Question

A light meter rod has two-point masses each of $2\, kg$ fixed at its ends. If the system rotates about its centre of mass with an angular speed of $0.5\, rad \cdot s ^{-1}$, 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, $m=(2+2) kg =4\, kg$
Angular speed, $I=0.5\, rads ^{-1}$
Length of rod $=1\, m$

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