Question

A mass is whirled in a circular path with a constant angular velocity and its angular momentum is L. If the length of string is now halved keeping the angular velocity same, the new angular momentum is

Easy

Solution

$L = {mr}^{2} ω; L \propto r^{2}; \frac{L_{1}}{L_{2}} = {(\frac{r_{1}}{r_{2}})}^{2} \Rightarrow L_{2} = \frac{L}{4}$

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