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physics 6.pdf - Chapter 2 Rolling Motion Angular Momentum...

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Unformatted text preview: Chapter 2 Rolling Motion; Angular Momentum 2.1 The Important Stuff 2.1.1 Rolling Without Slipping When a round, symmetric rigid body (like a uniform cylinder or sphere) of radius R rolls without slipping on a horizontal surface, the distance though which its center travels (when the wheel turns by an angle 6) is the same as the arc length through which a point on the edge moves: AxCM = s = R6 (2.1) These quantities are illustrated in Fig. 2.1. The speed of the center of mass of the rolling object, vCM = ”61% and its angular speed are related by ”CM 2 Raw (2.2) and the acceleration magnitude of the center of mass is related to the angular acceleration by: aCM 2 R06 (2.3) Figure 2.1: Illustration of the relation between A30, 8, R and 0 for a rolling object. 35 36 CHAPTER 2. ROLLING MOTION; ANGULAR MOMENTUM The kinetic energy of the object is: Km“ : yaw? + éMUéM . (2.4) The first term on the right side represents the rotational kinetic energy of the object about its symmetry axis; the second term represents the kinetic energy the object would have if it moved along with speed UCM without rotating (i.e. just translational motion). We can remember this relation simply as: Kmu 2 Km + Ktrans. When a wheel rolls without slipping there may be a frictional force of the surface on the wheel. If so, it is a force of static friction (which does no work) and depending on the situation it could point in the same direction or opposite the motion of the center of mass; in all cases it tends to oppose the tendency of the wheel to slide. 2.1.2 Torque as a Vector (A Cross Product) In the last chapter we gave a definition for the torque 7' acting on a rigid body rotating around a fixed axis. We now give a more general definition for “torque”; we define the torque acting on a single particle (relative to some fixed point O) when a force acts on it. Suppose the (instantaneous) position vector of a particle (relative to the origin 0) is r and a single force F acts on it. Then the torque T acting on the particle is 7’2er (2.5) If gt is the angle between the position vector r and the force F then the torque T has magnitude 7' = rFsinng 2.1.3 Angular Momentum of a Particle and of Systems of Particles There is yet more important quantity having to do with rotations that will be of help in solving problems involving rotating objects; just as the linear momentum p was of importance in problems with interacting particles, the angular momentum of objects which have motion about a given axis will be useful when these objects interact with one another. Admittedly, some of the first definitions and theorems will be rather abstract! But we will soon apply the ideas to simple objects which rotate around an axis and then the theorems and examples will be quite downitoiearth. We start with a fundamental definition; if a particle has position vector r and linear momentum p, both relative to some origin O, then the angular momentum of that particle (relative to the origin) is defined by: €=r><p=m(r><v) (2.6) . . 2 Angular momentum has units of kgsm . One can show that the net torque on a particle is equal to the time derivative of its angular momentum: 2.1. THE IMPORTANT STUFF 37 Zr 22—? (2.7) This relation is analogous to the relation 2 F = ”fl—It) from linear motion. For a set of mass points in motion, we define the total angular momentum as the (vector) sum of the individual angular momenta: L=€1+£2+£3+... When we consider the total angular momentum, we can prove a theorem which is a bit different in its content than Eq. 2.7. It’s a bit subtle; when the particles in a system all move around they will be acted upon by forces from outside the system but also by the forces they all exert on one another. What the theorem says is that the rate of change of the total angular momentum just comes from the torques arising from forces exerted from outside the system. This is useful because the external torques aren’t so hard to calculate. The theorem is: ZText : E (2.8) This tells us that when the sum of external torques is zero then L is constant (conserved). We will encounter this theorem most often in problems where there is rotation about a fixed axis (and then once again we will only deal with the 2 components of 7' and L). 