Giancoli's Physics: Principles with Applications, 6th Edition

8

Rotational Motion

Change chapter8-1: Angular Quantities

8-2 and 8-3: Constant Angular Acceleration; Rolling

8-4: Torque

8-5 and 8-6: Rotational Dynamics

8-7: Rotational Kinetic Energy

8-8: Angular Momentum

8-9: Angular Quantities as Vectors

Problem 41

A

$a = \dfrac{g(m_2-m_1)}{\dfrac{I}{R^2} + m_2 + m_1}$

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COMMENTS

By estherjlee93 on Tue, 3/26/2013 - 12:12 AM

Isn't the sum of the torques supposed to be CCW minus CW, and thus, Ft1-Ft2?

By andrewjhunt92 on Thu, 1/4/2024 - 9:57 PM

Same question as estherjlee93 -- any response?

By Mr. Dychko on Mon, 1/15/2024 - 6:25 PM

Hi andrewjhunt, thank you for your question. I guess I've had 11 year to think about it? ;) Wow it's been a long time since I've been making physics videos. In any case, of course you're correct about the convention where CCW torque is positive and CW torque is negative, in which case the video should show CCW torque minus CW torque. Things work out anyway since on the other hand, when we are finding the linear acceleration of a point on the edge we are concerned only about the <b>magnitude</b> of that acceleration. It's sign would depend on which side of the pulley you were looking at; it goes up on the left and down on the right. Had I correctly done CCW minus CW torque I would have found a negative answer for acceleration and then taken the magnitude of that answer anyway, and the solution would proceed as shown. In short, when finding the acceleration of a point on the edge of the pulley we're concerned only with the magnitude of that acceleration, not the sign of it, and the arrangement of CW minus CCW coincidentally gave the positive answer for that acceleration.

All the best,

Shaun

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