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Rotational and translational energy problem

  • #1
A solid sphere of radius R is placed at a height of 30cm on a 15 deg slope. It is released and rolls, without slipping to the bottom.

a) From what height should circular hoop of radius R be released in order to equal the sphere's speed at the bottom?

First i started with the sphere
mgh=.5mv2 +.5Iw2
I=.5mR2
w=v/r
so
mgh=.5mv2+.25mv2
gh=.75v2
h=30
v2 =40g

next the hoop
the only difference is we dont know h and I is different
I=mR2
so gh=v2
therefore setting them equal, h=40.

However the answer is 43. I did neglect to include R in the height because both objects have this height R. However when i included R in to h. I found 40+(1/3)R=h which leave me know where.

Any ideas? There is no mention of friction...
 

Answers and Replies

  • #2
4,662
5
Check your number for the sphere's moment of inertia. The value you are using is for a disc or solid cylinder.
 
  • #3
Thankyou! jeez and to think i spent an hour for a little mistake. For anyone else I for a sphere is 2/5MR2
 

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