Transcript Example

Chapter 11
Rolling, Torque, and Angular Momentum
In this chapter we will cover the following topics:
-Rolling of circular objects and its relationship 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 of particles
-Newton’s second law for rotational motion
-Conservation of angular momentum
Applications of the conservation of angular momentum
(11-1)
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 of a rolling bicycle wheel shown in the figure.
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 
(eq. 1). During t
dt
the bicycle rider sees the wheel rotate by an angle  about O so that
ds
d
s  R 
R
= (eq. 2). If we combine equation 1 with equation 2
dt
dt
we get the condition for rolling without slipping: vcom  R
(11-2)
vcom  R
We have seen that rolling is a combination of purely translational motion
with speed vcom and a purely rotational motion about the center of mass
vcom
with angular velocity  
. The velocity of each point is the vector sum
R
of the velocities of the two motions. For the translational motion the
velocity vector is the same for every point (vcom ,see fig. b). The rotational
velocity varies from point to point. Its magnitude is equal 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
(11-3)
vcom (r  0). Finally, the velocity of the top point T is equal to 2vcom .
vT
Rolling as Pure Rotation
A
vA
vO
B
vB
Another way of looking at rolling is shown in the figure.
We consider rolling as a pure rotation about an axis
of rotation that passes through the contact point P
between the wheel and the road. The angular
v
velocity of the rotation is   com .
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. For example, at point A the velocity vector v A is
perpendicular to the dotted line that connects point A with point B. 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 P, r  0 thus vP  0.
(11-4)
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 rolling
about the contact point P. The rolling object has mass M
and radius R.
1
The kinetic energy K is then given by the equation K  I P 2 . Here I P is the
2
rotational inertia of the rolling body about point P. We can determine I P using
1
I com  MR 2   2 .

2
1
1
1
1
1
2
2
2
2
2
2 2
K  I com  Mvcom
K   I com  MR    I com  MR 
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 second term is associated with the kinetic energy due to the translational
motion of every point with speed vcom . .
(11-5)
the parallel axis theorem. I P  I com  MR 2  K 
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
no frictional force acts there. If a net force acts on the
acom  0
rolling body it results in a nonzero 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 between 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
(11-6)
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 using Newton's second law for the
translational and rotational motion.
Newton's second law for motion along the x-axis: f s  Mg sin   Macom (eq. 1)
Newton's second law for rotation about the center of mass:   Rf 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
(eq. 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
(11-7)
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
(11-8)
Example: A uniform cylinder rolls down a ramp inclined at an
angle of θ to the horizontal. What is the linear acceleration of the
cylinder at the bottom of the ramp? Remember that: The friction
force is used to rotate the object.
Example: Consider a solid cylinder of radius R that rolls without
slipping down an incline from some initial height h. The linear velocity
of the cylinder at the bottom of the incline is vcm and the angular
velocity is ω.
We can also solve for angular velocity using the equation   v cm 
R
4 gh
3R 2
Example: A bowling ball has a mass of 4.0 kg, a moment of inertia of
1.6×10**(−2) kg ·m2 and a radius of 0.10 m. If it rolls down the lane
without slipping at a linear speed of 4.0 m/s, what is its total energy?
Example: A bowling ball has a mass of 4.0 kg, a moment of inertia of
1.6×10**(−2) kg ·m2 and a radius of 0.10 m. If it rolls down the lane
without slipping at a linear speed of 4.0 m/s, what is its total energy?
The Yo - Yo
Consider a yo-yo of mass M , radius R, and axle radius R0
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:
acom
Mg  T  Macom (eq. 1).
y
Newton's second law for rotation about the center of mass:
acom
  R0T  I com . Angular acceleration  
.
R0
We substitute  in the second equation and get
T  I com
acom
R0 2
Mg  I com
(eq. 2). We substitute T from equation 2 into equation 1 
acom
 Macom 
2
R0
acom 
g
.
I
1  com2
MR0
(11-9)
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. 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 shown in the figure both r and F 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 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
(11-10)
Torque Revisited
ˆi
ˆj
  r  F  (x  x o ) ( y  y o )
Fx
Fy
kˆ
z  z o 
Fz
Example: (T72_Q17.) A force F   2.0 ˆi  3.0 ˆj N is applied to an object
that is pivoted about a fixed axis aligned along the z-axis. If the force is
applied at the point of coordinates (4.0, 5.0, 0.0) m, what is the applied
torque (in N.m) about the z axis?
ˆi
  r  F  (4  0)
2
 
