Transcript Rolling, Torque, and Angular Momentum

AP Physics C
Montwood High School
R. Casao
•When
a wheel moves along a straight track, the
center of the wheel moves forward in pure
translation.
•A point on the rim of the wheel traces out a
complex path called a cycloid.

For a wheel passing at constant speed while
rolling smoothly (no sliding):
◦ the center of mass O of the wheel moves forward
at constant speed vcom.
◦ the point P where the wheel makes contact with the
surface also moves forward at speed vcom so that it
remains directly below the center of mass O.
During time interval t, both
O and P move forward by
a distance s. Observers
see the wheel rotate thru an
angle q about the center of
the wheel.

t1 = 0
t2 = t



The point on the wheel P that was touching the
surface at t = 0 s moves through an arc length
s.
Equations: s = R·q; vcom = ds/dt;  = dq/dt
Differentiating the arc length equation wrt time:
ds/dt = R·dq/dt to give us vcom = R·.
t1 = 0
t2 = t





The rolling motion of a wheel is a combination of
purely translational and purely rotational motions.
Every point on the wheel rotates about the center with
angular speed .
Every point on the outside edge of the wheel has linear
speed vcom.
Purely translational motion: every point on the wheel
moves forward with speed vcom.
Combination of rotational motion and translational
motion produces the rolling motion of the wheel.

The portion of the wheel at the bottom (point P)
is stationary and the top of the wheel is moving
at speed 2·vcom. The wheel is moving fastest
near the top than near the bottom because the
spokes are more blurred at the top than at the
bottom.

Consider the rolling motion about an axis thru
point P: K = 0.5·Ip·²
◦ Ip is the rotational inertia of wheel about axis thru point
P and  is the angular speed of the wheel.
◦ Apply the parallel-axis theorem to determine Ip:
Ip = Icom + M·h²; for the wheel, h = R

Substituting:

K = 0.5·Icom·² + 0.5·M·R²·²
Because vcom = R·,
K = 0.5·Icom·² + 0.5·M·vcom²
K  0.5  (Icom  M  R )  
2
2

A rolling object has two types of kinetic energy: a
rotational kinetic energy (0.5·Icom·²) due to its
rotation about its center of mass and a
translational kinetic energy (0.5·M·vcom²) due to
translation of its center of mass.
Forces of Rolling


If a wheel rolls at constant speed it has no tendency to
slide at the point of contact P and no frictional force acts
there; acom = 0 at point P.
If a net force acts on the rolling wheel to speed it up or
slow it down, the net force causes an acceleration of the
center of mass along the direction of travel.





The wheel also rotates faster or slower, so an
angular acceleration  occurs about the center
of mass.
These accelerations tend to make the wheel
slide at point P and a frictional force must act on
the wheel at P to oppose the tendency to slide.
If the wheel does NOT slide, the force is a static
frictional force fs and the motion is smooth
rolling motion.
Smooth rolling motion: acom = R·
If the wheel does slide when the net force acts
on it, the frictional force that acts at P is a
kinetic frictional force and the motion is not
smooth rolling motion.
For a uniform body of mass M and radius R
rolling smoothly down a ramp at angle q along
an x axis.
 The normal force N acts at
point P but has been shifted
up to the center of mass.
a

Static frictional force fs acts
at point P and is directed
up the incline. If the body
were to slide down the
incline, the frictional force
would oppose that motion.

com


Motion along the x-axis (up as positive):
fs – M·g·sin q = M·acom
Angular motion: Torque = I·;
◦ N and Fg·sin q pass thru the pivot at the center of
mass and do not produce a torque.
◦ fs applied perpendicularly a distance R at point P and
produces a torque.
◦ Torque = F·r = fs·R;
fs·R = Icom·
◦ -acom = R·; acom is negative because it is directed in
the negative direction on the x-axis. Unlike the linear
motion problems we have done in the past where we
took the direction of motion as positive, with the
rotation involved, we take the positive direction as the
positive direction of the rotation (ccw = positive; cw =
negative).

Solve for  and substitute:  =-acom/R
fs·R = Icom· -acom/R
 Icom  a com
fs 
R2
fs  M  g  sinθ  M  a com
 Icom  a com
 M  g  sinθ  M  a com
2
R
Icom  a com
M  a com 
 M  g  sinθ
2
R
Icom 

a com   M  2   M  g  sinθ
R 

 M  g  sinθ
 g  sinθ
a com 
; a com 
Icom
Icom
M 2
1
R
M  R2



As yo-yo rolls down its string from
height h, potential energy m·g·h is
converted to both rotational
(0.5·Icom·²) and translational
kinetic energy (0.5·M·vcom²). As it
climbs back up, kinetic energy is
converted to potential energy.
The string is looped around the axle
and when the yo-yo hits the bottom
of the string, an upward force on the
axle from the string stops the
descent.
The yo-yo spins with the axle inside
the loop with only rotational kinetic
energy.
acom
y