2.1.4 Angular Momentum for Rotation About a Fixed Axis An extended object is really a set of mass points, and it has a total angular momentum (vector) about a given origin. We will keep things simple by considering only rotations about an axis which is fixed in direction (say, the z direction), and for that case we only need to consider the component of L which lies along this axis, Lz. So, for rotation about a fixed axis the “angular momentum” of the rigid object is (for our purposes) just a number, L. Furthermore, one can show that if the angular velocity of the object is w and its moment of inertia about the given axis is I, then its angular momentum about the axis is L = [ca (2.9) Again, there is a correspondence with the equations for linear motion: px=muw (a) Lzlw 2.1.5 The Conservation of Angular Momentum In the chapter on Momentum (in Vol. 1) we used an important fact about systems for which there is no (net) external force acting: The total momentum remains the same. One can show a similar theorem which concerns net external torques and angular momenta. For a system on which there is no net external torque, the total angular momentum remains constant: L,- = Lf. This principle is known as the Conservation of Angular Momentum. 38 CHAPTER 2. ROLLING MOTION; ANGULAR MOMENTUM 2.2 Worked Examples 2.2.1 Rolling Without Slipping 1. An automobile traveling 80.0 km/hr has tires of 75.0 cm diameter. (a) What is the angular speed of the tires about the axle? (b) If the car is brought to a stop uniformly in 30.0 turns of the tires (without skidding), what is the angular acceleration of the wheels? (0) How far does the car move during the braking? [HRW5 12-31 (a) We know that the speed of the center of mass of each wheel is 80.0 km/hr. And the radius of each wheel is R = (75.0 cm)/2 = 37.5 cm. Converting the speed to 3 we have: 1h 103m km : km : m 80 h (80 h > (36008) <1km> 22.2 S From the relation between UCM and w for an object which rolls without slipping, we have: UCM = wR : w = 110% and we get (22.2 E) d = —S = 59.3 i w (0.375 m) s The angular speed of the wheel is 59.3 %. (b) As the car comes to a halt, the tires go through 30.0 turns. Thus they have an angular displacement of (with 60 = 0): 277 rad 1 rev 6 = (30.0 rev) < > = 188.5 rad . Also, when the wheel has come to a halt, its angular velocity is zero! So we have the initial and final angular velocities and the angular displacement. We can get the angular acceleration of the wheel from Eq. 1.8. From that equation we get: 2— 2 ofl2—593fli2 05:“ “0=( S) < S) = 9331““;d 2e 2(188.5rad) s The magnitude of the wheels’ angular acceleration is 9.33 :13. The minus sign in our result indicates that a goes in the sense opposite to that of the initial angular velocity (and angular displacement) of the wheel during the stopping. 2.2. WORKED EXAMPLES 39 (a) (b) Figure 2.2: (a) Constant horizontal applied to a rolling wheel in Example 3. (b) The forces acting on the wheel, with the points of application as indicated. (c) As we saw, the angular displacement of any wheel during the stopping was 188.5 rad. The radius of the wheel is R = 0.375 m, so from Eq. 2.1 the linear displacement of the wheel (i.e. its center) is: 330M 2 R6 = (0.375 m)(188.5 rad) 2 70.7m so the car goes 70.7m before coming to a halt. 2. A bowling ball has a mass of 4.0 kg, a moment of inertia of 1.6 X 10’2 kg-m2 and a radius of 0.10m. If it rolls down the lane without slipping at a linear speed of 4.0 ?, what is its total energy? [Ser4 11—5] The total (kinetic) energy of an object which rolls without slipping is given by Eq. 2.4. To use