Question

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 :

• A. $2m{L}^2{\pi }^2{n}^2$
• B. $\frac{1}{2}m{L}^2{\pi }^2{n}^2$
• C. $\frac{2}{3}m{L}^2{\pi }^2{n}^2$
• D. $\frac{1}{6}m{L}^2{\pi }^2{n}^2$

Answer 1

The correct option is C $\frac{2}{3}m{L}^2{\pi }^2{n}^2$

Moment of inertia of the rod about the axis passing through its end $I=\frac{m{L}^2}{3}$

Frequency of rotation $f=n$

Angular velocity of rod $w=2\pi f=2\pi n$

Kinetic energy of the rod $K.E=\frac{1}{2}I{w}^2$

$⟹K.E.=\frac{1}{2}\times \frac{m{L}^2}{3}\times (2\pi n{)}^2=\frac{2}{3}m{L}^2{\pi }^2{n}^2$