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Systems of Particles and Rotational Motion — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsSystems of Particles and Rotational Motion3 Marks
Question
A particle performs uniform circular motion with an angular momentum L. If the frequency of particle's motion is doubled and its K.E. is halved, what happens to its angular momentum?
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Answer

$\text{L}=\text{mvr}$ and $\text{v}=\text{r}\omega=\text{r}(2\pi\text{n})$ $\text{r}=\frac{\text{v}}{2\pi\text{n}}$ $\therefore\ \text{L}=\text{mv}\Big(\frac{\text{v}}{2\pi\text{n}}\Big)=\frac{\text{mv}^2}{2\pi\text{n}}$ As $\text{K.E.}=\frac{1}{2}\text{mv}^2,$ therefore, $\text{L}=\frac{\text{K.E.}}{\pi\text{n}}$ When K.E. is halved and frequency (n) is doubled, them, $\text{L'}=\frac{\text{K.E'.}}{\pi\text{n'}}=\frac{\frac{\text{K.E.}}{2}}{\pi(2)\text{n}}$ $=\frac{\text{K.E.}}{4\pi\text{n}}=\frac{\text{L}}{4}$ i.e., angular momentum becomes one fourth.

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