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Lecture 903, Slide 9: Two Identical pucks pulled by strings

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Strings are wound around two identical pucks: puck 1 around its outer rim (larger radius); the puck 2 around its axle (smaller radius). If both are pulled from rest with the same force F across a frictionless surface, which puck has greater center-of-mass speed?

 

The answer is both have the same C.O.M speed. I am trying to understand why. Am I right in saying the following?

Puck 2 will have a greater rotational speed do to its smaller radius. However, since most of the force applied goes into the rotational speed, it has less translational speed. The opposite is true for Puck 1. Somehow they end up with the same CoM speed.

Am I right? Is there an equation I'm missing/forgetting?

asked Dec 5, 2013 in Intro College Physics by inertialoha (180 points)

1 Answer

+1 vote
When dealing with both rotational and translational motion, we need to separate the two. For translational motion, ||ma=F_{net}|| and for rotational motion ||\alpha=I\tau||.

So in this case, the external forces applied on both are identical, which should result in identical COM acceleration.

The difference in this question is that the two forces will create different torque. The string wrapped around the outer rim will create a larger torque causing a larger angular acceleration. So after a while, both pucks will have same translational speed but the one got wrapped around outer rim will have larger angular speed (rotating faster).

In this question, the rotational and translational motions are independent of each other.
answered Dec 5, 2013 by PhysicsDude (21,750 points)
Thank you! It was bothering me.
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