Rotational Motion Problem - 12

  • Thread starter coldblood
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  • #1
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Hi friends,
Please help me in solving this problem, I'll appreciate the help.

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


Thank you all in advance.
 

Answers and Replies

  • #2
haruspex
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You don't need to break it into two stages. Just look at work conservation from moment of release to maximum extension.
 
  • #3
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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?
 
  • #4
haruspex
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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.
 
  • #5
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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
 
  • #6
haruspex
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https://fbcdn-sphotos-f-a.akamaihd.net/hphotos-ak-ash3/1488253_1462165574010548_1361638838_n.jpg
Well, well. Another error in the book! Your answer is correct.
 

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