Motion of ring/body down an incline

[konichiwa2x]

A body of mass 'm' slides down an incline and reaches the bottom with a velocity 'v'. If the same mass were in the form of a ring which rolls down the incline, what would have been the velcity of the ring?

(A)$$v$$

(B)$$\sqrt{2}v$$

(C)$$\frac{1}{\sqrt{2}}v$$

(D)$$\frac{\sqrt{2}}{\sqrt{5}}v$$

How do I do this? please help.

 

[OlderDan]

Rolling implies rotational energy as well as translational energy. The knietic energy of the ring involves two terms, one for translation and one for rotation.

 

[konichiwa2x]

ok

$$mgh = \frac{mv^2}{2} + \frac{I\omega^2}{2}$$

solving, $$velocity = \frac{v}{\sqrt{2}}$$

correct? thanks for your help.

 

[OlderDan]

$$velocity = \frac{v}{\sqrt{2}}$$

Looks good.