A disc is rolling without slipping on a straight surface. ratio of its translational kinetic energy

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6.System of Particles and Rotational Motion

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A disc is rolling without slipping on a straight surface. The ratio of its translational kinetic energy to its total kinetic energy is

A

$\frac{2}{3}$

B

$\frac{1}{3}$

C

$\frac{2}{5}$

D

$\frac{3}{5}$

(AIIMS-2009)

Solution

$T_{KE} = \frac{1}{2}\,m{v^2}$

$R_{KE} = \frac{1}{2}\,I{\omega ^2}$

$\omega=v/R$

$ \Rightarrow \frac{{T_{KE}}}{{T_{KE} + R_{KE}}} = \frac{2}{3}$

Standard 11

Physics

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(NEET-2017)