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```					                       Chapter 6   Rotation and Angular Momentum.

Chapter 6            Rotation and Angular Momentum

Basic Requirements：
1. Understand the concept of translation and rotation;
2. Master the kinematic equations for constant angular acceleration;
3. Master the relationship between the linear and angular variables;
4. Master kinetic energy of rotation;
5. Master the calculation of the rotational inertia;
6. Master the parallel-axis theorem;
7. Learn to apply Newton's second law for rotation;
8. Master the work-kinetic energy theorem for rotation.
9. Understand the concept of rolling;
10. Master the kinetic energy of rolling;
11. Master the forces of rolling;
12. Master angular momentum;
13. Learn to apply Newton's second law in angular form;
14. Understand the angular momentum of a system of particles;
15. Understand the angular momentum of a rigid body rotating about a fixed axis;
16. Master the conservation law of angular momentum.

Review and Summary
Static Equilibrium A rigid body at rest is said to be in static equilibrium. For such a
body , the vector sum of the external forces acting on it is zero:


Fnet  0      (balance of forces )                          (6-3)

If all the forces lie in the xy plane, this vector equation is equivalent to two
component equations:

Fn e, t  0
x     and      Fnet , y  0   (balance of forces)     (6-7,6-8)

Static equilibrium also implies that the vector sum of the external torques acting on the

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Chapter 6     Rotation and Angular Momentum.
body about any point is zero, or


 net  0                        (balance of torques)      (6-5)

If the forces lie in the xy plane, all torque vectors are parallel to the z axis, and
Eq.6-5 is equivalent to the single component equation

 net , z  0                    (balance of torques).     (6-9)

Angular Position To describe the rotation of a rigid body about a fixed axis, called
the rotation axis, we assume a reference line is fixed in the body, perpendicular to
that axis and rotating with the body. We measure the angular position  of this line
relative to a fixed direction. When  is measured in radians,
s
r
Where s is the arc length, of a circular path of radius r and angle  . Radian
measure is related to angle measure in revolutions and degrees by

1r e v 3 6 0 2 r a d .
                                                  (6-11)

Angular Displacement        A body that rotates about a rotation axis , changing its

angular position from 1 to  2 , undergoes an angular displacement

   2  1 ,                                 (6-13)

Where  is positive for counterclockwise rotation and negative for clockwise
rotation.
Angular Velocity and Speed  If a body rotates through an angular displacement

 in a time interval t , its average angular velocity  avg is


 avg        .                                   (6-14)
t
The (instantaneous) angular velocity            of the body is
d
          .                                      (6-15)
dt
Both    avg and  are vectors, with directions given by the right hand rule. they

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Chapter 6   Rotation and Angular Momentum.

are positive for counterclockwise rotation and negative for clockwise rotation.
The magnitude of the body’s angular velocity is the angular speed.
Angular Acceleration     If the angular velocity of a body changes from 1 to 2 in
a time interval t  t1  t 2 , the average angular acceleration       avg of   the body is

2  1        
 avg                                                  (6-16)
t2  t1        t
The (instantaneous) angular acceleration           of a body is
d
       .                                              (6-17)
dt
Both    avg and  are vectors.
The Kinematic Equations for Constant Angular Acceleration Constant angular
acceleration (  =constant) is an important special case of rotational motion .The
appropriate kinematic equations are

  0  at                                                (6-18)

1
  0  0t   t 2                                       (6-19)
2

 2  02  2 (  0 )                                   (6-20)

1
   0  (  0 )t                                       (6-21)
2
1
  0  t   t 2                                        (6-22)
2
Linear and Angular Variables Related A point in a rigid rotating body , at a
perpendicular distance r from the rotation axis , moves in a circle with radius r . If
the body rotates through an angle  , the point moves along an arc with length s
given by

s   r (radian measure),                                 (6-23)

Where  is in radians .
The liner velocity v of the point is tangent to the circle, the point’s liner speed is

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Chapter 6    Rotation and Angular Momentum.
given by
v  r              (radian measure),            (6-24)
Where    is the angular speed (in radians pre second) of the body .
The liner acceleration a of the point has both tangential and radial components.
The tangential component is

at   r           (radian measure),            (6-28)

Where  is the magnitude of the angular acceleration (in radians per

second-squared ) of the body . The radial component of a is

v2
ar         2r      (radian measure),          (6-28)
r
If the point moves in uniform circular motion. the period T of the motion for the
point and the body is
2 r 2
v    
Rotational Kinetic Energy and Rotational Inertia         The kinetic energy K of a rigid
body rotating about a fixed axis is given by
1 2
2
In which I is the rotational inertia of the body, defined as

I   mi ri
2
(6-39)

for a system of discrete particles and as

I   r 2 dm                                     (6-43)

for a body with continuously distributed mass. The r and ri in these expressions
represent the perpendicular distance from the axis of rotation to each mass element in
the body.
The Parallel-Axis Theorem          The parallel-axis theorem relates the rotational inertia
I of a body about any axis to that of the same body about a parallel axis through the
center of mass:

I  I com  Mh 2                                           (6-42)

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Chapter 6   Rotation and Angular Momentum.

Here h is the perpendicular distance between the two axes.

