Rotational Kinematics-YoYo Dropping from Ceiling

1. Sep 13, 2008

Darkalyan

1. The problem statement, all variables and given/known data
Here's the problem in a JPG, with the image:

(the 25 points is b/c it's a practice test)

2. Relevant equations

3. The attempt at a solution

A)KE=0, PE=0 (defined to be at cylinder's initial position)

B)PE=Mg(-h)

C)Final KE=.5*M*(2gh)^2 (To get this, I used the v^2=vi^2 + 2aX equation, plugged it in for v^2 in KE=.5mv^2)

However, I don't think my equations are right because if my KE was that, there would have to be extra energy put into the system for the yo-yo to rotate at all, thus there would be no rotational kinetic energy, which doesn't seem to jive with the original provlem. Where am I going wrong?

Last edited by a moderator: May 3, 2017
2. Sep 13, 2008

Shooting Star

You are quite correct about you last eqn.

The final KE = ½Mv² + ½Iω²., where ω is related to v. This should take take of your problem.

3. Sep 13, 2008

Darkalyan

Okay, so I know that I=(MR^2)/2, but you said that w is related to v. I don't see the relation?

4. Sep 13, 2008

Shooting Star

Without giving you a direct answer, think of the angular displacement of any point on the disk and its relationship to the distance travelled by the centre. What is the relationship between the v and ω of a wheel rolling without sliding? I am sure you can take it from here.

5. Sep 13, 2008

Darkalyan

Hmm. Okay. So, if there's rolling w/out slipping, then v=wR. I then plugged that into the fact that -Mgh=½Mv² + ½Iω², and plugged into the w (that w=v/R). I did a bunch of algebra, and got that v=sqrt(-4gh/3), which works out to be a positive # b/c h is defined to be a negative distance fallen. Is that the right answer for the velocity of the center of mass?

6. Sep 13, 2008

Shooting Star

Looks good, but I fail to see where the "lot" of algebra came in.

[-Mgh = ½Mv² + ½Iω² = (3/4)Mv², by putting ω=v/R , cannot be counted as a lot of algebra...]

However, looking at the original question, you are supposed to give the KE-rot in terms of R, M and v. That would just be (1/4)Mv².

I think that covers everything. Good job.

7. Sep 13, 2008

Darkalyan

Sorry, I'm kind of lazy when it comes to simplification, so that's why I thought it was a 'lot' of algebra :). Thank you very, very much for your help.