PH 221-3A Fall 2010
ROLLING, TORQUE, and
ANGULAR MOMENTUM
Lectures 18-19
Chapter 11
((Halliday/Resnick/Walker,
y
, Fundamentals of Physics
y
8th edition))
1
Chapter 11
Rolling, Torque, and Angular Momentum
In this chapter we will cover the following topics:
-Rolling
g of circular objects
j
and its relationshipp with friction
-Redefinition of torque as a vector to describe rotational problems that
are more complicated than the rotation of a rigid body about a fixed
axis
-Angular Momentum of single particles and systems or particles
-Newton’s second law for rotational motion
-Conservation
C
i off angular
l Momentum
M
-Applications of the conservation of angular momentum
2
t1 = 0
t2 = t
Rolling as Translation and Rotation Combined
Consider an object with circular cross section that rolls
along a surface without slipping. This motion, though
common, is complicated. We can simplify its study by
treating it as a combination of translation of the center of
mass and rotation of the object about the center of mass
Consider the two snapshots
p
of a rollingg bicycle
y wheel shown in the figure.
g
An observer stationary with the ground will see the center of mass O of the wheel
move forward with a speed vcom . The point P at which the wheel makes contact
with the road also moves with the same speed. During the time interval t between
ds
the two snapshots both O and P cover a distance s. vcom 
(eqs.1) During t
dt
the bicycle rider sees the wheel rotate by an angle  about O so that
ds
d
R
=R (eqs.2) If we cambine equation 1 with equation 2
dt
dt
we get the condition for rolling without slipping. vcom  R
3
s  R 
vcom  R
We have seen that rolling is a combination of purely translational motion
with speed
p
vcom and a purely
p
y rotaional motion about the center of mass
vcom
. The velocity of each point is the vector sum
R
off the
th velocities
l iti off the
th two
t motions.
ti
For
F the
th translational
t
l ti l motion
ti
the
th

velocity vector is the same for every point (vcom ,see fig.b ). The rotational
with angular velocity  
velocityy varies from poi
p nt to ppoint. Its magnitude
g
is equal
q to  r where r is
the distance of the point from O. Its direction is tangent to the circular orbit
(see fig.a). The net velocity is the vector sum of these two terms. For example
the velocity of point P is always zero. The velocity of the center of mass O is


4
vcom (r  0). Finally the velocity of the top point T is wqual to 2vcom .
Problem 2. An automobile traveling at 80.0 km/h has tires of 75.0 cm diameter.
(a) What is the angular speed of the tires about their axes?
(b) If the car is brought to a stop uniformly in 30.0 complete turns of the tires
(without skidding), what is the magnitude of the angular acceleration of the wheels?
(c) How far does the car move during the braking?
The initial speed of the car is
v   80 km/h  (1000 m/km)(1 h/3600 s)  22.2 m/s .
The tire radius is R = 0.750/2 = 0.375 m.
(a) The initial speed of the car is the initial speed of the center of mass of the tire, so
0 
vcom0
R

22.2 m/s
 59.3 rad/s.
0.375 m
(b) With  = (30.0)(2)
(30 0)(2) = 188 rad and  = 0,
0 Eq.
Eq 10-14
10 14 leads to
(59.3 rad/s)2
    2   
 9.31 rad/s2 .
2 188 rad 
2
2
0
(c) R = 70.7 m for the distance traveled.
5
vT
A
vO
vA
Rolling as Pure Rotation
Another way of looking at rolling is shown in the figure
W consider
We
id rolling
lli as a pure rotation
t ti about
b t an axis
i
B
vB
of rotation that passes through the contact point P
between the wheel and the road. The angular
g
velocity of the rotation is  
vcom
R
In order to define the velocity vector for each point we must know its magnitude
as well as its direction. The direction for each point on the wheel points along the

tangent to its circular orbit.
orbit For example at point A the velocity vector v A is
perpendicular to the dotted line that connects pont A with point P. The speed
of each point is given by: v  r. Here r is the distance between a particular
point and the contact point P. For example at point T r  2 R.
Thus vT  2 R  2vcom . For point O r  R thus vO   R  vcom
For point
i P r  0 thus
h vP  0
6
The Kinetic Energy of Rolling
Consider the rolling object shown in the figure
It is easier to calculate the kinetic energy of the rolling
body by considering the motion as pure rotation
about the contact point P.
P The rolling object has mass M
and radius R.
1
I P 2 . Here I P is the
2
rotational inertia of the rolling body about point P. We can determine I P using
The kinetic energy K is then given by the equation: K 
1
I com  MR 2   2