Torque Torque is a turning or twisting action on a body about a rotation axis due to a
                                                                       
force F . If F is exerted at a point given by the position vector r relative to the
axis, then the magnitude of the torque is
  (r )( F sin  )
                        
Where Ft is the component of F perpendicular to                        r , and  is the angle
        
between r and F . The quantity r                   is the perpendicular distance between the

rotation axis and an extended line running through the F vector. This line is called

the of action of F . Similarly, r is the moment arm of Ft .
The SI unit of torque is the Newton-meter ( N  m ). A torque is positive if it tends to
rotate a body at rest counterclockwise and negative if it tends to rotate the body in the
clockwise direction.
Newton’s Second Law in Angular Form                    The rotational analog of Newton’s second
law is

 n e t I ,                                            (6-45)

Where  net is the net torque acting on a particle or rigid body, I is the rotational
inertia of the particle or body about that rotation axis. and           is the resulting angular
Work and Rotational Kinetic Energy                 The equations used for calculating work and
power in rotational motion correspond to equations used for translational motion and
are
f
W    d                                                 (6-56)
i

When    is constant, Eq. 6-56 reduces to
W   ( f  i )                                            (6-57)

dW
And                       P                                                          (6-58)
dt
The form of the work-kinetic energy theorem used for rotating bodies is
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Chapter 6   Rotation and Angular Momentum.
1 2 1 2
K  K f  Ki          I  f  I i  W               (6-60)
2        2
Rolling Bodies For a wheel of radius R that is rolling smoothly (no sliding),

 com  R                                  (6-67)

Where com is the linear speed of the wheel’s center and  is the angular speed of

the wheel about its center. The wheel may also be viewed as rotating instantaneously
about the point P of the “road” that is in contact with the wheel. The angular speed
center. The rolling wheel has kinetic energy
1           1
K      I com 2  M 2com                            (6-70)
2           2
Where I com is the rotational moment of the wheel about its center and M is the

mass of the wheel. If the wheel is being accelerated but is still rolling smoothly, the

acceleration of the center of mass acom is related to the angular acceleration          

acom  R                                          (6-71)

If the wheel rolls smoothly down a ramp of angle  , its acceleration along an              x
axis extending up the ramp is

g sin 
acom , x                                  (6-75)
1  I com / MR 2

Torque as a Vector        In three dimensions, torque  is a vector quantity defined
relative to a fixed point (usually an origin); it is
        
  r F,                                           (6-76)
                                              
Where F is a force applied to a particle and r is a position vector locating the

particle relative to the fixed point (or origin). The magnitude of  is given by

  rF sin   rF  r F ,                    (6-77,6-78,6-79)

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Chapter 6     Rotation and Angular Momentum.
                                                
Where  is the angle between F and r , F is the component of F
                                                                       
perpendicular to r , and r is the moment arm of F . The direction of  is given
by the right-hand rule for cross products.

Angular Momentum of a particle                      The angular momentum  of a particle with
                                                 
linear momentum p , mass m , and linear velocity  is a vector quantity defined
relative to a fixed point (usually origin); it is
            
  r  p  m(r  )                                   (6-80)
The magnitude of            is given by
 rm sin                                           (6-81)
 rp  rm                                         (6-82)
 r p  r m                          (6-83)
        
Where  is the angle between r and p , p  and   are the components of
                           
p and  perpendicular to r , and r is the perpendicular distance between the

fixed point and the extension of p . The direction of is given by the right-hand
rule for cross products.
Newton’s Second Law in Angular Form                               Newton’s second law for a particle can
be written in angular form as

d
 net          ,                                     (6-85)
dt

Where  net is the net torque acting on the particle, and                      is the angular momentum
of the particle.

Angular Momentum of a System of Particles                              The angular momentum L of a
system of particles is the vector sum of the angular momentum of the individual
particles:
n
L       1      2            n           i                     (6-88)
i 1

The time rate of change of this angular momentum is equal to the net external torque
on the system (the vector sum of the torques due to interactions of the particles of the

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Chapter 6    Rotation and Angular Momentum.
system with particles external to the system):

        dL
 net                 (system of particles)                     (6-91)
dt
Angular Momentum of a Rigid Body             For a rigid body rotating about a fixed axis is
L  I          (rigid body, fixed axis).                (6-93)

Conservation of Angular Momentum               The angular momentum L of a system
remains constant if the net external torque acting on the system is zero:

L  acons tan t            (isolated system)                         (6-94)
    
Or               Li  L f                   (isolated system) .                       (6-95)
This is the law of conservation of angular momentum. It is one of the fundamental
conservation laws of nature, having been verified even in situations (involving
high-speed particles or subatomic dimensions) in which Newton’s laws are not
applicable.

Examples

Example 1 .Show that the moment of inertia of a uniform hollow cylinder of inner
1
radius R1 ,outer radius R2 ,and mass M , is I              M ( R12  R2 ) ,as stated in the
2

2
figure, if the rotation axis is through the center along the axis of symmetry.
2
Solution：We know that the moment of inertia of a thin ring of radius R is m R . So

we divide the cylinder into thin concentric cylindrical rings
or hoops of thickness dR , one of which is indicated in
Fig.6-1. If the density (mass per unit volume ) is  ,then

dm  dV ,

Where dV is the volume of the thin ring of radius R ,
thickness dR , and height h . Since

dV  (2R)(dR)(h)                            Fig. 6-1   Example 1

We have
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Chapter 6    Rotation and Angular Momentum.

dm  2hR dR
Then the moment of inertia is obtained by integrating (summing) over all these hoops :

I   R 2 dm

R1                  R 4  R14 
  2hR 3 dR  2h  2         ，
 4 
R2

Where we are given that the cylinder has uniform density ,   constant .(If this were

not so ,we would have to know         as a function of R before the integration could

be carried out) The volume V of this hollow cylinder is V  (R2  R1 )h ,so its
2         2

mass M is

M  v   ( R2  R12 )h.
2

Since ( R2  R1 )  ( R2  R1 )( R2  R1 ) ,we have
4      4       2      2      2     2

h                                1
I         ( R2  R12 )( R2  R12 ) 
2           2
M ( R12  R2 )
2
2                                2
As stated in figure. As a check ,note that for a solid cylinder , R1  0 and we obtain,

with R2  R0 :

1
I       2
MR0 ,
2
Which is that given in Fig.10-21c for a solid cylinder of mass M and radius R 0 .

Example 2      what will be the speed of a solid sphere of mass M and radius R 0
when it reaches the bottom of an incline if it starts from rest at a vertical height H
and rolls without slipping ? see Fig.6-2. Ignore losses due to dissipative forces ,and
compare your result to that for an object sliding down a frictionless incline.

