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For a ball of mass $m$, on a string/rope, set in a horizontal circular motion at the top of the circle the ball will make when the velocity is big enough, I understand that the minimum tangential velocity at the top of the circular motion $v$ is given by $$v=\sqrt{gr}$$ since the centripetal force $\geq$ the force of gravity acting on the ball.

However, when the string/rope is replaced by a rigid rod, I found out that $v \geq 0$ is the new criterium for the ball to complete its path.

I am not sure what accounts for this difference and would like to find some explanation especially through physical concepts/intuition. Thanks in advance to all who tried to give me an answer.

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  • $\begingroup$ That certainly can't be right, can it? Where does the mass start from? If it starts from the bottom, and if the condition is that $v\geq 0 \implies$ the mass+rod always completes a circular path, and so every rigid rod would magically start rotating! ;) Is it likely that they are talking about a mass released from the "top" of the circle? $\endgroup$ – Philip Aug 13 '20 at 15:24
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    $\begingroup$ @Philip I agree and edited the question. $\endgroup$ – Deschele Schilder Aug 13 '20 at 15:43
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The rod can support the ball by pushing upwards on it when it is at the top, and also by pulling when the ball is at the bottom. The string, on the other hand, can only pull along its length, not push. So with the string the ball has to move fast enough that no pushing up is required from the string when the ball is at the top, but this is not true in the case of the rod.

At the top of a circular trajectory the ball momentarily moves in the horizontal direction while it accelerates downwards in the vertical direction. This downwards acceleration is provided by the combination of gravity and whatever force the string or the rod provides. Let's first consider the case of a string, when the ball is moving just slowly enough that the force from the string is zero when the ball is at the top of its trajectory. In this situation Newton's second law ("f=ma") reads $$ mg = m a $$ and for circular motion we always have $$ a = v^2 / r $$ so you get $v = \sqrt{gr}$. Obviously you already knew that, but now I will show the case where instead of a string one has a rod. In this case Newton's second law is $$ mg - f_{\rm rod} = ma $$ where $f_{\rm rod}$ is the force provided by the rod. In this case the ball can go around the circle at pretty much any speed, including zero, because one finds that the force provided by the rod can provide whatever amount of force is needed: $$ f_{\rm rod} = m(g - a) $$ so for $a \rightarrow 0$ we get $f_{\rm rod} \rightarrow mg$ and that is a perfectly viable solution.

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Welcome to physics SE!
If the mass is tied to a rigid rod the ball without tangential velocity the ball finds itself in an unstable equilibrium. The condition to let the motion start is that we give the ball a velocity of $v\gt 0$. If we compare the situation with the ball attached to a rope/string it's obvious that if the velocity at the top is zero, the ball will just fall down.
To let an equilibrium of forces develop at the top the tangential velocity must be such that the centrifugal force equals the gravitational force, leading to the velocity you mentioned in your question.

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