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?
✓
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.
$\text{r}=\frac{\text{v}}{2\pi\text{n}}$
$\therefore\ \text{L}=\text{mv}\Big(\frac{\text{v}}{2\pi\text{n}}\Big)=\frac{\text{mv}^2}{2\pi\text{n}}$
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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