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Chapter 6   Rotation and Angular Momentum.
Solution ： We use the law of conservation of
energy, and we must now include rotational
kinetic energy. The total energy at any point a
vertical distance y above the base of the
incline is
1         1
Mv 2  I CM  2  Mgy ,
2         2
Where v is the speed of the CM . We equate                          Fig. 6-2   Example 2

the total energy at the top （ y  H and v    0) to the total energy at the bottom

( y  0 );

1       1
0  0  MgH         Mv 2  I CM w 2  0 .
2       2
The moment of inertia of a solid sphere about an axis through its COM is
2
ICOM  MR0 .
2

5
Since the sphere rolls without slipping , the speed , v , of the center of mass with
respect to the point of contact (which is momentarily at rest at any instant) is equal to
the speed of a point on the edge relative to the center. We therefore have w  v / R .
Hence

12      v          
2
1
Mv 2   MR02  2            MgH .
2       25      R0



Dividing out the M , s and R , s, we obtain

1 1 2
  v  gH
 2 5
So

10
7
Note first that v is independent of both the mass M and the radius R of the sphere .
Also , we can compare this result for the speed of a rolling sphere to that for an object

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Chapter 6   Rotation and Angular Momentum.
1
sliding down a plane without rotating and without friction (       Mv 2  mgH ),
2
in which case v         2 gH , which is greater. An object sliding without friction
transforms its initial potential energy into translational kinetic energy (none into
rotational kinetic energy ), so its speed is greater.

Example 3 Suppose a 60-kg person stands at the
edge of a 6.0-m-diameter circular platform, which is
mounted on frictionless bearings and has a moment of
2
inertia of 1800kg. m . The platform is at rest initially,
but when the person begins running at a speed of 4.2m/s
(with respect to the ground).around its edge the platform
begins to rotate in the opposite direction as in figure 6-3.
Calculate the angular velocity of the platform.                       Fig. 6-3 Example 3

Solution： The total angular momentum is zero initially .Since there is no net
torque, L is conserved and will remain zero ,as in Fig.6-3.The person’s angular

momentum is L per  (mR )(v / R) and we take this as positive .the angular
2

momentum of the platform is Lplat   I .Thus

L  Lper  Lplat

v
0  mR 2    I 
R
So

m      .
m R v ( 6 0 k g) ( 3 . 0 ) ( 4m 2 s / )
                           2
 . 4 r a d /.s
I           1800g mk .

The    frequency of       rotation   is     f  w / 2  0.067rew / s and the period

T  1 / f  15s per revolution.                                         (Answer)

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Chapter 6    Rotation and Angular Momentum.

Example 4 A uniform rod of length l and mass m is placed horizontally by
putting its left end B on the edge of a table and holding its right end A with your
hand . Then you release the end A suddenly . find , at the instant end A is
released

(a) the acceleration a of the rod’s center of mass ?

(b) the force F exerted on the rod at end B ?
Solution: Draw the free-body diagram of the rod .set the coordinate system as in
the figure. At the moment the end A is released there are two forces acting on the
      
rod: the gravitational force F  mg exerted at the
g

center of the rod , downward. The force F
n     B                       A
exerted at end B from the table. Apply
the Newton’s second law for the center of mass
of the rod.                                                                   Fig. 6-4 Example 4
    
 F  ma
i        c

In    direction:                 F  mg  ma
              c
(1)

vc 2
In n direction:             Fn  mac  m                                                 (2)
l/2
According to the theorem of rotation we write:
l
mg       I z                                                   (3)
2
from the relationship between the linear and angular quantities we have
l
ac                                                                  (4)
2
1
and                 I z  ml 2                                               (5)
3
At the moment end A is just released .the speed of the rod’s center of mass v  0          c

thus v  0
c

Substituting v  0 into Eq. (2), yields F  0
c                                        N

Then solving equations (3), (4) and (5) for a yields         c

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Chapter 6       Rotation and Angular Momentum.
1                          1
ac   g               .a        gˆ
3                          3
Substituting ac into Eq. (1) we obtain

1                      1
F  mg               F      ˆ
4                      4

Example 5 A uniform solid ball of radius r and mass m rolls down a circular
track from rest without sliding .The track, radius R , is in the vertical plane. At the
beginning the ball is in the same height as the center of the circular track , Find
(a) the ball’s speed as it rolls down to the bottom of the track,
(b) the normal force on the track from the ball at the same moment as in (a).
Solution：(a) Taking the circular track as the reference
frame , the system of the problem includes the ball ,                             FN
the track and the Earth. Draw the free-body diagram
FS
of the ball in the process of rolling down the track as
Fg
shown in figure 6-5. Three forces , the gravitational

force Fg , the normal force F              and the static               Fig. 6-5 Example 5
                     N

frictional force F are acting on the ball, But only the gravitational force Fg , being
S

a conservative force, does work in the rolling process .Thus the mechanic energy of
the system conserved. When the ball is at the bottom of the track, its center is chosen
as the reference of the zero potential energy. From the conservation law of mechanic
energy , we have

mg ( R  r )  (1/ 2)mv2  (1/ 2) I c 2                        (1)

Where I   c
is the rotational inertia of the ball about a axis through its center, v is the
speed of the center of mass (COM) of the ball , Then we have

I c  (2 / 5)mr 2                                           (2)

v  r                                                     (3)
Solve these three equations for v yield

v  (10 / 7) g ( R  r )
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Chapter 6   Rotation and Angular Momentum.

