# Rotational Motion Problem - 12

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

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

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

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?

haruspex
Homework Helper
Gold Member
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?
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.

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.

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

haruspex