2
1
1
1
1
1
2
2
2
2 2
2
2
K   I com  MR    I com  MR 
K  I com  Mvcom
2
2
2
2
2
The expression for the kinetic energy consists of two terms. The first term
corresponds to the rotation about the center of mass O with angular velocity .
the parallel axis theorem.
theorem I P  I com  MR 2  K 
The second term is associated with the kinetic energy due to the translational
motion of evey point with speed vcom
7
P ro b le m 9 . A so lid c ylin d e r o f ra d iu s 1 0 c m a n d m a ss 1 2 k g sta rts fro m re st a n d ro lls w ith o u t
slip p in g a d ista n c e L = 6 .0 m d o w n a ro o f th a t is in c lin e d a t th e a n g le  = 3 0  .
( a ) W h a t is th e a n g u la r sp e e d o f th e c ylin d e r a b o u t its c e n te r a s it le a v e s th e ro o f?
(b ) T h e ro o ff'ss e d g e is a t h e ig h t H = 5 .0 m . H o w fa r h o riz o n ta lly fro m th e ro o ff'ss e d g e
d o e s th e c ylin d e r h it th e le v e l g ro u n d ?
(a) We find its angular speed as it leaves the roof using conservation of energy. Its initial
kinetic energy is Ki = 0 and its initial potential energy is Ui = Mgh where
h  6.0 sin 30  3.0 m (we are using the edge of the roof as our reference level for
computing U). Its final kinetic energy (as it leaves the roof) is
Kf 
1
2
Mv 2  21 I 2 .
Here we use v to denote the speed of its center of mass and  is its angular speed — at
the moment it leaves the roof. Since (up to that moment) the ball rolls without sliding we
can set v = R = v where R = 0.10 m. Using I  21 MR 2 (Table 10-2(c)), conservation of
energy leads to
Mgh 
1
1
1
1
3
Mv 2  I  2  MR 2 2  MR 2 2  MR 2 2 .
2
2
2
4
4
The mass M cancels from the equation, and we obtain

1
R
4
1
gh 
3
0.10 m
c
hb
g
4
9.8 m s2 3.0 m  63 rad s .
3
8
( ) Now this becomes a projectile
(b)
p j
motion of the type
yp examined in Chapter
p 4. We put
p the
origin at the position of the center of mass when the ball leaves the track (the “initial”
position for this part of the problem) and take +x leftward and +y downward. The result
of part (a) implies v0 = R = 6.3 m/s, and we see from the figure that (with these positive
direction choices) its components are
v0 x  v0 cos 30  5.4 m s
v0 y  v0 sin 30  3.1 m s.
The projectile motion equations become
x  v0 x t and
y  v0 y t 
1 2
gt .
2
We first find the time when y = H = 5.0 m from the second equation (using the quadratic
formula, choosing the positive root):
t
v0 y  v02y  2 gH
g
 0.74s.
b
gb
g
Then we substitute this into the x equation and obtain x  5.4 m s 0.74 s  4.0 m.
9
Friction and Rolling
When an object rolls with constant speed (see top figure)
it has no tendency to slide at the contact point P and thus

acom  0
no frictional force acts there. If a net force acts on the

rolling body it results in a non-zero
non zero acceleration acom
for the center of mass (see lower figure). If the rolling
object accelerates to the right it has the tendency to slide

at point P to the left. Thus a static frictional force f s
opposes the tendency to slide. The motion is smooth
rolling as long as f s  f s ,max
The rolling condition results in a connection beteen the magnitude of the
acceleration acom of the center of mass and its angular acceleration 
vcom   R We take time derivatives of both sides  acom
acom  R
dvcom
d

R
 R
dt
dt
10

P roblem 7. A constant horizontal force Fapp of m agnitude 10N is applied to a w heel of m ass
10 kg and radius 0.30m . T he w heel rolls sm oothly on the horizontal surface, and
the acceleration of its cent er of m ass has m agnitude 0.60 m /s 2 .
( a ) In unit-vector notation, w hat is the frictional force on the w heel?
(b) W hat is the rotational inertia of the w heel about the rotation axis through its center of m ass?
(a) Newton
Newton’ss second law in the x direction leads to


Fapp  f s  ma  f s  10N  10kg  0.60 m s  4.0 N.
2

In unit vector notation, we have f s  (4.0 N)iˆ which points leftward.
(b) With R = 0.30 m, we find the magnitude of the angular acceleration to be
|| = |acom| / R = 2.0 rad/s2,

The only force not directed towards (or away from) the center of mass is f s , and the
torque it produces is clockwise:
 I