 2 kˆ N  m
ˆj
(5  0)
3
kˆ
0  0
0
Example: (T71_Q19.). At an instant, a particle of mass 2.0 kg has a
2
position r  9.0 ˆi  15.0 ˆj m of and acceleration of a  3.0 ˆi m/s. What
is the net torque on the particle at this instant about the point having the
position vector:r  9.0 ˆi m ?


o
 


Example: (T71_Q19.). At an instant, a particle of mass 2.0 kg has a
2
position r  9.0 ˆi  15.0 ˆj m of and acceleration of a  3.0 ˆi m/s. What
is the net torque on the particle at this instant about the point having the
position vector:r  9.0 ˆi m ?


o


 
ˆi
ˆj
ˆi
kˆ
jˆ
kˆ
  r  F  m r  a  m (x  x o ) ( y  y o )  z  z o   2 (9  9) (15  0)  0  0 
ax
ay
az
3
3
0
 
 
 90 kˆ N  m
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
B
 rmv sin  ,
where  is the angle between r and p. From triangle
OAB we have: r sin   r   r mv.
Note: Angular momentum depends on the choice of the origin O. If the origin
is shifted, in general we get a different value of .
SI unit for angular momentum: kg  m2 / s. Sometimes the equivalent J  s is used.
 r  p  mr  v 
 r mv
(11-11)
Example: (T052 Q#18) A stone attached to a string is whirled at 3.0 rev/s
around a horizontal circle of radius 0.75 m. The mass of the stone is 0.15
kg. The magnitude of the angular momentum of the stone relative to the
center of the circle is:
L  mvr  mr   0.15   0.75    3  2   1.6 kg.m 2 / s
2
2
Newton's Second Law in Angular Form
dp
. Below we
dt
will derive the angular form of Newton's second law for a particle.
Newton's second law for linear motion has the form: Fnet 
 mr  v  
 net


d
 m  r  a    r  ma   r  Fnet   net
dt
d
dp

. Compare with: Fnet  .
dt
dt
v v  0 
Thus:  net
d
d
dv dr


 m r  v   m  r    v   m r  a  v  v 
dt
dt
dt dt


d

dt
(11-12)
The Angular Momentum of a System of Particles
z
m1
ℓ1
m2
ℓ2
x
We will now explore Newton's second law in
angular form for a system of n particles that have
ℓ3
m3
O
angular momentum
1
,
2
, 3 ,..., n .
y
The angular momentum L of the system is L 
1

2
dL
The time derivative of the angular momentum is
=
dt

3
 ... 
n
n
  i.
i 1
n
d i
.

i 1 dt
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
  net
Thus Newton's second law for a system in angular form takes the form:
dt
(11-13)
Angular Momentum of a Rigid Body Rotating About a Fixed Axis
We take the z -axis to be the fixed rotation axis. We will determine
the z -component of the net angular momentum. The body is
divided into n elements of mass mi that have a position vector ri .
The angular momentum
Its magnitude is
iz
of
i
is
iz

i
of the ith element is
i
 ri  pi .
 ri pi  sin 90  = ri mi vi . The z -compoment
i
i
sin    ri sin   mi vi   ri mi vi .
The z -component of the angular momentum Lz is the sum:
n
Lz  
i 1
 n

  ri mi vi  ri mi  ri       mi ri 2 
i 1
i 1
 i 1

n
iz
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
(11-14)
Conservation of Angular Momentum
For any system of particles (including a rigid body) Newton's
second law in angular form is
dL
  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
tim
e
t
f
i

 

In equation form:
Li  L f
Note: If the component of the external torque along a certain
axis is equal to zero, then the component of the angular momentum
of the system along this axis cannot change.
(11-15)
Example: The figure shows a student seated
on a stool that can rotate freely about a
vertical axis. The student, who has been set into
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
(11-16)
Sample Problem 11-7:
I wh  1.2 kg  m 2
wh  2  3.9 rad/s
I b  6.8 rad/s
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
(11-17)
Analogies Between Translational and Rotational Motion
Translational Motion
Rotational Motion
x  
v
a
 
 
p

mv 2
K
2
m
F  ma


I 2
K
2
I
  I
F


P  Fv

P  
dp
dt
p  mv

 net 

L  I
Fnet 

d
dt
(11-18)