Question 533: Centre of Mass and Rotational Motion

Question

A solid cylinder of mass ' $M$ ' and radius ' $R$ ' rolls down an inclined plane of height ' $h$ '. When it reaches the foot of the plane, its rotational kinetic energy is ( $\mathrm{g}=$ acceleration due to gravity)

Accepted Answer

From the law of conservation of energy, we have
Potential energy = Translational kinetic energy + Rotational kinetic energy
or $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I} \omega^2$
or $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^2 \omega^2+\frac{1}{2}\left(\frac{1}{2} \mathrm{mr}^2ight) \omega^2=\frac{3}{4}{m r^2 \omega^2}^2$
or $\quad \omega^2=\frac{4 \mathrm{gh}}{3 \mathrm{r}^2}$
Now the rotational kinetic energy $=\frac{1}{2} \mathrm{I} \omega^2$
$	herefore \quad$ Substituting for $\omega^2$ and I, we have,
Rotational kinetic energy $=\frac{1}{2}\left(\frac{1}{2} \mathrm{mr}^2ight) \frac{4 \mathrm{gh}}{3 \mathrm{r}^2}$
$$=\frac{\mathrm{Mgh}}{3} \quad\ldots(\because M=m)$$

Answer

$\frac{\mathrm{Mgh}}{3}$

Answer

$\frac{\mathrm{Mgh}}{6}$

Answer

$\frac{\mathrm{Mgh}}{4}$

Answer

$\frac{\mathrm{Mgh}}{2}$

Centre of Mass and Rotational Motion

A solid cylinder of mass ' M ' and radius ' R ' rolls down an inclined plane of height ' h '. When it reaches the foot of the plane, its rotational kinetic energy is ( \mathrm{g}= acceleration due to gravity)

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