Consider a ball of radius $r$ moving(surfaces are frictionless) inside a hemispherical bowl of radius $R$ and mass $m$. Now if I were to find the angular acceleration ($α$ of the ball I analyse the motion in two ways. I assume the angle is small so the small angle approximation holds. However, The two answers I receive are not consistent

Method 1:

$F=ma$ on Centre of mass in tangential direction $mgθ=ma$

$gθ=a$ and then I write the acceleration of centre of mass ($a$) as $a=α(R-r)$ therefore angular acceleration of centre of mass $$α=\frac{gθ}{(R-r)}$$

Method 2:

$τ=Iα$ on the sphere about the centre of the hemisphere $m g θ (R-r) = ( \frac{2}{5} m r^2 + m (R-r)^2 ) α$


$$ α = \frac{ g θ (R-r)}{\frac{2}{5} r^2 + (R-r)^2} $$

I know that the second method is a more fundamental method but where am I wrong in the first one?

  • $\begingroup$ I added math formatting. Please check to see if I inadvertently changed the intent or the logic of the question. In the future, please use math formatting with dollar signs around math expressions $...$ for inline expressions, and double dollar signs $$...$$ for centered paragraph math. $\endgroup$ – ja72 Jan 9 at 15:23
  • $\begingroup$ Hint: If there is no friction there is no rolling so there shouldnt be a $\tfrac{2}{5} m r^2$ term which is the MMOI of a rolling sphere. $\endgroup$ – ja72 Jan 9 at 15:26
  • $\begingroup$ @ja72 The OP is using the relation $\tau=I\alpha$ about the center of the bowl and determining $I$ using the parallel axis theorem, so the $\frac25mr^2$ is fine. I am unsure about where the $16$ comes from though $\endgroup$ – Aaron Stevens Jan 9 at 15:33
  • $\begingroup$ I believe there might be something in the difference between treating the sphere somewhat as a point particle in the first method, since notice when $r\ll R$ the results are the same (except for the 16, which I don't think should be there). But I am still thinking through a full answer $\endgroup$ – Aaron Stevens Jan 9 at 15:42
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    $\begingroup$ The question does not make it clear which angular acceleration is meant. There is angular acceleration of the ball with respect to inertial frame, and there is angular acceleration in the sense of second derivative of $\theta$. Which one are you trying to find? $\endgroup$ – Ján Lalinský Jan 9 at 16:30

The angular acceleration denoted $\alpha$ in method 1) is second derivative of $\theta$. It describes how radius vector of the ball center rotates, but it is not influenced by ball's rotation in any way.

It can be found using both methods, but only the first one is done correctly in OP.

In the method 2), OP is trying to use the torque equation with torque and inertia moment of the little ball rotating around a bowl center, but this would be legitimate only if the ball was rotating around that point as a rigid body. However, by assumption the little ball does not rotate at all.

Thus the proper moment of inertia to use in the torque equation is that of a mass point, $I=m(R-r)^2$. Then, we get

$$ \tau = I\alpha $$

$$ mg\theta (R-r) = m(R-r)^2 \alpha $$ which gives the same result as method 1:

$$ \alpha = \frac{g\theta}{R-r}. $$

  • $\begingroup$ Sorry to keep bothering you, but I still don't think I buy this. Why can we not consider the ball as rotating about the center of the bowl? $\endgroup$ – Aaron Stevens Jan 9 at 17:06
  • $\begingroup$ The torque equation $\tau = I\alpha$ for some reference point is derived from the general mechanical laws under the assumption that the body is actually rotating around that reference point, or that the point concides with center of mass. But here it is not either, because when you take two points on the ball, they do not move in concentric circles centered at common point. The ball is in translation motion, not rotating motion. So the only way in which to apply the torque equation correctly here is to replace the ball by a mass point at the ball's center. $\endgroup$ – Ján Lalinský Jan 9 at 17:12
  • $\begingroup$ Such mass point then can be regarded as in rotation motion around the bowl's center for the purpose of derivation of torque equation, because it is a single point moving in circle. $\endgroup$ – Ján Lalinský Jan 9 at 17:14
  • $\begingroup$ So if we had a similar scenario of a physical pendulum of a sphere at the end of a rod of negligible mass, we would still use $I=m(R-r)^2$ to determine things like the period of oscillation? $\endgroup$ – Aaron Stevens Jan 9 at 17:21
  • $\begingroup$ No, in case of a pendulum the ball on the end of the rod or string would actually rotate around the bowl's center, so one would use the moment of inertia around that center (thus one would use the parallel axis theorem). $\endgroup$ – Ján Lalinský Jan 9 at 17:51

In the first method you forget the reaction of the bowl on the sphere which has a non-zero tangential component.

The two situations are not equivalent. In the second method, you consider the sphere as a pendulum with a link axis without mass up to the axis of rotation. You calculate the moment of inertia around the axis by Huygens' theorem. But in this case the sphere must turn so that the point of contact with the bowl is always the same point of the sphere. It would be the link axis, rigid, that would cause this rotation. The rigid link axis would exert a tangential action on the sphere.

In the first method, as there is no friction, the sphere is in translation (not rectilign): it does not rotate. A diameter always keep the same direction. The point of contact with the bowl is never the same.

  • $\begingroup$ forgot to mention there is no friction so normal is the only contact force $\endgroup$ – Harsh Somani Jan 8 at 18:23
  • $\begingroup$ is it necessary for the sphere to turn about its own axis for applying huygen's theorem? $\endgroup$ – Harsh Somani Jan 8 at 20:50
  • $\begingroup$ Yesterday, I did not look in detail at your calculation. For Huygens' theorem, I would rather write ${{J}_{\Delta O}}={{J}_{\Delta G}}+m{{(R-r)}^{2}}$. I hope I understood correctly your approach ! Otherwise, when you apply the kinetic moment theorem around Oz (and use the moment of inertia around Oz), you suppose that the system is spinning around Oz $\endgroup$ – Vincent Fraticelli Jan 9 at 5:31
  • $\begingroup$ Vincent, the ball is rotating around the center of the bowl though, then though it is not physically connected to the center of the bowl $\endgroup$ – Aaron Stevens Jan 9 at 15:41

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