Rotational and translational energy problem

[call-me-kiko]

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 don't 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 into h. I found 40+(1/3)R=h which leave me know where.

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

 

[Bob S]

Check your number for the sphere's moment of inertia. The value you are using is for a disc or solid cylinder.

 

[call-me-kiko]

Thankyou! jeez and to think i spent an hour for a little mistake. For anyone else I for a sphere is 2/5MR²