Rotational Kinematics and Dynamics

STOP RIGHT THERE - take a moment to view this motivational ROTATING PENCIL VIDEO!


Okay, now that you've gotten that out of your system, you're all motivated, and you see how SUPER COOL rotation can be, we can begin. All motion is some combination of translation and rotation. If you’ve ever been on a roller coaster, you’ve experienced a combination of both of these first hand: screaming your way down the hill, you experienced translation (linear motion), and as you flew around the loops, you experienced rotation. The best thing about rotation is this: it's super easy! However, before you can understand it, you have to know the vocabulary.

Reference Line
Since we only deal with rigid bodies in AP Physics, we can draw a line along the radius of the rotating body (a disc for example) and know that all points along this reference line will move ROTATIONALLY in the exact same way.
Angular Displacement
This is exactly the same as linear displacement (net distance traveled) aside from the way it is denoted; it is not Δx. Instead, we call it Δθ, and we generally measure it in radians.
Arc Length
This is sort of a measure of angular displacement; it is measured in degrees and lets us know how far an object has traveled along a circle. It is denoted Δs.
Angular Velocity
Again, this has a linear counterpart, but a different name. We call it ω (omega), and we measure it in radians per unit of time.
Angular Acceleration
This is a tricky concept. First of all, this is NOT cetripetal acceleration ($\frac{v^2}{r}$), and this is NOT tangential accerlation (linear acceleration). This is simply angular accerlation. It is it's own thing. Specifically, it is the time-derivative of ω. We call it α (alpha).
Rotational Inertia
This is the angular counterpart of mass. It is denotec I (how original), and in the angular forms of linear equations, it is used in place of mass. A good thing to remember is that the inertia of a point mass is $mr^2$. If you absolutely need the technical definition, inertia is an objects tendency to stay in motion (or at rest) when it is already in motion (or at rest).
This is the angular counterpart of force. It is denoted τ, and again, if you must be technical, this is the effectiveness of a force in producing rotational acceleration.
Radius Division
Alright, I made this one up. But it is a good thing to remember: If you need the angular counterpart of velocity or acceleration, divide the linear measure by the radius of rotation.


None of the following equations are on the AP Physics C equation sheet (with the exception of two nondescript torque equations), but their linear analogs ARE. If you have a good handle on the relationships between linear and rotational variables, you should be alright as far as memorization goes.

Calculus Applications and Variable Interrelations

Instantaneous Measures

$\omega =\frac{d \theta}{dt}$

$\alpha = \frac{d \omega}{dt}$

Interrelated Variables

$r = \frac{\Delta s}{\Delta \theta}$

$\omega = \frac{v_{linear}}{r}$

$\alpha = \frac{a_{linear}}{r}$

Changes in s, θ, ω, and α

Memorization tip: Each of these equations has a linear analog. If you already have it memorized (which you should =]), then just remember which angular variable goes with which linear variable, and you should be able to correctly "guess" these equations.

Finding Angular Displacement

$\Delta \theta = \omega_{ave} \Delta t = \omega_{0} \Delta t + \frac{1}{2} \alpha (\Delta t)^2$

Finding Angular Velocity and Acceleration

$\Delta \omega = \alpha \Delta t$

$\omega^2 = \omega_{0}^2 + 2 \alpha \Delta \theta$

Torque and Rotational Inertia

These require a little explanation; if you only want the equations, see the equation summary at the end.


Imagine a rod with a pivot point at one end. Intuitively, you know that neither pulling on it from the other end, pushing on it from the other end, nor pushing it at the pivot point will cause it to rotate. However, applying an upward or downward force F to the free end (opposite the pivot point), would induce rotation. That is because the force you would apply has torque.

Now, let $r$ be the distance from the pivot (axis of rotation) to the point where you applied that force F, and let $\theta$ be the angle between vectors r and F.

We get that $\tau = rF sin \theta$

In many cases, you apply what is called a tangential force, and in that case, s$sin \theta$ is 1, making $\tau = rF$. If you would rather remember $\tau = rF$, then you can make sure that $sin \theta$ will always be one.

Making sure that $sin \theta$ is ALWAYS 1

We can find the perpendicular distance to any force by extending that force as an infinate line in both directions. We then draw a line $l$ (which is called the lever arm or moment arm) from the pivot point to the extended vector F so that the angle between $l$ and F is a right angle. In this case, $\tau = lF$, and you never, EVER have to worry about $sin \theta$!

And by the way, anytime you see $r_{\perp}$, that's just a fancy way of saying $l$, which means that $\tau = r_{\perp}F$, is just a fancy way of saying $\tau = lF$.

So there! You know every single torque equation that anybody is ever going to throw at you!

Rotational Inertia


Rotational intertia is the way it is because we really super want a rotational analog of Newton's Secon Law $F_{net} = ma$, and we have rotational analogs of translational acceleration ($\alpha$), and force (torque), but no analog for mass.

Imagine a small body on the outside of a circle, the center of which is the axis of rotation. It is being acted upon by a tangential force F. We want the net force on the body.

We begin with $F_{net} = ma$, but when we realize that the body will soon be rotating, we substitute $a = r \alpha$ to get $F_{net} = mr \alpha$. Finally, we remember that the force is tangential, so the angle between r and F is right, making $sin \theta = 1$ and $\tau = rF$ (because in this case, $l = r$), so we multiply both sides by $r$ to get $rF = \tau = mr^2 \alpha$.

This is why the rotational interia of a point mass at a distance $r$ from the axis of roataion is define as $mr^2$. Fortunately, inertia is additive, so if you have multiple point masses, you can add up their inertias to obtain $I_{net}$.

Inertia varies when you are not dealing with a point mass; all other inertia equations are available in the equation summary.

The Parallel Axis Theorem (PAT)

Also important when considering inertia, is the question: how do we calculate inertia when the axis of rotation does not pass through the center of the body in question?

We make sure that the new axis is parallel to the one that passes through the center of mass, and then we can add the inertia about the center of mass to the the inertia from the center of mass to the other axis.

In other words, when $x$ is the distance between the two axes, and $I$ is the inertia around the center of mass, we have $I_{tot} = I + mx^2$.

Outside Application: This is useful in calculating center of mass as well. Remeber that center of mass is $\frac {1}{m_{tot}} \sum{m_{i}x_{i}}$ where $x$ has the same value as in the PAT.

Equation Summary

$\tau = rF sin \theta$

When $sin \theta = 1$, $\tau = rF$

Similarly, $\tau = r_{\perp}F$

When $l$ is the lever arm of F relative to the pivot, $\tau = lF$

For a point mass, $I = mr^2$

For a hoop about a central axis, $I = mr^2$

For a hoop about a diameter, $I = \frac{1}{2} mr^2$

For a disk (or solid cylinder) about a central axis, $I = \frac{1}{2} mr^2$

For a disk (or solid cylinder) about a central diameter, $I = \frac{1}{4} mr^2 + \frac{1}{12}mh^2$ where $h$ is the height of the cylinder.

For a solid sphere (or thin spherical shell) about any diameter, $I = \frac{2}{3} mr^2$

Important: Inertia is additive.

The parallel axis theorem: $I_{tot} = I_{cm} + mx^2$

Try a Few Examples

Here are a few typical AP-style questions about rotational kinematics and dynamics.
Remember: this topic alone is about about 10% of the exam!

A block of mass $m$ is hung from a pully of radius $R$ and of mass $M$ and allowed to fall. What is the acceleration of the block?

What is the net torque on a cylinder of radius 12 m acted upon by the following forces?
Tangential Force $F_1 = 100N$, and $F_2 = -80N$ and is applied at a distance two thirds from and perpendicular to the axis of rotation.

An object, originally at rest, begins spinning under uniform angular acceleration. In 10 s, it completes an angular displacement of 60 rad. What is the numerical value of the angluar acceleration?

Special Cases

Another good thing about rotation is this: there are no special cases. Tricky problems, sure, but special cases, no. You can do everything you will need to do on the AP Physics Exam with the equations listed here. There may be special cases in the real world, but do not worry your pretty little head about them on the exam.

Problem Solving Applications


Here are a few buzz words that should tip you off that you're trying to solve a rotation problem:

  • Any direct mention of torque, intertia, or angular ANYTHING
  • Pendululm problems
  • Masses (or penguins) hung from/balanced on a hoizantal stick
  • Any mention of obvious circular motion (ie: tightening a bolt)
  • An object rolling without slipping
  • Any pulley with mass M (we can no longer assume it is massless…IT HAS MASS M! giving it torque)

AND YES - this picture has everything to do with rotation. What do you think the people do when they sit inside the cups?!

External Links


Don't worry. The worst question they've ever put on the AP Physics C exam was "describe the universe in 200 words and give three examples." No sweat.

I'm sorry, I was just kidding. Maybe this joke will cheer you up…

There is this farmer who is having problems with his chickens. All of the sudden, they are all getting very sick and he doesn't know what is wrong with them. After trying all conventional means, he calls a biologist, a chemist, and a physicist to see if they can figure out what is wrong. So the biologist looks at the chickens, examines them a bit, and says he has no clue what could be wrong with them. Then the chemist takes some tests and makes some measurements, but he can't come to any conclusions either. So the physicist tries. He stands there and looks at the chickens for a long time without touching them or anything. Then all of the sudden he starts scribbling away in a notebook. Finally, after several gruesome calculations, he exclaims, 'I've got it! But it only works for spherical chickens in a vacuum.'

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