AP Physics
Rotational Equilibrium and Dynamics
In this Chapter we learn how to apply Newton's laws of motion to rotating, rigid
bodies. The major new concepts include torque, which is the rotational analogue of
force, center of gravity, and moment of iner
AP Physics
Another Condition for Equilibrium
For a body to be in equilibrium, Newton's laws require the sum of the forces to be
zero:
(3)
= 0 (ForceBalance).
If the body is an extended object, then the forces don't necessarily act at the same
point. Thus,
AP Physics
Relation Between Torque and Angular
Acceleration
Consider a mass m moving in a circle of radius r , acted on by a tangential force Ft as
shown in Figure 8.2.
Figure 8.2: Torque and angular acceleration
Using Newton's second law to relate Ft to
AP Physics
Rotational Kinetic Energy
The kinetic energy of rotation of a rigid body is obtained by first dividing it up into a
collection of smaller masses, and then summing up the kinetic energies due to the
tangential velocities of the individual masses
AP Physics
Problem Solving Strategies
1.
Draw a diagram of the entire system, including all forces.
2.
Decide whether the net acceleration (angular and linear) is zero or not.
Determine the unknown quantity that you need to find.
3.
Draw force diagrams fo
A 10 m ladder weighs 50 N and balances against a smooth wall at 60 o to the
horizontal. If the ladder is just on the verge of slipping, what is the static coefficient of
friction between the floor and the ladder?
Solution:
The force diagram for the ladder
Calculate the x and y positions of the center of mass of the following three objects
where m1 = 1.0 kg , m2 = 2.5 kg , and m3 = 4.0 kg . See Figure 8.5.
Figure 8.5: Problem 8.3
Solution:
xcm =
=
= 1.6 m
[2mm]ycm =
=
= 0.27 m.
Calculate the moments of iner
a) About the axis through O (center) perpendicular to the page.
b) About an axis joining two of the masses.
Note: The moment of inertia depends strongly on the axis chosen.
Solution:
a)
IO
=
=
=
=
mi ri 2
2m1(22) + 2m2(32)
8m1 + 18m2
34 kg m 2.
b)
We cons
= N  mgcos
[2mm]
.
Fy = 0
where fs is the magnitude of the static frictional force. We must use the static
force because we assume that the object roles without slipping.
For the rotational motion we have:
=  fsr
= IO
fs =
.
The angular acceleration of
Do the above problem (part b) using energy conservation.
Solution:
Wnc = 0 =
KEt +
KEr +
PEg
=
mvcm2  0) + (
(
I
 0) + (0  mgh)
0=
mvcm2 + I
 2mgh.
where h is the vertical distance moved by the object. If L is the distance moved along
the ramp then:
h
=
( I + I)

I
=
I
=  16.7 J

I
where the negative sign indicates that the rotational kinetic energy has
decreased.
A student stands on the center of a rotating platform that has frictionless bearings. He
has a 2.0 kg object in each hand, held 1.0 m fro