this equation we have everything we need except the angular speed of the ball. From Eq. 2.2 it is related to the linear velocity of the ball by UCM = Ru}, so the angular speed is ”CM (4.0 g) d = — = —S = 40.0 i w R (0.10m) s and then the kinetic energy is Kroll = %ICMW2 + %MUéM 2 are x 10-2 kg . m2)(40.0 %‘)2 + %(4.0kg)(4.0 g)? = 44.8J The total kinetic energy of the ball is 44.8 J. 3. A constant horizontal force of 10N is applied to a wheel of mass 10kg and radius 0.30m as shown in Fig. 2.2. The wheel rolls without slipping on the horizontal surface, and the acceleration of its center of mass is 0.60 8%. (a) What are the magnitude and direction of the frictional force on the Wheel? (b) What is the rotational inertia of the wheel about an axis through its center of mass and perpendicular to the plane of the wheel? [HRW5 12—9] 40 CHAPTER 2. ROLLING MOTION; ANGULAR MOMENTUM (a) The forces which act on the wheel along with where these forces are applied are shown in Fig 2.2 (b). In addition to the applied force of 10N which points to the right, there is a force of static friction between the surface and the wheel (of magnitude f3), which for now we draw pointing to the left (we can ask: Does it really point that way?). There are vertical forces acting on the wheel (from gravity and the normal force of the surface) but these clearly cancel out and for now we don’t need to worry about them. Even though the wheel will be rolling during its motion, Newton’s 2nd law still holds, and the sum of the horizontal forces gives may, Here the wheel is clearly accelerating to the right and so with the choice of directions given in the figure, we find: 2ng = 10.0N — f5 = max = (10kg)(0.60 5%) = 6.0N so that f, = 10.0N — 6.0N : 4.0N and since this is positive, the frictional force does indeed point to the left, as we guessed. Actually, it wasn’t so hard to guess that, since only a leftward frictional force could make the wheel rotate clockwise ias we know it must herei but for some problems in rolling motion, the direction of the static friction force may not be so evident. (b) Rotational inertia is related to net torque and angular acceleration by way of 7' 2 la. It is true that in this problem the rotating object is also accelerating but it turns out this relation still holds as long as we choose the center of mass to be the rotation axis. Considering the four forces in Fig. 2.2 (b), the applied force and the (effective) force of gravity are applied at the center of the wheel, so they give no torque about its center. The normal force of the surface is applied at the rim, but its direction is parallel to the line which joins the application point to the center so it too gives no torque. All that remains is the friction force, applied at a distance R from the center and perpendicular to the line joining this point and the axis; this force gives a clockwise rotation, so if we take the clockwise direction as the positive sense for rotations, then the net torque on the wheel is 7 = +f57“ = (4.0N)(0.30m) = 1.2N . m From Eq. 2.3 we know the angular acceleration of the wheel; it is _ CLCM (0.60 8%) — — = — = 2.0 fl . 0‘ R (0.30m) 52 and then from 7' 2 la we get the rotational inertia: 1.2N -=:=(—dm)=0.60kg-m2 oz (2.0 2%) 4. A round, symmetrical object of mass M, radius R and moment of inertia I rolls without slipping down a ramp inclined at an angle 6. Find the acceleration of its center of mass. 2.2. WORKED EXAMPLES 41 Figure 2.3: Round, symmetrical object with mass M, radius R and moment of inertia I rolls down a ramp sloped at angle 0 from the horizontal. The problem is diagrammed in Fig. 2.3. We show the forces acting on the object and where they are applied. The force of gravity, Mg is (effectively) applied at the center of the object. As usual we decompose this force into its components down the slope and perpendicular to the slope. The slope exerts a normal force N at the point of contact. Finally there is a force of static friction f5 from the surface; this force points along