Question

A uniform solid disk and a uniform hoop are placed side by side at the top of an incline of height h. If they are released from rest and roll without slipping, which object reaches the bottom first?

Answer 1

Answer:

The disk will reach the bottom first.

Explanation:

As we know:

change in kinetic energy = change in potential energy

                                   ΔK.E = -ΔP.E

                             $\frac{1}{2} mv^{2} + \frac{1}{2}Iw^{2} = Mgh$

                                $Mv^{2} +I\frac{v^{2} }{R^{2} } = 2Mgh$

                           $v=\sqrt{\frac{2Mgh}{M+\frac{I}{R^{2} } } }$

For the disk:

$I=\frac{MR^{2} }{2}$

For Hoop:

$I=MR^{2}$

Hence

Velocity for the disk:

$v=\sqrt{\frac{4}{3}gh }$

Velocity for the Hoop:

$v=\sqrt{gh}$