Q101 / 150 · MHT CET 2026 (11 April Shift 1) · Physics
A flywheel rotating about a fixed axis has a kinetic energy of $500$ joule, if its angular frequency is $5$ Hz, then calculate the moment of inertia of the wheel about the axis of rotation. ($\pi^2=10$)
- A$0.6 \text{ kg m}^2$
- Correct answerB$1 \text{ kg m}^2$
- C$0.75 \text{ kg m}^2$
- D$0.15 \text{ kg m}^2$
Solution
Given kinetic energy $K = 500$ J and angular frequency $f = 5$ Hz.
Angular velocity $\omega = 2\pi f = 2\pi \times 5 = 10\pi$ rad/s.
The rotational kinetic energy is given by $K = \dfrac{1}{2} I \omega^2$.
Substituting the values:
$500 = \dfrac{1}{2} I (10\pi)^2$
$500 = \dfrac{1}{2} I (100\pi^2)$
Given $\pi^2 = 10$:
$500 = \dfrac{1}{2} I (100 \times 10)$
$500 = 500 I$
$I = 1 \text{ kg m}^2$
Answer: $1 \text{ kg m}^2$
Angular velocity $\omega = 2\pi f = 2\pi \times 5 = 10\pi$ rad/s.
The rotational kinetic energy is given by $K = \dfrac{1}{2} I \omega^2$.
Substituting the values:
$500 = \dfrac{1}{2} I (10\pi)^2$
$500 = \dfrac{1}{2} I (100\pi^2)$
Given $\pi^2 = 10$:
$500 = \dfrac{1}{2} I (100 \times 10)$
$500 = 500 I$
$I = 1 \text{ kg m}^2$
Answer: $1 \text{ kg m}^2$
Hence, the correct answer is option: 'B'.