Question

A light meter rod has two-point masses each of $2kg$ fixed at its ends. If the system rotates about its centre of mass with an angular speed of $0.5rad\cdot s^{-1}$, its rotational $K.E$. is

• A. 0.123 erg
• B. 1.25 erg
• C. 1.25 J
• D. 0.125 J

Answer 1

Answer: (D)

Total mass of light meter rod including two point masses, $m=(2+2)kg=4kg$
Angular speed, $I=0.5rad{s}^{-1}$
Length of rod $=1m$

Moment of inertia about centre of mass, $I=2{\left(\frac{1}{2}\right)}^2+2{\left(\frac{1}{2}\right)}^2=\frac{2}{4}+\frac{2}{4}=1kg-m^2$
$\therefore$ Rotational kinetic energy
$K_{\text{rot }}=\frac{1}{2}I{\omega }^2=\frac{1}{2}\times 1\times (0.5{)}^2=\frac{0.25}{2}=0.125J$