Thread: Rotational energy problem

Question:
http://i.imgur.com/qWrXh.jpg

(Correct) Answers:
(a) 0.146938776m
(b) 0.102857143m

My work:

First off, my problem likely lies with the fact that it is rotating instead of being translated like in regular translations (like a ball being thrown etc). I remember my teacher saying something of the sort and I checked my notes and it seems to be about moment of inertia which has me lost. I have a feeling that in the following my mistake lies with the initial kinetic energy.

Ok so what I did for (a) which gives me half the right answer is:
Ui = 0, Ki = 1/2 * m * v^2, Kf = 0, Uf = mgh
Ef = Ei
mgh = 1/2 * m * v^2
gh = 1/2 * v^2
h = (1/2 * v^2)/g
h = 1/2 * 1.2^2 / 9.8
h = 0.073469388

as for (b), it seems to be very similar.

Any help would be GREATLY appreciated!
Thanks in advance!

If a body slips,it has only translational Kinetic energy

and its

If it doesnot slip than it has Rotational energy and translational energy as well

the

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I for ring is (mr^2)/ 2

for hollow sphere its (2mr^2)/3

And for solid sphere its 2(mr^2)/5

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you will get correct answers now..feel free to ask incae you still find it difficult to digest.....