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  4. A solid sphere is rolling. What is the fraction of rotational kinetic energy associated with it to its total kinetic energy ?

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Q. A solid sphere is rolling. What is the fraction of rotational kinetic energy associated with it to its total kinetic energy ?

UP CPMTUP CPMT 2009

Solution:

Rotational kinetic energy $K_{R}=\frac{1}{2}I\omega^{2}$
For a solid sphere $I=\frac{2}{5}MR^{2}$
$\therefore K_{R}=\frac{1}{2}\times \frac{2}{5}MR^{2}\omega^{2}$
Translational kinetic energy $K_{T}=\frac{1}{2} Mv^{2}$
Total kinetic energy $K=K_{R}+K_{T}$
$K=\frac{1}{5}MR^{2}\omega^{2}+\frac{1}{2}Mv^{2}$
$=\frac{1}{5}MR^{2} \omega^{2}+\frac{1}{2}MR^{2}\omega^{2}$
$(\because v=R\omega)$
$=\frac{7}{10}MR^{2}\omega^{2}$
$\therefore \frac{\text{Kinetic energy of rotation} K_{R}}{\text{Total energy K }}$
$=\frac{\frac{1}{5}MR^{2}\omega^{2}}{\frac{7}{10}MR^{2}\omega^{2}}$
$=\frac{2}{7}$