Question 219: Work, Energy and Power & Circular Motion

Question

A particle of mass ' $m$ ' is performing uniform circular motion along a circular path of radius ' $r$ '. Its angular momentum about the axis passing through the centre and perpendicular to the plane is ' $L$ '. The kinetic energy of the particle is

Accepted Answer

$$\begin{aligned}
&\begin{aligned}
& \mathrm{E}=\frac{1}{2} \mathrm{I} \omega^2 \
& \mathrm{~L}=\mathrm{I} \omega \Rightarrow \mathrm{~L}^2=\mathrm{I}^2 \omega^2 \
	herefore\quad& \mathrm{E}=\frac{1}{2} \frac{\mathrm{~L}^2}{\mathrm{I}}
\end{aligned}\
&	ext { But } \mathrm{I}=\mathrm{Mr}^2\
	herefore\quad&\mathrm{E}=\frac{1}{2} \frac{\mathrm{~L}^2}{\mathrm{Mr}^2}=\frac{\mathrm{L}^2}{2 \mathrm{Mr}^2}
\end{aligned}$$

Answer

$\frac{L^2}{2 Mr^2}$

Answer

$\frac{2 \mathrm{~L}^2}{\mathrm{mr}^2}$

Answer

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

Answer

$\frac{2 \mathrm{~L}^2}{3 \mathrm{mr}^2}$

Work, Energy and Power & Circular Motion

A particle of mass ' m ' is performing uniform circular motion along a circular path of radius ' r '. Its angular momentum about the axis passing through the centre and perpendicular to the plane is ' L '. The kinetic energy of the particle is

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