# Rotational Motion Problem - 12

1. Dec 16, 2013

### coldblood

Hi friends,

The problem is as:

https://fbcdn-sphotos-b-a.akamaihd.net/hphotos-ak-prn1/q71/s720x720/1520824_1461728107387628_750491406_n.jpg

Attempt -

https://fbcdn-sphotos-b-a.akamaihd.net/hphotos-ak-frc3/q71/s720x720/1459220_1461728404054265_66267352_n.jpg

2. Dec 16, 2013

### haruspex

You don't need to break it into two stages. Just look at work conservation from moment of release to maximum extension.

3. Dec 17, 2013

### coldblood

According to this, K.E. →Max→at mean position (I think it won't be the release point)
and
P.E.→Max→ at ext. position.

1/2. mv2 + 1/2 Iω2 = 1/2. kx2 (since ω= v/r)

=> 1/2. mv2 + 1/2 (1/2.mr2)(v/r)2 = 1/2. kx2
=> 3/4. mv2 = 1/2. kx2
x2 = (3/2)(mv2)/k
Is that correct?

4. Dec 17, 2013

### haruspex

It starts at rest, and it will be instantaneously at rest at maximum extension, so there's no KE to consider. There's just the lost gravitational PE and the gained spring PE.

5. Dec 17, 2013

### coldblood

https://fbcdn-sphotos-f-a.akamaihd.net/hphotos-ak-ash3/1488253_1462165574010548_1361638838_n.jpg

6. Dec 17, 2013