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System of Particles and Rotational Motion
Kinetic Energy of Rolling Motion
1 question, with its answer.
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A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is K. If radius of the ball be R, then the fraction of total energy associated with its rotational energy will be :For rolling without slipping, $v = R\omega$ and $I = mK^2$.
Total kinetic energy: $T.K.E. = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2 = \dfrac{1}{2}mv^2\left(1 + \dfrac{K^2}{R^2}\right)$
Rotational kinetic energy: $R.K.E. = \dfrac{1}{2}I\omega^2 = \dfrac{1}{2}mK^2\dfrac{v^2}{R^2} = \dfrac{1}{2}mv^2\left(\dfrac{K^2}{R^2}\right)$
Fraction $= \dfrac{R.K.E.}{T.K.E.} = \dfrac{K^2/R^2}{1 + K^2/R^2} = \dfrac{K^2}{K^2 + R^2}$
That is every question in Kinetic Energy of Rolling Motion.
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