow, cylindrical spool has inside radius R/2, outside radius R , and mass M (Fig. P10.68). It is mounted so that it rotates on a massless horizontal axle. A mass m is connected to the end of a string wound around the spool. The mass m falls from rest through a distance y in time t. Show that the torque due to the frictional forces between spool and axle is R[m(g 2y/t 2) M(5y/4t 2)] f An electric motor can accelerate a Ferris wheel of moment of inertia I 20 000 kg m2 from rest to 10.0 rev/min in 12.0 s. When the motor is turned off, friction causes the wheel to slow down from 10.0 to 8.00 rev/min in 10.0 s. Determine (a) the torque generated by the motor to bring the wheel to 10.0 rev/min and (b) the power that would be needed to maintain this rotational speed. 70. The pulley shown in Figure P10.70 has radius R and moment of inertia I. One end of the mass m is connected to a spring of force constant k, and the other end is fastened to a cord wrapped around the pulley. The pulley axle and the incline are frictionless. If the pulley is wound counterclockwise so that the spring is stretched a distance d from its unstretched position and is then released from rest, find (a) the angular speed of the pulley when the spring is again unstretched and (b) a numerical value for the angular speed at this point if I 1.00 kg m2, R 0.300 m, k 50.0 N/m, m 0.500 kg, d 0.200 m, and 37.0°. R m k θ Figure P10.70 71. Two blocks, as shown in Figure P10.71, are connected by a string of negligible mass passing over a pulley of radius 0.250 m and moment of inertia I. The block on the frictionless incline is moving upward with a constant acceleration of 2.00 m/s2. (a) Determine T1 and T2 , the tensions in the two parts of the string. (b) Find the moment of inertia of the pulley. 72. A common demonstration, illustrated in Figure P10.72, consists of a ball resting at one end of a uniform board 2.00 m/s2 M T1 T2 15.0 kg m1 m m 2 20.0 kg R/2 R/2 y Figure P10.68 37.0° Figure P10.71 326 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis of length , hinged at the other end, and elevated at an angle . A light cup is attached to the board at rc so that it will catch the ball when the support stick is suddenly Cup rc this limiting angle and the cup is placed at rc θ Hinged end Figure P10.72 removed. (a) Show that the ball will lag behind the falling board when is less than 35.3° ; and that (b) the ball will fall into the cup when the board is supported at 3 cos (c) If a ball is at the end of a 1.00-m stick at this critical angle, show that the cup must be 18.4 cm from the moving end. 73. As a result of friction, the angular speed of a wheel changes with time according to the relationship d /dt Support stick 2 0e t where 0 and are constants. The angular speed changes from 3.50 rad/s at t 0 to 2.00 rad/s at t 9.30 s. Use this information to determine and 0 . Then, determine (a) the magnitude of the angular acceleration at t 3.00 s, (b) the number of revolutions the wheel makes in the first 2.50 s, and (c) the number of revolutions it makes before coming to rest. 74. The hour hand and the minute hand of Big Ben, the famous Parliament tower clock in London, are 2.70 m long and 4.50 m long and have masses of 60.0 kg and 100 kg, respectively (see Fig. P10.26). (a) Determine the total torque due to the weight of these hands about the axis of rotation when the time reads (i) 3:00, (ii) 5:15, (iii) 6:00, (iv) 8:20, and (v) 9:45. (You may model the hands as long thin rods.) (b) Determine all times at which the total torque about the axis of rotation is zero. Determine the times to the nearest second, solving a transcendental equation numerically. ANSWERS TO QUICK QUIZZES 10.1 The fact that is negative indicates that we are dealing with an object that is rotating in the clockwise direction. We also know that when and are antiparallel, must be decreasing — the object is slowing down. Therefore, the object is spinning more and more slowly (with less and less angular speed) in the clockwise, or negative, direction. This has a linear analogy to a sky diver opening her parachute. The velocity is negative — downward. When the sky diver opens the parachute, a large upward force causes an upward acceleration. As a result, the acceleration and velocity vectors are in opposite directions. Consequently, the parachutist slows down. 10.2 (a) Yes, all points on the wheel have the same angular speed. This is why we use angular quantities to describe rotational motion. (b) No, not all points on the wheel have the same linear speed. (c) v 0, a 0. (d) v R /2, a a r v 2/(R/2) R 2/2 ( at is zero at all points because is constant).(e) v R , a R 2. 10.3 (a) I MR 2. (b) I MR 2. The moment of inertia of a system of masses equidistant from an axis of rotation is always the sum of the masses multiplied by the square of the distance from the axis. 10.4 (b) Rotation about the axis through point P requires more work. The moment of inertia of the hoop about the center axis is I CM MR 2, whereas, by the parallelaxis theorem, the moment of inertia about the axis through point P is IP I CM MR 2 MR 2 MR 2 2MR 2 . ...
View Full Document

This note was uploaded on 11/23/2010 for the course CIVIL 105 taught by Professor No during the Spring '10 term at Lovely Professional University.

Ask a homework question - tutors are online