 0.30 m  4.0 N   I  2.0 rad
s2 
which yields the wheel
wheel’ss rotational inertia about its center of mass: I  0.60 kg  m2 .
11
Rolling Down a Ramp
Consider a round uniform body of mass M and radius R
rolling down an inclined plane of angle  . We will
calculate the acceleration acom of the center of mass
acom
along the x-axis
a is using
sing Newton's
Ne ton's second la
law for the
translational and rotational motion
Newton's second law for motion along the x-axis: f s  Mg sin    Macom (eqs.1)
Newton's second law for rotation about the center of mass:   Rff s  I com

acom
R
f s  I com
I com
We substitute  in the second equation and get: Rf s  I com
acom
R2
acom

R
(eqs.2) We substitute f s from equation 2 into equation 1 
acom
 Mg sin    Macom
2
R
acom 
g sin 
I com
1
MR 2
12
g sin 
| acom |
I
1  com2
MR
acom
Cylinder
Hoop
MR 2
I1 
2
g sin 
a1 
1  I1 / MR 2
I 2  MR 2
g sin 
a1 
1  MR 2 / 2MR 2
g sin 
a1 
1  1/ 2
2 g sin 
a1 
 (0.67) g sin 
3
g sin 
a2 
1  MR 2 / MR 2
g sin 
a2 
11
g sin 
a2 
 (0.5) g sin 
2
a2 
g sin 
1  I 2 / MR 2
13
The Yo-Yo
Consider a yo-yo of mass M , radius R, and axle radius Ro
rolling down a string . We will calculate the acceleration
acom of the center of its mass along the y -axis using Newton's
second law for the translational and rotational motion as we did
in the previous problem
Newton's second law for motion along the y -axis:
Mg  T  Macom (eqs.1)
acom
y
Newton's second law for rotation about the center of mass:
a
  RoT  I com . Angular acceleration   com
Ro
We substitute  in the second equation and get:
T  I com
acom
2
Ro
M  I com
Mg
(eqs.2) We substitute T from equation 2 into equation 1 
accom
om
 Macom 
2
Ro
acom 
g
I
1  com 2
MRo
14
Torque Revisited
In chapter 10 we defined the torque  of a rigid body rotating about a fixed axis
with each particle in the body moving on a circular path. We now expand the
definition of torque so that it can describe the motion of a particle that moves

along any path relative to a fixed point.
point If r is the position vector of a particle


  
on which a force F is acting, the torque  is defined as:   r  F


In the example
p shown in the figure
g
both r and F lie in the xyy -plane.
p
Usingg the

right hand rule we can see that the direction of  is along the z -axis.
The magnitude of the torque vector   rF sin  , where  is the angle


between r and F . From triangle OAB we have: r sin   r 
  r F , in agreement with the definition of chapter 10.
 
  r F

B
15
Problem 21. In unit-vector notation, what is the net torque about the origin on a flea


ˆ
located at coordinates (0, -4.0m, 5.0 m) when forces F  (3.0 N )k and F  ( 2.0 N ) ˆj
1
actt on the
th flea?
fl ?
2
iˆ
 
 
a  b  b  a  a x
ˆj
ay
bx
by
kˆ
ay
a z  iˆ
by
bz
az
a
 ˆj x
bz
bx
az ˆ ax
k
bx
bz
ay

by
 ( a y bz  by a z )iˆ  ( a z bx  bz a x ) ˆj  ( a x by  bx a y )kˆ
 




If we write r  x i + yj + zk, then we find r  F is equal to
d yF  zF ii + bzF  xF gj + d xF  yF ik.
z
y
x
z
y
x
With (using SI units) x = 0, y = – 4.0, z = 5.0, Fx = 0, Fy = –2.0 and Fz = 3.0 (these latter
terms being the individual forces that contribute to the net force)
force), the expression above
yields
  
ˆ
  r  F  (2.0N  m)i.
16
B
Angular Momentum


The counterpart of linear momentum p  mv in rotational
motion is a new vector known as angular momentum.
  
