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
Solution
<p>$K{E_R} = {1 \over 2}l{w^2}$</p>
<p>$= {1 \over 2} \times {2 \over 5} \times {\omega ^2} \times (m{R^2})$</p>
<p>$K{E_{total}} = {1 \over 2} \times {7 \over 5} \times m{R^2} \times {\omega ^2}$</p>
<p>$\therefore$ ${{K{E_R}} \over {K{E_{total}}}} = {2 \over 7}$</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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