Am I missing something?  on #49/#59, how can we calculate this without an R?   Can we assume a pulley's I=1/2 MR^2?   If we do, and solve for R, I don't get answers compatible with the answer given.  Thanks!

Rich - are these problems in the presentation, classwork/homework?

yes.

Rich, the trick here is that it is not a perfect pulley due to the torque caused by the friction of the axle on the pulley. I will not equal 1/2 MR^2. So, you don't need R, you're just calculating I for this "imperfect pulley." Use Torque = I times alpha. Alpha is delta omega over time. The sum of the torques is found by subtracting the friction torque from the constant torque in the counterclockwise direction. Then you'll get the answers in the answer key.  John

But the givens aren't two Torques... It's a Frictional Torque and a tension (with no R to multiply by).

Rich - I wonder if you have an old version - the online version for question 49 says the following:

. A large pulley of mass 5.21 kg (its mass cannot be neglected) is rotated by a constant torque of 19.6 N-m in the
counterclockwise direction. The rotation is resisted by the frictional torque of the axle on the pulley. The frictional torque is a constant 1.86 N-m in the clockwise direction. The pulley accelerates from 0 to 27.2 rad/s in 4.11 s. Find the moment of inertia of the pulley.

Yes, I had one from 2015, I guess:
49.	A large pulley of mass 5.21 kg (its mass cannot be neglected) is rotated by a constant Tension force of 19.6 N in the counterclockwise direction.  The rotation is resisted by the frictional torque of the axle on the pulley.  The frictional torque is a constant 1.86 N-m in the clockwise direction.  The pulley accelerates from 0 to 27.2 rad/s in 4.11 s.  Find the moment of inertia of the pulley.  with ANSWER: 49.	1.05 kg-m2

Rich, yes, we're always making changes based on our reviews or feedback from our users, so if you printed a problem set out a couple of years ago, it may have been revised to eliminate problems. If you scroll up in this message chain, you can see that Craig Shockley noticed this problem and it was fixed by Yuriy, changing force to torque.  Thanks! John

Why are almost all of the answers missing from the rotational motion Presentation practice problems (PDF w/ answers)?

Which presentation are you referring to? Can you post the name exactly as it appears on the website - or the link directly to it. - Melissa

1 slide per page w/ answers file
Home>Courses>Science>AP Physics 1>Rotational Motion>1 slide per page w/ answers

Richard - thank you for letting me know which file you were talking about. I have now corrected the error - you will see all of the answers on the 1 slide per page with answers PDF now. - Melissa

I noticed that too — If you click on the QR Code, the answer is dragged over and explained via video — FYI.

Genevieve - I have corrected all of the answer pull tabs now in the 1 slide per page with answers PDF. - Melissa

Thank You!

Hello John, How would you approach answering #87, parts (c) and (d). I have completed the others but am stuck on this one. Much appreciated.

Manuel, The key is to use the FBD for Block B and Block C. The friction force pulling block C is equal in magnitude to the friction force that C acts on B as explained by Newton's Third Law. Use the C equation to find the friction force, and input it into the B equation, where a = 2, and solve for mu.  Once you have mu, you can solve part (d) by using the FBD for block C. I'll email you the complete solution to check to see if my hint was any good!  John

Thank you John. I appreciate it.