Question 378: Centre of Mass and Rotational Motion

Question

A particle performing uniform circular motion has angular frequency is doubled & its kinetic energy halved, then the new angular momentum is

Accepted Answer

We know Rotational Kinetic Energy$$={1 \over 2}I{\omega ^2},$$
Angular Momentum $$L = I\omega  \Rightarrow I = {L \over \omega }$$
$$	herefore$$ Initial $$K.E. = {1 \over 2}{L \over \omega } 	imes {\omega ^2} = {1 \over 2}L\omega $$
Final $$K.E'$$ = $${{K.E} \over 2}$$ = $${1 \over 2}{L'} 	imes 2\omega $$
$$	herefore$$ $${{K.E} \over {K.E'}} = {{L 	imes \omega } \over {L' 	imes \omega '}} $$
$$\Rightarrow {{K.E} \over {{{K.E} \over 2}}} = {{L 	imes \omega } \over {L' 	imes 2\omega }}$$
$$	herefore$$ $$L' = {L \over 4}$$

Answer

$${L \over 4}$$

Answer

$$2L$$

Answer

$$4L$$

Answer

$${L \over 2}$$

Centre of Mass and Rotational Motion

A particle performing uniform circular motion has angular frequency is doubled & its kinetic energy halved, then the new angular momentum is

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