A ball with mass $m$ and radius $r$ rolls without sliding inside a cylinder with radius $R (R>>r)$, with $\theta <<1$. Find the angular frequency $\omega$

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What I Know: There are two movements involved: the rotation of the center of mass around the center of the cylinder $C$ and the rotation around the center of mass. These two movements are coupled because there is no sliding. So, if $\theta_{1}(t)$ is the angular displacement associated to the rotation around the center of mass and $\theta_{2}(t)$ is the angular displacement associated to the rotation of the center of mass, then:

$R\dot\theta_{2} = r\dot\theta_{1}$

If the movement were merely the rotation of the center of mass, then: $ma_{\theta} = -mg\sin(\theta) \therefore mR\ddot\theta = -mg\sin(\theta)$ (the tangent axis is oriented to increasing values of $\theta$) This would give $\ddot\theta + \frac{g}{R} \theta = 0$, considering $\sin \theta \approx \theta $. In this case, we would simply have $\omega = \sqrt{\frac{g}{R}}$.

I don't know how to account for the fact that the ball is also rolling. Could someone give me a hint to solve the original problem, considering the existence of two movements?

  • $\begingroup$ This site is going to be of help for you. Also, think what distance makes travels the ball when it makes a complete rotation in its own center-of-mass. Take a point of the top of the ball and think what distance along the cylinder bottom travels the ball until the point returns to the top of the ball. $\endgroup$
    – Sofia
    Mar 8, 2015 at 22:56

1 Answer 1


When a ball rolls down a ramp, it will accelerate more slowly than a sliding object (without friction). This suggests that you could treat the motion as that of a sliding object with a larger apparent inertial mass.

I believe you will be able to solve your problem if you take this hint and apply it to your situation. You might find some inspiration in this earlier answer I wrote

  • $\begingroup$ using exactly the same method you used in your earlier answer, I found that $\ddot\theta_{1} = \frac{5g}{7r}\theta_{1}$, where $\theta_{1}$ is the angular displacement associated with the rotation around the center of mass. As $\theta_{1}r = \theta_{2}R$ (because there is no sliding), I would get $\ddot\theta_{2} = \frac{5g}{7r}\theta_{2}$ and $\omega = \sqrt{\frac{5g}{7r}}$. Is this correct? $\endgroup$
    – El Cid
    Mar 9, 2015 at 2:13
  • $\begingroup$ It doesn't look right - your frequency must depend on R since that defines the "potential well". You made a mistake going from $\theta_1$ to $\theta_2$ (your equation looked right but you didn't use it). $\endgroup$
    – Floris
    Mar 9, 2015 at 3:04
  • 1
    $\begingroup$ I used an energy argument and I finally got the result. Thank you by your suggestion $\endgroup$
    – El Cid
    Mar 13, 2015 at 21:18
  • $\begingroup$ @ElCid it would be interesting to see what the final result was that you obtained - it will help future visitors if you could write it down (either by editing your question, or writing your own answer). $\endgroup$
    – Floris
    Mar 13, 2015 at 21:40

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