AP Physics C – Rotational Dynamics Review
Torques change angular momentum (and thus usually also angular velocity). The symbol for torque is
the Greek letter . Torque is given by this equation:
 = r × F or in terms of magnitudes,
  rF sin (this is what you will usually use)
r is the distance to the center of spin from where the force is applied. This
variable is often called the lever arm.
F sin  is the force component that is perpendicular to the lever arm.
F sin 
F

r
If the angle  is 90, then the force is perpendicular to the lever arm, the sine is one, and the equation for
torque is simply:
  Fr
You can see that the unit for torque is going to be a newton meter (nm). We leave it like that. This
looks very similar to the unit for work, the joule, but it is quite different. So energy and work are in
joules and torque is left in newton meters.
Torque is a vector quantity, meaning it has a direction. This means that torques can cancel (like when
they would cause rotations in opposite directions)..

125 N is applied to a nut by a wrench. The length of the wrench is 0.300 m. What is the torque?
  Fr  125 N  0.300 m  

A torque of 857 Nm is applied to flywheel that has a radius of 45.5
cm. What is the applied force?
  Fr

37.5 Nm
F

r


1
 857 Nm 
 
 0.455 m 
1880 N
You push on the door as shown in the drawing. What is the torque?
  rF sin
330 N
  330 N 1.5 m sin55.0

1.5 m
410 Nm
55.0
Torques Due to Wrapped Strings:
Frequently, the AP exam includes problems in which a string is wrapped around an object (kind of like a
yo-yo) and then is pulled. If the string does not slip, the torque is the tension at that point times the
radius of the object.
T
r
τ = Tr
Multiple Torques:
What happens if two or more torques act on an object at the same time?
Two forces are applied to the object in the drawing to
the right. The object is free to rotate about the spin
axis. Both cause a torque.
F1 causes a CCW (counter clockwise) rotation around
the axis.
F1
spin axis
r1
r2
F2 causes a CW (clockwise) rotation around the axis.
You are free to make either CCW or CW the positive
F2
direction.. In problems where an object will rotate
(e.g., a rolling object), it is usually most convenient to
choose torques to be positive when in the direction of the rotation. In the problem at right, if CW is
positive then, the sum of the two torques would be:
Equilibrium and Torque:
  =1   2
 F2r2  F1r1
If an object is in rotational equilibrium, it is either at rest or it is
rotating at a constant angular velocity (ω) and with constant angular momentum (L):
If object is in rotational equilibrium, the net torque about any
axis is zero.
This means that the sum of the torques acting on the object must be zero.
=0
Static equilibrium exists when an object has no motion, either linear or angular. There are two
conditions which must exist in order to have your good old static equilibrium:
The net force must be zero and the net torque must be zero.
F=0
=0

Two metal orbs are attached to a very lightweight rigid wire. They are suspended from a rigid point
on the overhead as shown. The system does not move. Calculate the distance from the suspension
line to the center of gravity on the right sphere.
Since the system is at rest, the sum of the torques and the
sum of the forces must be zero.
Let’s look at a FBD and a drawing showing the two torques:
Without using the torque equilibrium, we could not solve
4.0 kg
1.0 kg
F
45.0 cm
r1
?
r2
m1 g
m2 g
the problem. The sum of forces would simply tell us that the
m1 gr1
m2 gr2
upward force would be equal to weight of the two balls. Using
torque, however, allows us to solve the problem. All we have
to do is add up the torques. There is no torque due to the support force, F, since it acts at the location of
the pivot (so the lever arm is zero). Meanwhile, since the other torques are due to the force of
gravity/weight, which always acts at an object’s center of mass, we have:
1   2  0
r2  
m1gr1  m2 gr2  0
1.0 kg  0.45 m 
4.0 kg

