Question 604: Centre of Mass and Rotational Motion

Question

A solid cylinder of radius $$r$$ and mass $$M$$ rolls down an inclined plane of height $$h$$. When it reaches the bottom of the plane, then its rotational kinetic energy is ($$g=$$ acceleration due to gravity)

Accepted Answer

When the solid cylinder reaches the bottom of the plane, its rotational kinetic energy is given by
$$K_r=\frac{1}{2} \omega^2$$
As, $$\quad I=M k^2$$ and $$\omega=\frac{V_{\mathrm{CM}}}{R}$$
$$\Rightarrow K_r=\frac{1}{2} M k^2 	imes\left(\frac{v_{\mathrm{CM}}}{R^2}ight)^2=\frac{1}{2} M v_{\mathrm{CM}}^2\left(\frac{k^2}{R^2}ight)$$
At bottom, velocity, $$v_{\mathrm{CM}}=\sqrt{\frac{2 g h}{1+\frac{k^2}{R^2}}}$$
Also, for solid cylinder, $$\frac{k^2}{R^2}=\frac{1}{2}$$
$$	herefore K_r=\frac{1}{2} M\left[\frac{2 g h}{\left(1+\frac{1}{2}ight)}ight]\left(\frac{1}{2}ight)=\frac{4 M g h}{4 	imes 3}=\frac{M g h}{3}$$

Answer

$$\frac{M g h}{4}$$

Answer

$$\frac{M g h}{2}$$

Answer

$$M g h$$

Answer

$$\frac{M g h}{3}$$

Centre of Mass and Rotational Motion

A solid cylinder of radius $r and mass M rolls down an inclined plane of height h. When it reaches the bottom of the plane, then its rotational kinetic energy is (g=$ acceleration due to gravity)

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