I am modelling the motion of a wheel that should rotate at a constant speed over time. Think of a Ferris wheel or a car wheel.

I am trying to determine at what angle the wheel will be at about the center at any given time.

Rotating in a friction free, gravity-less perfect environment, such that r and m etc. are irrelevant.

My first guess would be something like this, but I have a feeling this won't hold true for all quadrants of the rotation.


Can anybody help? It's driving me wild and I won't be in front of a computer to help me figure it out for a good few days!

Furthermore, I then want to work out an equation to accelerate and decelerate the angle of the wheel rotating. So any help on that would be great too!

Many thanks in advance :)

  • $\begingroup$ What information do you have? For instance, does the wheel rotate at a constant angular speed? Do you know that angular speed? $\endgroup$ – Dan Rust Sep 28 '13 at 11:30

Do you have angular speed of your wheel? If yes (let's note it $\omega$), then the angle is $$\phi=\phi_0+\omega t,$$ or, if you want a value in $[0,2\pi)$, $$\phi=2\pi\left\{\frac{\phi_0+\omega t}{2\pi}\right\},$$ where $\{x\}$ denotes the fractional part of $x$.

Edit for arbitrary case

Let's suppose we have a wheel of radius $R$ rolling on a straight line without sliding (i.e. without wheelslip). Suppose also that the coordinate of center of our wheel is $x(t)$. For simplicity we take $x(0)=0$ and we take the angle at $t=0$ to be zero.

Note that the angle of rotation is a function of coordinate: indeed, if the wheel rolls a length $l$, then its angle changes by $\frac{l}{R}$. Thus the angle is $$\phi = \frac{x}{R}.$$ As previously, we can take it in $[0,2\pi)$:$$\phi=2\pi\left\{\frac{x}{2\pi R}\right\}.$$

The speed of our wheel is $\dot x$, hence local angular speed is $\frac{\dot x}{R}$. If you take zero acceleration (i.e. constant speed), you will reduce the formulas to the case above. By derivation with respect to time you can obtain expressions for angular acceleration as a function of $\ddot x$, etc.

  • $\begingroup$ Thanks, seems simple for linear speeds. Then just change w for acc/dec? Bit confused about where x comes in? Thanks $\endgroup$ – adam Sep 28 '13 at 12:24
  • $\begingroup$ @adam see edit. $\endgroup$ – TZakrevskiy Sep 28 '13 at 14:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.