m2 gr2  m1gr1
0.11 m or 11 cm
r2  
m1 gr1
m2 g
Moment of Inertia:
The rotational analog to mass is called moment of inertia (I). Just as torques
are measured relative to some point (usually the axis of rotation), so are moments of inertia. Here is the
formula to find an object’s moment of inertia.
I =  mr2 or I = ∫ 𝒓 2 dm
Most of the time, moments of inertia will be provided to you for objects other than point masses, and
their moments of inertia are easily calculated as mr2. If you have several items that are rigidly
connected (like a block attached to a rod attached to a disk), the total moment of inertia is simply the
sum of the moments of inertia of each individual item. Here are a couple of common moments of inertia
you might want to know.
½ mr2
2
1⁄
12 mL
1⁄ mL2
3
mr2
mr2
Uniform Disk About Axis Through Center:
Thin Rod About Its Center:
Thin Rod About An End:
Point Mass About Radius r:
Thin, Hollow Ring of Radius r:
The parallel-axis theorem allows us to calculate the moment of inertia of an object through an axis
parallel to, and a distance D from, an axis through its center of mass as follows:
(parallel axis theorem) I = Icm + mD2
Non-Equilibrium and Torque:
If an object has a non-zero net torque, then it experiences an
angular acceleration (α) given by,
  = Iα
This is the rotational analog to Newton’s 2nd Law (Σ F = ma). If the angular acceleration is constant,
you can apply rotational kinematics equations that are just like those for constant acceleration linear
motion but with linear quantities replaced with rotational analogs (see end).
Work Due to Torques:
Just as work is done when a force is applied over a displacement, work
is also done when torques are applied over an angular displacement.
W = ∫ 𝝉 𝒅𝜽
Power can also be calculated analogous to linear motion (where P = Fv). For rotations,
P = τω
Kinetic Energy of Rotating Objects:
not also translating, its kinetic energy is given by,
KErot = ½ Iω2
When an object is rotating about a fixed axis and is
Angular Momentum:
The rotational analog to linear momentum is angular momentum (L).
The general equation for angular momentum is,
L = r × p or in terms of magnitudes,
L = mvr sin θ
(this is what you will usually use)
Note that angular momentum is always measured relative to some reference point. In the above
equation, r sin θ is the perpendicular distance between a line along the direction of motion and the
reference point/axis of rotation. For an object moving in a circular path of radius r, its angular
momentum is simply L = mvr. For rigid objects rotating about a fixed axis (with no translation of the
center of mass),
L = Iω
If an object is both rotation and translating, its angular momentum is due to both.
L = Lcm + Lrot = mvcmr sinθ + Icmω
Torque and angular momentum are related as follows:
 τ = dL/dt
Conservation of Momentum:
When the net torque on an object is zero, its angular
momentum is conserved (constant). This is why, for example, an ice skater spins faster when he/she
pulls in arms to decrease moment of inertia, I. The angular momentum, Iω is constant, so if I goes
down, angular velocity (ω) must go up.
Likewise, if the net torque on a system of objects is zero, then the total angular momentum of the system
is conserved. This allows us to solve problems involving collisions. For example, if a person runs and
jumps on a merry-go-round that is initially at rest, the initial angular momentum relative to the axis of
the merry-go-round is due only to the running person. After the collision, the angular momentum is
conserved and is due to both the person and the merry-go-round rotating around the axis.
Li = Lf (so long as net torque is zero; this is true for collisions)
Rolling Objects:
When an object rolls, it has both translational (linear) motion and rotational
motion. Ordinarily, you divide this motion into the translational motion of the center of mass and the
rotational motion around the center of mass (particularly if the object rotates about an axis through the
center of mass).
KE = KEtrans + KErot = ½ mv2cm + ½ Icmω2
L = Lcm + Lrot = mvcmr + Icmω
(a) Rolling without slipping: If an object rolls without slipping, the linear motion of its center
of mass is related to the rotation motion of the object as follows:
vcm = rω and acm = rα and Δs = rΔθ
When an object rolls without slipping, it either experiences no friction or static friction.
Typically, whether an object can roll without slipping will depend on whether there is sufficient
friction to prevent slipping.
(b) Rolling with slipping: An object will roll with slipping if it rotates either too fast or too
slow relative to the linear motion of its center of mass (so the non-slip conditions above do not
hold). An object rolling with slipping will typically be subject to kinetic friction which will tend
to move the object closer to non-slip conditions (so if it was rotating too fast, the friction will
cause a torque that will tend to slow the rotational rate until slipping stops).
Linear Motion
(mass) m
(velocity) v
(acceleration) a
(position) x
(force) F
(momentum) p
↔
Rotational Motion
I (moment of inertia)
ω (angular velocity)
α (angular acceleration)
θ (angle)
τ (torque)
L (angular momentum)
Other key relationships: 𝜔 = dθ/dt
α = dω/dt
τ = dL/dt
2
2
2
For constant α: ω = ω0 + αt, Δθ = ω0t + ½ αt , ω = ω0 + 2αΔθ