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Q.

A thin rod of length $L$ is suspended from one end and rotated with $n$ rotations per second. The rotational kinetic energy of the rod will be :

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a

$2 m L^{2} π^{2} n^{2}$

b

$\frac{1}{2} m L^{2} π^{2} n^{2}$

c

$\frac{1}{6} m L^{2} π^{2} n^{2}$

d

$\frac{2}{3} m L^{2} π^{2} n^{2}$

answer is C.

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Detailed Solution

Frequency of rotation $= n H z$

so, $ω = 2 πn$

and kinetic energy, $K = \frac{1}{2} I ω^{2}$

so, $K = \frac{1}{2} \times \frac{m L^{2}}{3} \times (4 π^{2} \times n^{2})$

$\Rightarrow K = \frac{2}{3} m L^{2} π^{2} n^{2}$

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A thin rod of length L is suspended from one end and rotated with n rotations per second. The rotational kinetic energy of the rod will be :