Download Ch 11 Rolling, Torque and Angular Momentum

Ch 11
Rolling, Torque
and Angular Momentum
Question 10
Problems: 3, 5, 9, 11, 17, 19, 23, 27,
31, 37, 41, 45, 53, 58, 69
Rolling motion – combination of translational motion
and rotational motion.
Chapter will cover objects that roll ‘smoothly’ along a surface.
The object does not slip or bounce along the sufrace.
For rolling motion, the center of the object moves in a
line parallel to the surface. Other points on the object
do not follow straight paths.
See figure 11-2 to show the different paths taken by
the two different points on a rolling wheel.
Relation between arc length and angle of rotation.
s= R
When the object rotates through the angle , a point
at a distance R from the rotation axis moves through
a distance of s.
vcom
vcom
s
s
s= R
The arc length s is the same as the distance that the
wheel translates.
The linear (translational) speed, vcom, of the wheel is ds/dt.
The angular speed of the wheel is
So:
ds d
R
dt dt
vcom
R
= d /dt
Rolling motion is the combination of purely rotational motion
and purely translational motion.
pure rotation
+
vcom
pure translation
vcom
vcom
vcom
vcom
=
rolling motion
v= 2vcom
vcom
v= -vcom+vcom =0
The velocity of a point at the top of the rolling wheel is twice
that of the center of the wheel.
vtop = (2R) = 2( R) = 2vcom
Fig 11-5 shows a bicycle wheel, the spokes are more blurred at
the top. They are moving faster than the those at the bottom.
Kinetic Energy of Rolling
As an object rolls, the point at the very bottom, the
contact point with the surface, is instantaneously
stationary.
We will call this point P and we can treat the rolling as
rotation about this point.
K = ½ IP 2
IP is the rotational inertia about the point P
Parallel axis theorem says: IP = Icom + MR2
K
K
K
1
IP 2
2
1
I com 2
2
1
I com 2
2
1
MR2
2
1
2
Mvcom
2
2
The kinetic energy of a rolling object comes
from the rotational kinetic energy and
translational kinetic energy.
Forces of rolling
If a wheel rolls smoothly, there is no sliding at the contact point so
there is no friction.
However, if there is an external force that produces an acceleration,
there will be an angular acceleration, . The acceleration will make
the wheel want to slide at the contact point.
Then a frictional force will act on the wheel to oppose the tendency
to slide.
If the wheel does not slide the force is static frictional force.
If the wheel were to slide, the force would by kinetic frictional
force. However, this would not be smooth rolling motion.
Direction of static frictional force.
If the wheel, moving to the right, were to accelerate,
the bottom of the wheel would want to move to the
left compared to the surface. Thus the static friction
force is to the right.
If the same wheel were to slow down, the direction of
the acceleration and angular acceleration would
switch, and the static friction force will now be pointing
to the left.
Rolling down a ramp
The direction of the static friction force is the confusing
part here. It points up along the ramp. If the wheel
were to slide down the ramp, the friction opposing the
sliding would be pointing up.
N
x
fs
Fg sin
Fg cos
fs – Mg sin = M acom
Fg
=I
Only force on the wheel that produces torque is the friction.
R fs = Icom
will need to make use of: acom = R
(a is down the ramp, negative x-direction, but the wheel rolls
counterclockwise so is positive)
acom
R
So we can solve for fs: f s
a com
acom
I com 2
R
g sin
I com
1
MR2
Yo-yo
A Yo-yo is behaves similar to the wheel rolling down a ramp.
1) Instead of rolling down ramp
of angle , the yo-yo follows an angle of 900with the horizontal.
2) The yo-yo rolls down a string
on an axle of radius R.
T
Ro
3) Instead of friction, the tension
slows the yo-yo.
acom
g
I com
1
MR02
Ro
R
Fg
Angular Momentum
linear momentum p = mv
Rotational analogue is angular momentum.
Occurs when momentum is offset from some reference point.


 
r p
 
m( r v )

rmv sin

rp
rmv

r p
r mv
See fig 11-12.
where is the smallest angle between r and v.
O
p
r
Newton’s 2nd Law in Angular Form

Fnet

dp
dt
For rotations

net

d
dt
Proof:



d
dt

d
dt

d
dt

d
dt
 
m( r v )


d
v
d
r


m( r
v)
dt dt

 
m( r a v v )
 
m( r a )
 
r Fnet


r ma

net
Angular Momentum of a System of Particles
Just like we earlier added up the individual momentums of a system of
particles, we can do the same for angular momentums.

L
 
1  2

3 ...

i
i

dL
dt

dL
dt
i

d i
dt

net
The net external torque on a system of particles is equal to the time
derivative of the total angular momentum of a system of particles.
For a rigid body, all the particles that make up the
body have the same angular velocity.
L=I
To get the direction of L, use the right hand rule. L
will be in the same direction as .
Direction will be perpendicular to plane of rotation.
Conservation of Angular Momentum
Just like linear momentum is always conserved,
angular momentum is a conserved quantity.
Can see from Newton’s 2nd for rotations, if
there is no net external torque, the angular
momentum is constant.
So for any isolated system: Li = Lf
Ii i = If f
Gyroscope
A simple gyroscope is a rotating wheel that is on an axle
where one end of the axle is on a support.
When you let go of the axle, the gyroscope want to fall down
because of gravity.
The torque from the weight of the axle changes the angular
momentum of the wheel.
For a rapidly spinning gyroscope, the angular momentum is
fixed by L=I .
The torque will only change the direction of the angular
momentum vector.
Thus the gyroscope rotates in the horizontal plane. This
rotation is called precession.
L=I
 
dL dt
dL = dt Mg r dt
If the gyroscope precesses through an angle, d .
d

L (t )
d

dL
dt

L (t dt )
d
dt
dL
L
( Mgr)dt
I
Mgr
I
Problems:
8, 14, 24, 26, 48, 61