(b) When the ball rolls down to the bottom of the circular track, two forces,
     
F and F .exerted on the ball , Write Newton’s second law in the normal direction
n

Fn  mg  mv2 / ( R  r )

Therefore

FN  mg  mv / ( R  r )  (17 / 7)mg

This is the normal force on the ball from the track, Write Newton’s third law it is

known that the normal force FN on the track from the ball of the bottom


FN   FN            
FN  ( 1 7 / 7 ) g

Example 6         A thin rod of mass m and length l is connected with a small ball of
the same mass m at the rod’s one end, and the other end of the rod is pivoted on a
frictionless hinge as shown in the figure ,The rigid body is held at rest horizontally and
then released. What is (a) the rotational inertia of the rigid
body about the hinge ? (b) the angular speed of the rigid
body as its rod forms an angle  with the vertical line ?
(c) the angular acceleration of the rigid body at the same
instant as in(b)? (d) the normal acceleration of the center of
mass of the rigid body at the same instant as in (b)?
Solution: (a) the rotational inertia of the rigid body
about the hinge is                                                        Fig. 6-6    Example 6

I0  (1/ 3)ml 2  ml 2  (4 / 3)ml 2                                        (1) (Answer)

(b) During the rotation of the rigid body only the gravitational force does work so the
process obeys the conservation law of mechanic energy . But first of all we should find
out the center of mass of the rigid body by

M1 X1  M 2 X 2 3
X COM                    l                                                (2)
M1  M 2      4

When the rod forms an angle  with the vertical line its center of mass C                 com
has
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Chapter 6        Rotation and Angular Momentum.

descent a distance

h  (3 / 4)l cos                                                     (3)

Applying the conservation law of mechanic energy. We write

(2m) gh  (1/ 2) I 0 2

Substituting (1), (2) and (3) into above equation and solve for              , yields
  (3 / 2) ( g / l ) cos                                         (Answer)

(c) To find out the angular acceleration of the rigid body at the same instant as in (b)
       
we use Newton’s second law for rotation   I

0          (2mg )( 1 l sin  ) 9 g
3
                                     sin                      (Answer)
I           4
3 ml 2           8l

(d) The normal acceleration of the center of mass of the rigid body at the same instant
as in (b) is
3
v 2t     ( l ) 2    3      3 9        27
an              4
3         l 2  g cos      g cos                   (Answer)
r         4
l       4      4 4        16

Example 7 As show in the figure, the masses of wheels A and B are m1 and
m2 and the radii of them are r1 and r2 respectively, there is a thin rope rolling
around the two wheels and connecting the wheels as shown in the figure, where wheel
A rotates about the fixed O axis. (1) when wheel B drops what is the acceleration
of the center of the wheel? (2) what is pulling force of the rope?
Solution: We consider the problem as the combined motion of the two wheels. That
is the rotation of wheels A and the plane motion of wheel B . Then we can apply the
law of rotation (the Newton’s second law for rotation) and the Newton’s second law to
get the answer. As always we begin with a free-body diagram as shown in the
figure Wheel A rotates about the fixed axis O (perpendicular to the page). From the
law of rotation
 0  I 0 A

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Chapter 6   Rotation and Angular Momentum.

Fig. 6-7   Example 7
1
FT r1  m1r12 A                                              (1)
2
Wheel B rotates about the instantaneous axis C (perpendicular to the page) while its
,
center of mass C is moving down translationally. According to Newton s second
law for the motion of the center of mass of wheel B
m2 g  FT  m2 ac                                               (2)
From the law of rotation for wheel B
1
FT r2      m2 r22 B                                               (3)
2
By the relationship between the linear quantities and the angular quantities we have

aA                 aB
A        ,       B                                                (4)
r1                 rB

Where a A and a B are accelerations of a point at the edge of wheel A and B
respectively.
There are two more relations which help the problem solving
a A  aC  a B                                                 (5)
FT  FT                                                       (6)
Solving above six equations for a C and FT , obtaining

- 150 -
Chapter 6   Rotation and Angular Momentum.

2m1  m2 
3m1  2m2

m1m2
FT  FT,               g                                 (Answer)
3m1  2m2

Example 8 A thin rod of mass m1 and length  can rotate about a frictionless
axis O freely. A small ball of mass m2 is suspended at the end of a massless cord of
length  on the same axis as shown in the figure . At the beginning the rod rests in
vertical position and the ball is pulled with its cord makes an angle of  with the rod.
Then the ball is released and swing down making a elastic collision with the rod. The

rod deflects a maximum angle 60 . Write down enough equations to determine the
angle  .
Solution: We analyze the whole process by
dividing it into four successive parts as follows
and then use conservation law in each of them
(a) During the swing process of the ball, only the
gravitational force does work, so the mechanic
energy of the ball is conserved.
1
m2 g(1  cos  )       m2 v 2 (1) (Answer)
2
Where v is the speed of the ball just before it                Fig. 6-8   Example 8
collides with the rod.
(b) As the ball collides with the rod, there is no external torque about the reference
point O , thus the angular momentum of the system (the ball plus the rod) is
conserved
1
m2v  m2v  m1 2                                 (2)       (Answer)
3
Where v and v are the speed of the ball just before and after the collision,  is
the angular speed of the rod just after the collision.

- 151 -
Chapter 6   Rotation and Angular Momentum.

(c) Since the collision between the ball and the rod is elastic, the mechanic
energy of the system is conserved just before and after the elastic collision.