the surface and we can pretty quickly see that it must point up the slope because it is the friction force which gives the object an angular acceleration, which (here) is in the clockwise sense. Apply Newton’s 2nd law first: The forces perpendicular to the slope cancel, so that N 2 Mg cos6 (but we won’t need this fact). If aCM is the acceleration of the GM of the object down the slope, then adding the forces down the slope gives Mgsind — f3 = MaCM (2.10) Now we look at the net torque on the rolling object about its center of mass. The force of gravity acts at the center, so it gives no torque. The normal force of the surface acts at the point of contact, but since it is parallel to the line joining the pivot and the point of application, it also gives no torque. The force of friction is applied at a distance R from the pivot and it is perpendicular to the line joining the pivot and point of application. So the friction force gives a torque of magnitude 7' = Rfs sin90° = Rfs . and if we take the clockwise sense to be positive for rotations, then the net torque on the object about its CM is Tnet : Rfs From the relation 7' = 105 we then have T=RfS=Ioz (2.11) but we can also use the fact that for rolling motion (without slipping) the linear acceleration of the CM and the angular acceleration are related by: CLCMZTOK :> 04:— 42 CHAPTER 2. ROLLING MOTION; ANGULAR MOMENTUM and using this in Eq. 2.11 gives laCM l(ICM R 8 _ j 8 — 2.12 where we choose to isolate f, (the magnitude of the friction force) so that we can put the result into Eq. 2.10. When we do that, we get: [aCM MgsinQ— R2 = MaCM Now solve for aCM: I I Mgsind =MaCM—l— 23M 2 (M+—> aCM Our result is sensible in that if I is very small then (LCM is nearly equal to gsinQ, the result for a mass sliding with no rolling motion. 5. A uniform sphere rolls down an incline. (a) What must be the incline angle if the linear acceleration of the center of the sphere is to be 0.10 g? (b) For this angle, what would be the acceleration of a frictionless block sliding down the incline? [HRW5 12—7] (a) We will use the formula for CLCM (for rolling without slipping down a slope) in a previous problem. Note that we are not given the mass of the sphere! But it turns out that we don’t need it, because for a uniform sphere, we have I _ 2 MR2 _ 5 and as we can see from our earlier result, _ gsin6 aoM — —<1 + M11?) 7 (LCM just depends on the combination [/ (MR2). Solving this equation for sin 6 and plugging in the given numbers, we get: l | sin9=aC—M<1+ I ) <0'109)(1+§)=(0.10) = 0.14 g MR2 9 OWIKI 2.2. WORKED EXAMPLES 43 Which gives us 6 = sin—1(0.14) = 8.00 (b) As we saw in the Chapter on forces (Volume 1) when a mass slides down a frictionless incline its linear acceleration is given by a = gsinQ. For the slope angle found in part (a), this is a = gsinQ = (9.80 8%) sin 8.00 = 1.4 8% We can also calculate a/g: = Sing 2 Sin8.0O = 0.14 ‘QIQ so for the frictionless case we have a = 0149. 2.2.2 Torque as a Vector (A Cross Product) 6. What are the magnitude and direction of the torque about the origin on a plum (!) located at coordinates (—2.0m,0,4.0m) due to force F whose only component is (a) F36 : 6.0N, (b) F93 : —6.0 N, (0) F2 : 6.0N, and (d) FZ : —6.0N ? [HRW5 12—21] (a) The (vector) torque on a point particle is given by 'r = r x F I find it easiest to set up the cross product in determinant notation, discussed in Chapter 1 of Volume 1. We note that the units of the result must be N - m; then the cross product of r and F is i j k er= —2.0 0.0 4.0 N-m=(+(4.0)(6.0)N-m)j=(+24.0N-m)j 6.0 0.0 0.0 The torque 7' has magnitude 24.0N ' m and points in the +y direction. (b) Is is fairly clear that if we had had Fm = —6.0 N in part (a), we would have gotten ’7': (—24.0N-m)j so the torque would have magnitude 24.0 N - m and point in the —y direction. (c) When the only component of F is FZ = 6.0 N, then we have i j k er= —2.0 0.0 4.0 N-m=(—(—2.0)(6.0)N-n1)j=(+12.0N-m)j 0.0 0.0 6.0 so the torque would have magnitude 12.0 N - m and point in the +y direction. ((1) If instead we have only Fz = —6.0 N then the sign of the result in part (c) changes, and the torque would have magnitude 24.0 N - m and point in the —y direction. 