JEE MAIN - Physics (2022 - 26th June Evening Shift - No. 4)
A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is
$${2 \over 5}$$
$${2 \over 7}$$
$${1 \over 5}$$
$${7 \over 10}$$
Explanation
$$K{E_R} = {1 \over 2}l{w^2}$$
$$ = {1 \over 2} \times {2 \over 5} \times {\omega ^2} \times (m{R^2})$$
$$K{E_{total}} = {1 \over 2} \times {7 \over 5} \times m{R^2} \times {\omega ^2}$$
$$\therefore$$ $${{K{E_R}} \over {K{E_{total}}}} = {2 \over 7}$$
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