1         1          11      
m2 v 2  m2 v , 2   m1l 2  2                     (3)      (Answer)
2         2          23      

(d) In the consequential swing process of the rod the mechanic energy of the rod also
is conserved

11
23
2 2      

 m1l   m1 g 1  cos 60

        2

Solving above four equations for  we can get the answer.
Note that the most common mistake in solving this problem is apply the
conservation law of linear momentum during the collision between the ball and the rod
m2v  m2v  m1 
instead of equation (2). This is wrong because during the collision the horizontal
component of the force on the rod from the axis can not be neglected. This does not
satisfy the condition of applying the conservation law of linear momentum.

Problem Solving
1 (12) The wheel in Fig.6-9 has eight equally spaced
spokes and a radius of 30cm . It is mounted on a fixed
axle and is spinning at 2.5rev / s . You want to shoot a
20  cm long arrow parallel to this axle and through the
wheel without hitting any of the spokes. Assume that the
arrow and the spokes are very thin. (a) What minimum
speed must the arrow have? (b) Does it matter where
between the axle and rim of the wheel you aim? If so,
what is the best location?                                         Fig. 6-9   Problem 1
Solution: The angular speed of the rotating wheel is
  (2.5rev / s)(2 rad / rev)  5 rad / s
The angle between the maximum angular displacement when a arrow passes throug
the wheel, is

- 152 -
Chapter 6   Rotation and Angular Momentum.
8   4
The time required for a spaced spoke turns around the
angle  is
t                 s
 5 rad / s 20
During this time interval the 20cm  long arrow must pass though the wheel with
the required speed

l  0.2cm
v            4m / s                                     (Answer)
t (1/ 20) s
(b) Since in equal time interval the wheel turns a equal angle it does not matter where
between the axle and rim of the wheel you aim.                              (Answer)

2 (16) An early method of measuring the speed of light makes use of a rotating
slotted wheel. A beam of light passes through one of the slots at the outside edge of the
wheel, as in Fig.6-10, travels to a distant mirror, and returns to the wheel just in time
to pass through the next slot in the wheel. One such slotted wheel has a radius of 5.0
㎝ and 500 slots around its edge. Measurements taken when the mirror is L=500m
from the wheel indicate a speed of light of 3.0×105 ㎞／s.(a)What is the (constant)
angular speed of the wheel? (b)What is the linear speed of a point on the edge of the
wheel?
Solution: The distant for light beam traveling
through one slot to the reflecting mirror and
backing through the next slot in the wheel is
d  2L  2  500m  1000m
The time for the light bean travel that distance is
d    1000m      1
t                  105 s
v 3.0 10 m / s 3
8

During this time interval the wheel just
turns one slot, the angular displacement is
2
           rad                                     Fig. 6-10 Problem 2
500
So the angular speed of the wheel is


 ( 2 / 5 0r0a)d
                5
 r
 3 . 8 13 0 a d /
t        
(1 / 3 ) 1 0
- 153 -
Chapter 6   Rotation and Angular Momentum.
The linear speed of a point on the edge of the wheel is

v  R  (0.005m)(3.8 103 rad / s)  1.9 102 m / s                         (Answer)

3 (20) In Fig.6-11, a cylinder having a mass of 2.0 ㎏ can rotate about its central
axis through point O. Forces are applied as shown F1  6.0 N , F2  4.0 N ,
F3  2.0 N and F4  5.0N .Also, r  5.0cm and R  12cm . Find the
(a)magnitude and(b)direction of the angular acceleration of the cylinder.(During the
rotation, the forces maintain their same angles relative to the cylinder.)
Solution: We apply the Newton’s second law for
rotation to find out the angular acceleration of the

cylinder. Since F4 passes through the rotational
axis it produce no torque about the same axis
therefore
                        
       F1 R  F2 R  F3 r  I

Taking the counterclockwise direction as
positive for torques and the angular
Acceleration of the cylinder, we have                          Fig. 6-11 Problem 3
                   
F1 R  F2 R  F3 r  I


 F  F  R  Fr   6.0 N  4.0 N   0.12m  2.0 N  0.05m
1     2

1                         1
 2.0Kg  0.12m 
2
mR 2
2                         2
The direction of the angular acceleration of the cylinder is counterclockwise.

4 (22) Figure 6-12 shows a rigid assembly of a thin
hoop (of mass m and radius R  0.150m ) and a thin
radial rod (of mass m and length L  2.00 R ). The
assembly is upright, but if we give it a slight nudge, it
will rotate around a horizontal axis in the plane of the rod
and hoop, through the lower end of the rod. Assuming
that the energy given to the assembly in such a nudge is
negligible, what would be the assembly’s angular speed               Fig. 6-12    Problem 4
- 154 -
Chapter 6   Rotation and Angular Momentum.
about the rotation axis when it passes through the upside-down (inverted)
orientation?
Solution: In the rotating process only the gravitational force does work on the rigid
assembly, so we can use the conservation law of mechanical energy to find out the
answer. But first we must find out the position of the center of mass of the assembly
my  my mR  m3R
yc                     2R
mm     mm
Next we should work out the rotational inertia of the rigid assembly about a horizontal
axis in the plane of the rod and hoop through the low end of the rod

1               2
I0  I0Rod  I0Hoop  m(2R) 2   mR 2  m  3R    0.244m  kgm 2 
1
3          2                
Choose point c as the reference level for U  0 .
From the conservation law of mechanical energy, we have
1
Mgyc       I 0 0
2

2
Solving for  0 yields

2  m  m  g  4R   2  2  9.8 m 2  4  0.15
I0                  0.224kg  m 2                  s

5 (23) A tall, cylindrical chimney falls over when its base is ruptures. Treat the
chimney as a thin rod of length 55.0m . At the instant it makes an angle of 35.0
with the vertical as it falls, what are (a) its angular speed, (b) the radial acceleration of
the top, and (c) the tangential acceleration of the top.(Hint: Use energy considerations,
not a torque.) (d) At what angle  is the tangential acceleration equal to g ?
Solution: In the system under consideration ( the chimney plus the earth ),only the
gravitational force does work. So we can apply the conservation law to get the
answer. Supposing the mass of the chimney is M and taking the position of the center
of mass of the chimney when it makes an angle of 35.0 with the vertical as it falls
to be the zero potential energy level.(a) Form the principle of conservation of
mechanic energy we have
L               1 1
Mg     (1  cos  )  ( ML2 ) 2
2               2 3

- 155 -
Chapter 6   Rotation and Angular Momentum.

