First slide

Rolling motion

Question

A body is rolling down an inclined plane. If kinetic energy of rotation is 40% of kinetic energy in translatory state, then the body is a

Moderate

A

ring
B

cylinder
C

hollow ball
D

solid ball

Solution

Rotational kinetic energy is
$K_{R} = \frac{1}{2} {Iω}^{2} = \frac{1}{2} {Mk}^{2} {(\frac{v}{R})}^{2} (I - {Mk}^{2} and v = Rω)$
= $\frac{1}{2} {Mv}^{2} (\frac{k^{2}}{R^{2}})$
Translational kinetic energy is
$K_{T} = \frac{1}{2} {Mv}^{2}$
As per question, $K_{R} = 40 % K_{T}$
$\frac{1}{2} {Mv}^{2} (\frac{k^{2}}{R^{2}}) = 40 % \frac{1}{2} {Mv}^{2} or \frac{k^{2}}{R^{2}} = \frac{40}{100} = \frac{2}{5}$
For solid sphere, $\frac{k^{2}}{R^{2}} = \frac{2}{5}$
Hence, the body is solid ball.

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