The yo-yo spins until you jerk on the
string to cause the string to catch on
the axle and allow the yo-yo to climb
back up.
The rotational kinetic energy of the
yo-yo at the bottom of the string can
be increased by throwing the yo-yo
downward so that it starts down the
string with initial speeds vcom and 
instead of rolling down from rest.
The yo-yo rolls on the axle of radius
Ro.
The yo-yo is slowed by the tension
force T from the string on the axle.
acom
y


Acceleration equation: a com
g

Icom
1
M  R2
the yo-yo has the same downward acceleration
when it is climbing back up the string because
the forces acting on it are still those shown in
the figure.
Torque Revisited


We previously defined torque for a rigid body
that rotated about a fixed axis in a circle.
To define the torque for an individual particle
that moves along any path relative to a fixed
point (rather than a fixed axis), the path doesn’t
have to be a circle and the torque has to be
written as a vector.



The figure shows a particle at point A in the xy
plane with a single force F acting on the particle.
The particle’s position relative to the origin O is
given by the position vector r.
Torque equation:
 
  rxF



To find the direction of the torque, slide the
force vector without changing its direction until
its tail is at the origin O so that it is tail to tail
with the position vector.
Use the right hand rule to rotate the position
vector r into the force vector F. The thumb
points in the direction of the torque.

The magnitude of the torque is given by:
  r  F  sinθ

q is the angle between the position vector r and
the force vector F.
Angular Momentum
The figure shows a particle of mass
m with linear momentum p = m·v
as it passes thru point a in the xy
plane. The angular momentum l of
the particle with respect to the
 
origin O is: 
 

l  r x p  m  r x v 





r is the position vector of the particle wrt O.
As the particle moves wrt O in the direction of
its momentum p, the position vector r rotates
around O.
To have angular momentum, the
particle itself does not have to
rotate around O.
Unit for angular momentum:
kg·m²/s = J·s
The direction of the angular
momentum vector is found by
sliding the vector p until its tail
is at the origin O.




Use the right-hand rule, rotating the vector r
into vector p. The thumb points in the direction
of the angular momentum vector.
The magnitude of the angular momentum
vector is: l = r·m·v·sin q, where q is the angle
between r and p when the two vectors are tail to
tail.
Angular momentum only has meaning wrt a
specific origin.
The direction of the angular momentum vector
is always perpendicular to the plane formed by
the position vector and the linear momentum
vector.




Fnet = dp/dt expresses the close relation
between force and linear momentum for a
single particle.
There is also a close relationship between
torque and angular momentum: Tnet = dl/dt
The vector sums of all the torques acting on a
particle is equal to the time rate of change in
the angular momentum of that particle.
The torques and angular momentum are both
defined wrt the same origin.

For a system of particles with respect to an
origin, the total angular momentum L of the
system is the vector sum of the angular momenta
n 
l of the individual particles:    
L  l1  l 2  l 3  ...   li
i 1

Over time, the angular momenta of individual
particles may change due to interactions within
the system between individual particles or
because of outside influences on the system.


n
dl i
dL

dt
i 1 dt





n 
d
L
Since Tnet = dl/dt, then
  Tnet ,i
dt
i 1
The rate of change of the system’s angular
momentum L is equal to the vector sum of the
torques on the individual particles.
These torques include internal torques due to
forces between particles and external torques
due to forces on the particles from bodies
outside of the system.
The forces between particles always occur in 3rd
law pairs so their torques cancel each other, so
the only torques that can change the total
angular momentum L of a system are external

torques.

Tnet
dL

dt


2nd law for rotation of a system of particles: the
net external torque Tnet acting on a system of
particles is equal to the time rate of change of
the system’s total angular momentum L.
The torque and the system’s angular
momentum must be measured from the same
origin.
Angular Momentum of a Rigid Body
Rotating About a Fixed Axis

For a system of particles that form a rigid
body that rotates about a fixed axis rotating
with constant angular speed : L = I·


If the net external torque acting on a system is
zero, the angular momentum L of the system
remains constant, no matter what changes take
place within the system: Li = Lf; Ii·wi = If·wf
Because momentum is a vector quantity , the
conservation of momentum has to be
considered in all three dimensions (x, y, z).
Depending on the torques acting on a system,
the angular momentum of the system might be
conserved in only 1 or 2 directions but not in all
directions.
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
Angular momentum at t f : L f  I f  f
Ii
Li  L f  I ii  I f  f   f  i
If
Ii
Since I f  I i   1   f  i
If
The rotation rate of the student in fig.b is faster
(11-16)