3g (1  cos  )   3(9.8m / s)(1  cos 35.0 )
                     
L                   55.0m

（b）The linear speed of the top of the chimney is

v  L  55.0m  0.311rad s  17.1 m s
So the radial acceleration of the top is

v 2 (17.1 m s)2
an                    5.32 m s
2
L     55.0m
(c) The tangential acceleration of the top can be obtained by Newton’s second law for
rotation  0  I 0
l         1
Mg sin   ( ML2 )
2         3
3g
The angular acceleration of the rotationally falling chimney is           sin 
2L
when   35.0

3(9.8 m s 2 )
               sin 35.0  0.153 rad s 2
2(55.0m)

So the tangential acceleration of the top is

a  L  (35.0m)(0.153 rad s 2 )  8.43 m s 2                   (Answer)

3g        3
(d) From equation a  L  L         sin   g sin          when a  g
2L        2
3
we have      sin   1 ，Then
2

6 (31) Four particles, each of mass 0.20 ㎏, are placed at the vertices of a square

- 156 -
Chapter 6   Rotation and Angular Momentum.
with sides of length 0.50ｍ.The particles are connected by rods of negligible
mass. This rigid body can rotate in a vertical plane about a horizontal axis A that
passes through one of the particles. The body is released from rest with rod AB
horizontal, as shown in Fig.6-13. (a)What is the rotational inertia of the body about
axis A? (b)What is the angular speed of the body about axis at the instant rod AB
swings through the vertical position?

Solution: (a) The rotation inertia of the body about axis A is

I A  ml 2  ml 2  m( l 2  l 2 ) 2

 4ml 2  4(0.2kg )(0.50m) 2  0.20 kg  m 2 (answer)
Note that the particle at vertex A does not have rotational
inertia about axis A.                                              Fig. 6-13 Problem 6
(b) During the rotation of the body only the gravitational force does work, so the body
is mechanic energy is conserved.
The center of the mass C of the body is located at the center of the square, at the
instant rod AB swings through the vertical position the center of mass C is a distance
l below its initial position. Thus the potential energy of the body in the initial
position is U  4mgl . At the vertical position the kinetic energy of the body is

1 2
k     I  . From the conservation law of mechanic energy we have
2
1 2
I   4mgl
2
Solving for      yields

8mgl   8(0.20kg )(9.8 m s 2 )(0.50m)
                                         6.26 rad s
I             0.20kg  m 2                                      (Answer)

7 (40) Figure 6-14 shows a rigid structure consisting of a circular hoop of radius
R and mass m , and square made of four thin bars, each of length R and mass m .
The rigid structure rotates at a constant speed about a vertical axis, with a period of
rotation of 2.5s . Assuming R  0.50m and m  2.0kg ,calculate (a) The structure’s
rotational inertia about the axis of rotation and (b) its angular momentum about that

- 157 -
Chapter 6   Rotation and Angular Momentum.

axis.
Solution: (a) The structure’s rotational inertia about the axis of rotation is
1             1             1
I  I s  I n  ( mR 2  mR 2  mR 2  0)  ( mR 2  mR 2 )
3             3             2
 3.17mR 2  3.17(2.0kg )(0.50m 2 )  1.6 2 kg  m 2                     (Answer)

(b) The angular speed of the rotating rigid structure is
                  2.51rad / s
T        2.5s
Thus the angular momentum of the rigid structure

L  I   (1.6kg  m2 )(2.51 rad s)
 4.0kg  m2 s                                 (Answer)             Fig. 6-14 Problem 7

8 (43) In Fig.6-15, a small 50g block slides down a frictionless surface through
height h  20cm and then sticks to a uniform rod of mass 100g and length
40cm .The rod pivots about point O through angle  before momentarily
stopping. Find  .
Solution : The whole process can be divided into three
parts :
(1), The small block slides down the frictionless surface
through height h , In this part only the gravitational
force , being a conservitive force ,does work, so the law
of conservation of mechanic energy holds
1
m1 gh           2
m1v1                        (1)
2
Where v1 is the speed of the block before it collides with               Fig. 6-15 Problem 8
the rod.
(2) The small block collides with the rod and sticks to it. During this interaction there
is no net torque acting on the block –rod system relative to the point O , the angular
momentum of the system is conserved (Note: since there is a net force acting on the
rod at point O by the pivot, the law of conservation of linear momentum does not

- 158 -
Chapter 6   Rotation and Angular Momentum.
hold! )

m1v1 L  I                                                     (2)

Where  is the angular speed of the system about point O just after the
collision. I is the rotational inertia of the block-rod system about point O ,which is
1
I  m2 L2  m1L2                                                (3)
3
(3) The block-rod system swings up until it momentarily stops , During this process
the mechanic energy of the system is conserved , we thus write
1 2
I  (m1  m2 ) ghcom                                        (4)
2
Where hcom is the height change of the center of mass of the block-rod system in

the swing up process. In the vertical position the center of mass of the system is
below point O at

m1 L  m2 (0.5L) (0.05kg )(0.4m)  (0.1kg )(0.2m)
Lcom                                                      0.2667m
m1  m2             0.05kg  0.10kg

So                    hcom  Lcom (1  cos  )                                   (5)