The new vector is defined as follows:   r  p


In the example shown in the figure both r and p
lie in the xy -plane. Using the right hand rule we

can see that the direction of  is along the z -axis.
The magnitude of angular momentum   rmv sin  ,


where  is the angle between r and p. From triangle
OAB we have: r sin   r    r mv
Note:
o e: Angular
gu momentum
o e u depends
depe ds on
o thee choice
c o ce o
of thee origin
o g O. If thee origin
o g

is shifted in general we get a different value of 
SI unit for angular momentum: kg.m 2 / s Sometimes the equivalent J.s is used
  
 
  r  p  mr  v 
  r mv
17

P r o b le m 2 9 . A t o n e in s ta n t, f o r c e F  4 .0 ˆj N a c ts o n a 0 .2 5 k g o b je c t th a t h a s p o s itio n


v e c to r r  ( 2 .0 iˆ  2 .0 kˆ ) m a n d v e lo c ity v e c to r v  (  5 .0 iˆ  5 .0 kˆ ) m /s . A b o u t th e o r ig in
a n d in u n it v e c to r n o ta tio n , w h a t a r e
ˆj kˆ
iˆ
( a ) th e o b je c t's a n g u la r m o m e n tu m a n d  
 
a y az
a az ˆ ax a y
a  b  b  a  a x a y a z  iˆ
 ˆj x
k

( b ) th e to r q u e a c tin g o n th e o b je c t.
by bz
bx by
bx bz
bx by bz
 ( a y bz  by a z )iˆ  ( a z bx  bz a x ) ˆj  ( a x by  bx a y )kˆ

 


(a) We use   mr  v , where r is the position vector of the object, v is its velocity
vector, and m is its mass. Only the x and z components of the position and velocity
 
vectors are nonzero, so Eq. 3-30 leads to r  v   xvz  zvz j. Therefore,
b
g

  m   xvz  zvx  ˆj   0.25 kg     2.0 m  5.0 m s    2.0 m  5.0 m s   ˆj  0.
 




(b) If we write r  x i  yj  zk,
Eq 33-30)
30) we find r  F is equal to
k then (using Eq.
d yF  zF i i  bzF  xF g j  d xF  yF i k .
z
y
x
z
y
x
With x = 2.0, z = –2.0, Fy = 4.0 and all other components zero (and SI units understood)
the expression above yields
 
  r  F  8.0 i  8.0 k N  m.

e
j
18
Newton's Second Law in Angular Form


dp
Newton's second law for linear motion has the form: Fnet  . Below we
dt
will derive the angular form of Newton
Newton'ss second law for a particle.




d
d  
 
   
  dv dr  
  mr  v  
 m r  v   m r    v   mr  a  v  v 
dt
dt
dt dt



d
 
 


 

 m  r  a    r  ma   r  Fnet   net
v v  0 
dt



d
dp

Thus:  net 
Compare with: Fnet 
dt
dt



d

 net 
dt
19
The Angular Momentum of a System of Particles
z
m1
ℓ1
m2
m3
ℓ2
x
We will now explore Newton's second law in
angular form for a system of n particles that have
  

angular momentum 1 , 2 , 3 ,..., n
ℓ3
O
y
n 


   
The angular momentum L of the system is L  1  2  3  ...  n   i

dL
The time derivative of the angular momentum is
=
dt

n
d i

i 1 dt
i 1

d i 
The time derivative for the angular momentum of the i-th particle
  net ,i
dt

Where  net ,i is the net torque on the particle. This torque has contributions
from external as well as internal forces between the particles of the system. Thus

n
dL



  net ,i   net Here  net is the net torque due to all the external forces.
dt
i 1
By virtue of Newton's third law the vector sum of all internal torques is zero.

dL 
Thus Newton's second law for a system in angular form takes the form:
  net ext
dt
20
Angular Momentum of a Rigid Body Rotating about a Fixed Axis
We take the z-axis to be the fixed rotation axis.
axis We will determine
the z-component of the net angular momentum. The body is

divided n elements of mass mi that have a position vector ri

  
The angular momentum i of the i-the element is: i  ri  pi
 i  ri pi  sin90  = ri mi vi The z-compoment
Its magnitude
 iz of  i is:  iz   i sin    ri sin   mi vi   ri mi vi
The z-component of the angular momentum Lz is the sum:
 n

Lz    iz   ri mi vi  ri mi  ri       mi ri 2 
i 1
i 1
i 1
 i 1

n
n
n
n
The sum
2

m
r
 i i is the rotational inertia I of the rigid body
i 1
Thus: Lz  I 
Lz  I 
21