44 CHAPTER 2. ROLLING MOTION; ANGULAR MOMENTUM 6'5 kg 2.2 m/s if 1.5 In all 2.8 m 0 Figure 2.4: Two masses and their motion relative to the point 07 as in Example 8. 2.2.3 Angular Momentum of a Particle and of Systems of Particles 7. The position vector of a particle of mass 2.0 kg is given as a function of time by r = (6.0i+5.0tj) m when t is given in seconds). Determine the angular momentum of the particle as a function of time. [Ser4 11—17] The angular momentum of a point mass is given by 2 = r X p. The velocity of our particle is given by d d . . - m V : fir : $63.01 + 5.0m) m — (5.0J) g and its momentum is p = mv = (2.0 kg)(5.0j) § = (10.0j) 1% Now take the cross product to get 3: i j k €=r><p= 6.0 5.015 0.0 5%:(50010hssg 0.0 10.0 0.0 8. Two objects are moving as shown in Fig. 2.4. What is their total angular momentum about point 0? [HRVV5 12—27] In this problem we must use the definition of the angular momentum of a particle (with respect to some origin 0): K = r x p First consider the 3.1 kg mass. Its momentum vector has magnitude P1 = 772101 2 (3.1kg)(3.6 §) 2 11.2 kgT-m 2.2. WORKED EXAMPLES 45 and is directed upward in this picture. The vector r which goes from point 0 to the particle has magnitude 2.8m and points to the right. By the rightihand rule for cross products, the vector r X p points up out of the page, which is along the +2 axis; and since the two vectors are perpendicular, the magnitude of r X p is 2 |r1 >< p1| 2 mm sin 90° = (2.8m)(11.2 1%) = 31.2 kg: and so the angular momentum of this particle about 0 is 31 2 (+312 flyim The 6.5 kg mass has a momentum vector of magnitude p2 = mm = (6.5kg)(2.2 g) = 14.3 kg?“ and is directed to the right. The vector r which goes from the point 0 to the particle has magnitude 1.5m and points straight up. By the rightihand rule for cross products, the vector r X p points into the page, which is along the —z axis; and since the two vectors are perpendicular, the magnitude of r X p is |r2 >< p2l = r2192 sin 90° 2 (1.5m)(14.3 1%) = 21.5 k331i and so the angular momentum of this particle about 0 is 32 = (—21.5 kg‘m2)k S The total angular momentum of the system is 3 = 31% = (31.2 — mm W = (+9.8 Wm. S S 2.2.4 Angular Momentum for Rotation About a Fixed Axis 9. A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 6.00m long, weighs 10.0 N, and rotates at 240 rev/min clockwise when seen from above. Calculate (a) the rotational inertia of the rod about the axis of rotation and (b) the angular momentum of the rod about that axis. [HRW5 12—45] (a) The mass of the rod is M = W/g = (10.0 N)/(9.80 8%) = 10.2 kg and its angular velocity in units of % is . > = 25.1 @ min 608 S w_240rev 27rrad (1min _ 1rev 46 CHAPTER 2. ROLLING MOTION; ANGULAR MOMENTUM We know the formula for the moment of inertia of a uniform rod rotating about an axis at one of its ends (see Chapter 1, Fig. 1) so we calculate I as: 2 2 2 1: gML = (10.2kg)(6.00m) = 12.2kg-m l 3 The rotational inertia of the rod (about the given axis) is 12.2 kg - m2. (b) The angular momentum of the rotating rod will be given by L = I to. We find that the magnitude of the angular momentum is: L = 1w = (12.2 kg - m2)(25.1 %) = 309 kg“? S The vector L would point upward (along the +2 axis if the rotation were counterclockwise as seen from above. That is not the case (it is clockwise) so the direction of L is downward. 2.2.5 The Conservation of Angular Momentum 10. Suppose that the Sun runs out of nuclear fuel and suddenly collapses to form a white dwarf star, with a diameter equal to that of the Earth. Assuming no mass loss, what would then be the Sun’s new rotation period, which currently is about 25 days? Assume that the Sun and the white dwarf are uniform, solid spheres; the present radius of the Sun is 6.96 X 108 m. [HRW5 12-55] In this ...
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