Substituting all the known values into equations (1)~(5), we can then get the answer as
the following steps:
Form Eq.(1)

v1  2 gh  2(9.8m / s 2 )(0.2m)  1.98m / s
From Eq.(3)
1                 0.1kg
I  ( m2  m1 ) L2  (        0.05kg )(0.4m) 2  0.0133kgm2
3                   3
From Eq.2 we have

  m1v1L I  (0.05kg )(1.98m / s)(0.4m) /(0.0133kgm2 )
From Eqs. (4) and (5), we have
1 2
I  (m1  m2 ) gLcom (1  cos )
2
- 159 -
Chapter 6   Rotation and Angular Momentum.
cos   0.85
Thus the angle    we look for is
9 (44) A certain gyroscope consists of a uniform disk with a 50cm radius mounted
at the center of an axle that is 11cm long and of negligible mass. The axle is
horizontal and supported at one end. If the disk is spinning around the axle at
1000rev / min , what is the precession rate?
Solution: According to formula of the procession rate of a Gyroscope
Mgr

I
Where the rotational inertia of the disk about the horizontal axle is
1
I     MR 2
2
The angular speed of the spinning dist is

 1000rev  2rad  1 min 
                             104.7rad / s
 1 min  1rev  60 s 
Therefore the precession rate of the gyroscope is


Mgr

2 gr

           
2 9.8m / s 2 0.11m 
1              R 2 0.5m 2 104.7rad / s 
MR 2
2

10 (45) Figure6-16 shows an overhead view of a ring that can rotate about its center
like a merry-go-round. Its outer radius R2 is 0.800m , its inner radius R1 is R2 / 2.00 ,
its mass M is 0.800kg , and the mass of the crossbars at its center is negligible. It
initially rotates at an angular speed of 8.00rad / s with
a cat of mass m  M / 4.00 on its outer edge, at
radius R2 . By how much does the cat increase the kinetic
energy of the cat-ring system if the cat crawls to the inner
Solution: During the process as the cat crawls to the
inner edge from the out edge of the rotating ring there is
not torque acting along the rotating axis, so the angular           Fig. 6-16   Problem 10
- 160 -
Chapter 6       Rotation and Angular Momentum.
momentum of the cat-ring system is conserved. From the conservation law of
angular momentum of a rigid body rotating about a fixed axis we have

I 22  I11                                               (1)

Where I 2 and I 1 are the rotational inertia of the cat-ring system with the cat on the

ring , s inner edge and on the out edge, respectively. Work out for them

I 2  I ring  I cat2 
1
2

M R12  R2  mR12
2


1
2
                                  
8.00 Kg 2 0.400m2  0.800m2  2.00 Kg 0.400m2
 3.2 Kgm 2  0.32 Kgm 2
 3.52 Kgm 2

I1  I ring  I cat1 
1
2
   2     2

M R12  R2  mR2  4.48Kgm 2                       (3)

From the problem, the initial angular speed of the cat-ring system       1  8.00rad / s

Substituting   1 , I1 and I 2 into equation (1) and solving for 2 lead to

I1    4.48Kgm 2
 2  1            8.00rad / s   10.18rad / s
I2    3.53Kgm 2
Next let us calculate the kinetic energy of the car-ring system both at the initial stage

K1 and at the final stage K 2 . The initial kinetic energy of the system is

I112   4.48Kgm2  8.00rad / s   143.4 J
1        1
K1 
2

2        2
The final kinetic energy of the system is

K2 
1
2
I 2 2  3.52 Kgm 2 10.18rad / s   182.4 J
2 1
2
                 2

Thus the amount of the increased kinetic energy of the cat-ring system as the cat
crawls to the inner edge from the outer edge of the ring is

K  K 2  K1  182.4J  143.4J  39 J                                 (Answer)

- 161 -
Chapter 6       Rotation and Angular Momentum.
11 (46)    A uniform wheel of mass 10.0kg and radius 0.400m is mounted
rigidly on an axle through its center (Fig.6-17). The radius of the axle is 0.200m , and
the rotational inertia of the wheel-axle combination about its central axis is
0.600kg  m2 . The wheel is initially at rest at the top of a surface that is inclined at
angle   30.0 with the horizontal; the axle rests on the surface while the wheel
extends into a groove in the surface without touching the surface. Once released, the
axle rolls down the surface by 2.00m , what are (a) its rotational kinetic energy and (b)
its translational kinetic energy?
Solution: The key idea here is that as the
wheel-axle combination rolls down the inclined
surface only the gravitational force does work. So
this process obey the conservation law of
mechanic energy. First we should find out the mass
of the axle m. Let M the mass of the wheel then
the    rotational        inertia       of   the       wheel-axle
combination is                                                         Fig. 6-17   Problem 11

I  I wheel  I axle 
1
2
      1

M R 2  r 2  mra2
2
Where R and r are the radius of the wheel and the axle, respectively. Substituting
the known values and solving for m . We get

m
2         1
 I  2 M Rw  ra 
2
  2 

ra2                     


2
0.200m      2

1.1Kgm  10 Kg 0.400m   0.200m  

2 1
2
2
       2 



 5.0 Kg
The potential energy of the rigid body before releasing is

U   M  m  gh sin 30  10.0 Kg  5.0 Kg   9.8m / s 2   2.0m  sin 30

 147J
From the conservation law of mechanic energy we have

K rot  K tran  U

- 162 -
Chapter 6       Rotation and Angular Momentum.