P ro b le m 3 4 . A p a rtic le is a c te d o n b y tw o to rq u e s a b o u t th e o r ig in :  1 h a s a m a g n itu d e

o f 2 .0 N m a n d is d ir e c te d in th e p o s itiv e d ir e c tio n o f th e x a x is , a n d  2 h a s a m a g n itu d e
o f 4 .0 N m a n d is d ir e c te d in th e n e g a tiv e d ir e c tio n o f th e y a x is . In u n it-v e c to r n o ta tio n ,


dl
fin d
, w h e r e l is th e a n g u la r m o m e n tu m o f th e p a r r tic le a b o u th e o r ig in .
dt
The rate of change of the angular momentum is

d  
  1   2  (2.0
(2 0 N  m)iˆ  (4
(4.00 N  m)ˆj.
j
dt

Consequently, the vector d dt has a magnitude
(2.0 N  m) 2   4.0 N  m   4.5 N  m
2
and is at an angle  (in the xy plane, or a plane parallel to it) measured from the positive x
axis, where
 4.0 N  m 
  tan 1 
  63 ,
2
2.0
0
N

m


the negative sign indicating that the angle is measured clockwise as viewed “from above”
(by a person on the z axis).
22
Problem 42. A disk with a rotational inertia of 7.00 kgm 2 rotates like a merry-go-round
while undergoing a torque given by   (5.00  2.00t ) Nm. At time t=1.00 s, its angular
momentum is 5.00
5 00 kgm 2 / s. What is its angular momentum at t=3.00s?
t=3 00s?
Torque is the time derivative of the angular momentum. Thus, the change in the angular
momentum is equal to the time integral of the torque. With   (5.00  2.00t ) N  m , the
angular momentum as a function of time is (in units kg  m 2 /s )
L(t )    dt   (5.00  2.00t )dt  L0  5.00t  1.00t 2
S
Since
g
constant is L0  1 . Thus,, the
L  5.00 kgg  m 2 /s when t  1.00
.00 s , the integration
complete expression of the angular momentum is
L(t )  1  5.00t  1.00t 2 .
At t  3.00 s , we have L(t  3.00)  1  5.00(3.00)  1.00(3.00) 2  23.0 kg  m 2 /s.
23
Conservation of Angular momentum
For any system of particles (including a rigid body) Newton's

dL 
second law in angular form is:
  net
dt

dL

If the net external torque  net  0 then we have:
0
dt

L  a constant
This result is known as the law of the
conservation of angular momentum. In words:
 Net angular momentum   Net angular momentum 


   at some later time t
at
some
initial
time
t
f
i

 

In equation form:
 
Li  L f
Note: If the component of the external torque along a certain
axis
i is
i equall to zero, then
h the
h componet off the
h angular
l momentum
of the system along this axis cannot change
24
Example: The figure shows a student seated
on a stool that can rotate freely about a
vertical
ti l axis.
i The
Th student
t d t who
h has
h been
b
sett into
i t
rotation at an initial angular speed i , holds
two dumbbells in his outstretched hands. His

angular momentum vector L lies along the
rotation axis, pointing upward.
The student then pulls in his hands as shown in fig.b. This action reduces the
rotational inertia from an initial value I i to a smaller final value I f .
No net external torque acts on the student-stool system. Thus the
angular momentum of the system remains unchanged.
Angular momentum at ti : Li  I ii
Li  L f  I ii  I f  f   f 
Ii
i
If
Angular momentum at t f : L f  I f  f
Since I f  I i 
The rotation rate of the student in fig.b is faster
Ii
 1   f  i
If
25
Sample Problem 11-7:
I wh  1.2
1 2 kg
kg.m
m2
wh  2  3.9 rad/s
I b  6.8 kg.m 2
b  ?
y-axis
Li  L f  Lwh   Lwh  Lb  Lb  2 Lwh
I bb  2 I whwh  b 
2 I whwh 2 1.2  2  3.9

 2  1.4 rad/s
Ib
6.8
26
Problem 60. A horizontal platform in the shape of a circular disk rotates on a frictionless
bearing about a vertical axle through the center of the disk. The platform has a mass of
150 kg, a radius of 2.0 m, and a rotational inertia of 300 kgm 2 about the axis of rotation.
A 60 kg student walks slowly from the rim of the platform towrd the center.
If the angular speed of the system is 1.5 rad/s when the student starts at the rim,
what is the angular speed when she is 0.5 m from the center?
The initial rotational inertia of the system is Ii = Idisk + Istudent, where Idisk = 300 kg  m2
((which,, incidentally,
y, does agree
g
with Table 10-2(c))
( )) and Istudent = mR2 where m = 60 kg
g
and R = 2.0 m.
The rotational inertia when the student reaches r = 0.5 m is If = Idisk + mr2. Angular
momentum conservation leads to
I disk  mR 2
I i i  I f  f   f   i
I disk  mr 2
which yields, for i = 1.5 rad/s, a final angular velocity of f = 2.6 rad/s.
27
Analogies between translational and rotational Motion
Translational Motion
Rotational Motion

v  
a  
x

p
 
mv 2
K
2
m

F  ma

F

P  Fv


dp
Fnet 
dt
p  mv




I 2
K
2
I
  I

P  

d

 net 
dt
L  I
28