MHT CET · Physics · Rotational Motion
The earth is assumed to be a sphere of radius ' \(R\) ', and mass ' \(M\) ' having period of rotation ' \(T\) '. The angular momentum of earth about its axis of rotation is
- A \(\frac{2 \pi \mathrm{MR}^2}{5 \mathrm{~T}}\)
- B \(\frac{4 \pi \mathrm{MR}^2}{5 \mathrm{~T}}\)
- C \(\frac{\mathrm{MR}^2 \mathrm{~T}}{2 \pi}\)
- D \(\frac{\mathrm{MR}^2 \mathrm{~T}}{4 \pi}\)
Answer & Solution
Correct Answer
(B) \(\frac{4 \pi \mathrm{MR}^2}{5 \mathrm{~T}}\)
Step-by-step Solution
Detailed explanation
Angular momentum,
\(\mathrm{L}=\mathrm{I} \omega\)
Moment of inertia of solid sphere is \(\frac{2}{5} \mathrm{MR}^2\)
\(\begin{aligned}
& \mathrm{L}=\frac{2}{5} \mathrm{MR}^2 \omega \\
& \mathrm{~L}=\left(\frac{2}{5} \mathrm{MR}^2\right)\left(\frac{2 \pi}{\mathrm{~T}}\right)=\frac{4 \pi \mathrm{MR}^2}{5 \mathrm{~T}}
\end{aligned}\)
\(\mathrm{L}=\mathrm{I} \omega\)
Moment of inertia of solid sphere is \(\frac{2}{5} \mathrm{MR}^2\)
\(\begin{aligned}
& \mathrm{L}=\frac{2}{5} \mathrm{MR}^2 \omega \\
& \mathrm{~L}=\left(\frac{2}{5} \mathrm{MR}^2\right)\left(\frac{2 \pi}{\mathrm{~T}}\right)=\frac{4 \pi \mathrm{MR}^2}{5 \mathrm{~T}}
\end{aligned}\)
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