2
I  M  m r 2  U
1 2 1
2
 
Substituting the known values and solving for     , the angular speed of the
combination when it moves down the surface by 2.00m , lead to   13.2rad / s
Therefore, the rotational kinetic energy of the wheel-axle combination is

K rot 
2
I  1.1Kgm 2 13.0rad / s   95.8J
1 2 1
2
              2

The translational kinetic energy of the wheel-axle combination is

K tra 
1
M  m V 2  1 M  m r 2
2               2

1
2
 52.3 J

12 (47)     In Fig.6-18, a constant horizontal force Fapp of magnitude 12N is
applied to a uniform solid cylinder by fishing line wrapped around the cylinder. The
mass of the cylinder is 0.10m , and the cylinder rolls smoothly on the horizontal
surface. (a) What is the magnitude of the acceleration of the center of mass of the
cylinder? （b）What is the magnitude of the angular acceleration of the cylinder about
the center of mass? (c) In unit-vector notation, what is the frictional force acting on the
cylinder?
Solution: this is the plane notion of a rigid body. There
are two forces exerted on the cylinder: the applyed
                                     
force Fapp and the frictional force F . Consider first
the translation motion of the center of mass of the
cylinder. From Newton’s second law, we have

Fapp  F  ma                                 (1)           Fig. 6-18   Problem 12

Where a c is the magnitude of the acceleration of the center of mass of the cylinder.
Next consider the rotational motion of the cylinder. From Newton’s second law for
rotation we write
       
FappR  FR  I                                                      (2)

Where R is the radius of cylinder, I is the rotational inertia of the cylinder.
- 163 -
Chapter 6   Rotation and Angular Momentum.
1
I       MR 2
2
(3)
There is a relation between the linear acceleration of the center of mass of the cylinder

a c and the angular acceleration of the rolling cylinder

a c  R                                                                  (4)

Solving above four equations. We obtain

ac  1.6m / s 2                                                         (Answer)

                    4 N
     
The frictional force F  Fapp  mac  12 N  10 Kg  1.6m / s
2

Choose the positive direction of x axis to the right. In unit-vector vector notation the
          
frictional force acting on the cylinder can be written as F  4i N                       (Answer)

13 (49) In Fig.6-19, a small 0.50kg block has a horizontal velocity v0 of
magnitude 3.0m / s when it slides off a table of height h  1.2m . Answer the
following in unit-vector notation for a coordinate system in which the origin is at the
edge of the table (at point O ), the positive x direction is horizontally away from the
table, and the positive y direction is up. What are the angular momentum of the block
about point A (a) just after the block leaves the table and (b) just before the block
strikes the floor? What are the torques on the block
about point A (c) just after the block leaves the
table and (d) just before strikes the floor?
Solution: Taking point A as the reference point,
      
rA and rB are the position vectors, when the block
just leaves the table and before it strikes the floor.
(a) The angular momentum of the block about point
A just after block leaves the table is
            
L0  rA  mV0                                   Fig. 6-19       Problem 13

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Chapter 6          Rotation and Angular Momentum.
Its magnitude is

L0  r0 mv0 sin 90  1.2m1.5kg 3.0m / s   1.8kgm2 / s                      ( Answer )

Its direction can be determined by the right hand rule, which is perpendicular to the
          
plane formed by rA and v 0 , entering into this page (  ).

(b) The angular momentum of the block about point A , just before the block strikes
the floor is
         
L  rB  mv

We should find out the velocity v of the block in that moment in horizontal direction

v x  v0  3.0m / s .

In vertical direction the block is in free fall motion, thus
1 2
h          gt
2
2h    21.2m
t                         0.5s
g    9.8m / s 2

         
v y  gt  9.8m / s 2 0.5s   4.9m / s

The horizontal displacement of the block is

R  v0 t  3.0m / s 0.5s   1.5m

So the magnitude of the velocity of block just before it strikes the floor is

v  vx  v y 
2     2
3.0m / s 2  4.9m / s 2    5.75m / s

The angle  made by velocity v with the x axis is
vy                 5.75m / s
  tg 1             tg 1              58.5 
vx                 3.0m / s
Thus the magnitude of angular momentum of the block in (b) is

L  rB mv sin   1.5m 0.5kg  5.75m / s   sin58.5   3.7kgm2 / s (Answer)

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Chapter 6    Rotation and Angular Momentum.
                                               
The direction of L is perpendicular to the plane formed by rB and v ,

(c) The torque on the block about point A just after the block leaves the table is
                               
 0  rA  Fg  rA  mg
            
Note that, at this moment, rA and g are in the same vertical line but opposite in


direction. so the torque  0 is zero.                                          (Answer)

(d) Just before the block strikes the floor the torque on the block about point A is

  rB  mg

         
Since now rB  g the magnitude of the torque acting on the block is

  rB mg  1.5m0.5kg 9.8m / s 2   7.4kgm 2 / s 2           (Answer)

         
The direction of the torque is perpendicular to the plane formed by rB and g

14 (50) An impulsive force F (t ) acts for a short time t on a rotating rigid body
of rotational inertial I . Show that

 dt  RF    avg   t  I ( f  i )

Where    is the torque due to the force, R is the moment arm of the force, Favg is

the average value of the force during the time it acts on the body, and  i and           f

are the angular velocities of the body just before and just after the forces acts (The

quantity    dt  RF      avg   t is called the angular impulse, analogous to Favg t , the

linear impulse).
Solution: The angular impulse, the time accumulation effect of a torque, is defined as

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Chapter 6       Rotation and Angular Momentum.

 dt  RF        avg   t

(1)

For a variable force F t  , the torque produced by it is a function of time

 t   RF t                                                       (2)

Substituting Eq.(1) and set t  t 2  t1 and                      avg  RFavg , we obtain

RF  t dt  RFavg t   avg t
t2
t1
(3)

While according to the Newton’s Second law for rotation and the definition of angular
acceleration, we have
2  1
 avg t  I  I                                                    (4)
t
Recast above equation obtaining

 avg  I 2  1                                                 (5)

Combining equation (2)、(3) and (5) we get

RF  t dt  RFavg t  I 2  1 
t2
t1