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A body is rolling down an inclined plane. If kinetic energy of rotation is $40\%$ of translational kinetic energy, then the body is a
Ring
Cylinder
Hollow ball
Solid ball
Solution
Rotational kinetic energy is
$\mathrm{K}_{\mathrm{R}}=\frac{1}{2} \mathrm{I} \omega^{2}=\frac{1}{2} \mathrm{Mk}^{2}\left(\frac{\mathrm{v}}{\mathrm{R}}\right)^{2}\left(\because \mathrm{I}=\mathrm{Mk}^{2} \text { and } \mathrm{v}=\mathrm{R} \omega\right)$
$=\frac{1}{2} \mathrm{Mv}^{2}\left(\frac{\mathrm{k}^{2}}{\mathrm{R}^{2}}\right)$
Translational kinetic energy is
$\mathrm{K}_{\mathrm{T}}=\frac{1}{2} \mathrm{Mv}^{2}$
As per questions, $\mathrm{K}_{\mathrm{R}}=40 \% \mathrm{K}_{\mathrm{T}}$
$\therefore \frac{1}{2} \mathrm{Mv}^{2}\left(\frac{\mathrm{k}^{2}}{\mathrm{R}^{2}}\right)=40 \% \frac{1}{2} \mathrm{Mv}^{2}$ or $\frac{\mathrm{k}^{2}}{\mathrm{R}^{2}}=\frac{40}{100}=\frac{2}{5}$
For solid sphere, $\frac{\mathrm{k}^{2}}{\mathrm{R}^{2}}=\frac{2}{5}$
Hence, the body is solid ball.
