Question 588: Centre of Mass and Rotational Motion

Question

A particle performs rotational motion with an angular momentum 'L'. If frequency of rotation is doubled and its kinetic energy becomes one fourth, the angular momentum becomes.

Accepted Answer

Kinetic energy $$k=\frac{1}{2} I \omega^2$$
$$\begin{aligned}
	herefore \frac{\mathrm{K}_2}{\mathrm{~K}_1} & =\frac{\mathrm{I}_2 \omega_2^2}{\mathrm{I}_1 \omega_1^2} \quad 	herefore \frac{1}{4}=\frac{\mathrm{I}_2}{\mathrm{I}_1} \cdot 4 \quad 	herefore \frac{\mathrm{I}_2}{\mathrm{I}_1}=\frac{1}{16} \
\frac{\mathrm{L}_2}{\mathrm{~L}_1} & =\frac{\mathrm{I}_2 \omega_2}{\mathrm{I}_1 \omega_1}=\frac{1}{16} 	imes 2=\frac{1}{8} \
	herefore \mathrm{L}_2 & =\frac{\mathrm{L}_1}{8}
\end{aligned}$$

Answer

L

Answer

$$\mathrm{\frac{L}{4}}$$

Answer

$$\mathrm{\frac{L}{8}}$$

Answer

$$\mathrm{\frac{L}{2}}$$

Centre of Mass and Rotational Motion

A particle performs rotational motion with an angular momentum 'L'. If frequency of rotation is doubled and its kinetic energy becomes one fourth, the angular momentum becomes.

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