Question 571: Centre of Mass and Rotational Motion

Question

A solid sphere rolls without slipping on an inclined plane at an angle $$	heta$$. The ratio of total kinetic energy to its rotational kinetic energy is

Accepted Answer

Moment of Inertia of a solid sphere, $$I=\frac{2}{5} \mathrm{MR}^2$$
Since there is no slipping,
$$\mathrm{v}=\mathrm{R} \omega$$
$$	herefore$$ Rotational kinetic energy
$$\begin{aligned}
\mathrm{E}_{\mathrm{rot}}=\frac{1}{2} \mathrm{I} \omega^2 \
=\frac{1}{2} 	imes \frac{2}{5} 	imes \mathrm{M} 	imes \mathrm{R}^2 	imes \omega^2 \
=\frac{\mathrm{MR}^2 \omega^2}{5} \
=\frac{\mathrm{MV}^2}{5} \quad 	ext{.... (i)}
\end{aligned}$$
Total kinetic energy
$$\begin{aligned}
E_{\mathrm{K}} & =\frac{1}{2} I \omega^2+\frac{1}{2} \mathrm{MV}^2 \
& =\frac{\mathrm{MV}^2}{5}+\frac{\mathrm{MV}^2}{2} \
& =\frac{7 \mathrm{MV}^2}{10} \quad 	ext{..... (ii)}
\end{aligned}$$
Dividing (ii) by (i), we get,
$$\frac{\mathrm{E}_{\mathrm{K}}}{\mathrm{E}_{\mathrm{rot}}}=\frac{\left(\frac{7 \mathrm{MV}^2}{10}ight)}{\left(\frac{\mathrm{MV}^2}{5}ight)}=\frac{7}{2}$$

Answer

$$\frac{7}{2}$$

Answer

$$\frac{5}{2}$$

Answer

$$\frac{7}{3}$$

Answer

$$\frac{5}{4}$$

Centre of Mass and Rotational Motion

A solid sphere rolls without slipping on an inclined plane at an angle $\theta$. The ratio of total kinetic energy to its rotational kinetic energy is

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