Q124 / 160 · AP EAMCET 2020 (18 Sep Shift 2) · Physics
A light meter rod has two-point masses each of \(2 \mathrm{~kg}\) fixed at its ends. If the system rotates about its centre of mass with an angular speed of \(0.5 \mathrm{rad} \mathrm{s}^{-1}\), its rotational \(\mathrm{KE}\) is
- A0.125 erg
- B\(1.25 \mathrm{erg}\)
- C\(1.25 \mathrm{~J}\)
- Correct answerD\(0.125 \mathrm{~J}\)
Solution
Total mass of light meter rod including two point masses, \(m=(2+2) \mathrm{kg}=4 \mathrm{~kg}\)
Angular speed, \(I=0.5 \mathrm{rad} \mathrm{s}^{-1}\)
Length of rod \(=1 \mathrm{~m}\)
Moment of inertia about centre of mass, \(I=2\left(\frac{1}{2}\right)^2+2\left(\frac{1}{2}\right)^2=\frac{2}{4}+\frac{2}{4}=1 \mathrm{~kg}-\mathrm{m}^2\) \(\therefore\) Rotational kinetic energy \(K_{\text {rot }}=\frac{1}{2} I \omega^2=\frac{1}{2} \times 1 \times(0.5)^2=\frac{0.25}{2}=0.125 \mathrm{~J}\)
Moment of inertia about centre of mass, \(I=2\left(\frac{1}{2}\right)^2+2\left(\frac{1}{2}\right)^2=\frac{2}{4}+\frac{2}{4}=1 \mathrm{~kg}-\mathrm{m}^2\) \(\therefore\) Rotational kinetic energy \(K_{\text {rot }}=\frac{1}{2} I \omega^2=\frac{1}{2} \times 1 \times(0.5)^2=\frac{0.25}{2}=0.125 \mathrm{~J}\)
Hence, the correct answer is option: 'D'.