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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 rigidobject 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 200m or
400m 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 righthand 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 rigidbody 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 righthand 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 473s 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 linearmotion 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 crosssectional 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 crosssectional 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 parallelaxis theorem, often simpliﬁes the calculation. Suppose the moment of inertia about an axis through the center of mass of
an object is ICM . The parallelaxis 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) Parallelaxis 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 ParallelAxis 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 parallelaxis 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 ParallelAxis 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 centerofmass axis and the
y axis is D L/2, the parallelaxis 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 onepiece 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 rigidbody 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 freebody 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. Freebody 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) Freebody diagrams for the blocks. (c) Freebody 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 3.00s 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 orderofmagnitude 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 6.00kg 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 singleengine 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 unitvector 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 (3.80cm 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 orderofmagnitude 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 500kg 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 parallelaxis theorem and Table 10.2 to ﬁnd the
moments of inertia of (a) a solid cylinder about an axis
parallel to the centerofmass 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 500kg 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 diskshaped 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 freebody 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 9.00cmdiameter 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 5.60cmdiameter 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 15.0kg mass and a 10.0kg 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 800N merrygoround 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 midfall (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 4.00m 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 crosssectional 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 1.00m 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 . ...
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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.
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