First slide

Rolling motion

Question

A body of radius R and mass m is rolling smoothly with speed v on a horizontal surface. It then rolls up a hill to a maximum height h. If h = $\frac{3 v^{2}}{4 g}$ , what might the body be?

Moderate

A

a solid circular cylinder
B

a hollow circular cylinder
C

a solid circular sphere
D

a hollow circular sphere

Solution

Interpreting h as the height increase for all the centre of mass of the body, then mechanical energy conservation, $K_{i} = U_{f}, leads to$
$\frac{1}{2} {mv}_{com}^{2} + \frac{1}{2} {Iω}^{2} = mgh \Rightarrow \frac{1}{2} {mv}^{2} + \frac{1}{2} I {(\frac{v}{R})}^{2}$
= mg( $\frac{3 v^{2}}{4 g}$ )
from which v cancels and we obtain I = $\frac{1}{2} {mR}^{2}$
This moment of inertia is the moment of inertia of solid cylinder.

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