Question 503: Centre of Mass and Rotational Motion

Question

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

Accepted Answer

$$ 	ext { Total K.E. }=\frac{1}{2} m v^2+\frac{1}{2} I \omega^2 $$
$$ \begin{aligned} & =\frac{1}{2} m v^2+\frac{1}{2}\left(\frac{1}{2} m R^2ight) \frac{v^2}{R^2} \ & =\frac{1}{2} m v^2+\frac{1}{4} m v^2=\frac{3}{4} m v^2 \end{aligned} $$
At foot of the plane, total K.E. = P.E.
$$ \begin{array}{ll} 	herefore & \frac{3}{4} \mathrm{mv}^2=\mathrm{mgh} \ \Rightarrow & \mathrm{mv}^2=\frac{4}{3} \mathrm{mgh} \end{array} $$
But, rotational K.E. $=\frac{1}{4} m v^2=\frac{m g h}{3}$

Answer

$\frac{\mathrm{mgh}}{3}$

Answer

$\frac{\mathrm{mgh}}{6}$

Answer

$\frac{\mathrm{mgh}}{2}$

Answer

$\frac{\mathrm{mgh}}{4}$

Centre of Mass and Rotational Motion

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

Choose the correct answer: