A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to the bottom.

a) From what height should a circular hoop of radius R bereleased on the same slope in order to equal the sphere's speed atthe bottom?
b) Can a circular hoop of different diameter be released froma height of 30 cm and match the sphere's speed at the bottom? If, so what is the diameter? If not, why not?

The height is  

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

   From the question we are told that

           The height is

            The angle of the slope is

According to the law of conservation of energy

     The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

Where I is the moment of inertia which is mathematically represented as this for  a sphere

  The angular velocity is mathematically represented as

So the equation for conservation of energy becomes

Considering a circular hoop

   The moment of inertial is different for circle and it is mathematically represented as

Substituting this into the conservation equation above

Where is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

Recall that  

            Substituting values

they should be similar or the same