Q199 / 200 · MHT CET - Medical 2026 (24 April Shift 1) · Physics
A flywheel used to prepare earthenware pots is set into rotation at 50 rpm. It is in the form of a disc of mass 5 kg and radius 0.2 m. A lump of clay (to be taken equivalent to a particle) of mass 0.8 kg falls on it and adheres to it at a certain distance 'x' from the centre. The value of 'x' if the wheel now rotates at 40 rpm, is
- A$1$
- B$\dfrac{1}{4}$
- Correct answerC$\dfrac{1}{\sqrt{32}}$
- D$\dfrac{1}{\sqrt{8}}$
Solution
The initial moment of inertia of the flywheel (disc) is $I_1 = \dfrac{1}{2}MR^2$.
Substituting the given values, $I_1 = \dfrac{1}{2} \times 5 \times (0.2)^2 = 0.1 \text{ kg m}^2$.
The initial angular velocity is $\omega_1 = 50 \text{ rpm}$.
When the lump of clay of mass $m = 0.8 \text{ kg}$ falls and adheres at a distance $x$, the new moment of inertia becomes $I_2 = I_1 + mx^2 = 0.1 + 0.8x^2$.
The new angular velocity is $\omega_2 = 40 \text{ rpm}$.
By the principle of conservation of angular momentum, $I_1\omega_1 = I_2\omega_2$.
$0.1 \times 50 = (0.1 + 0.8x^2) \times 40$
$5 = 4 + 32x^2$
$32x^2 = 1$
$x^2 = \dfrac{1}{32}$
$x = \dfrac{1}{\sqrt{32}} \text{ m}$
Answer: $\dfrac{1}{\sqrt{32}}$
Substituting the given values, $I_1 = \dfrac{1}{2} \times 5 \times (0.2)^2 = 0.1 \text{ kg m}^2$.
The initial angular velocity is $\omega_1 = 50 \text{ rpm}$.
When the lump of clay of mass $m = 0.8 \text{ kg}$ falls and adheres at a distance $x$, the new moment of inertia becomes $I_2 = I_1 + mx^2 = 0.1 + 0.8x^2$.
The new angular velocity is $\omega_2 = 40 \text{ rpm}$.
By the principle of conservation of angular momentum, $I_1\omega_1 = I_2\omega_2$.
$0.1 \times 50 = (0.1 + 0.8x^2) \times 40$
$5 = 4 + 32x^2$
$32x^2 = 1$
$x^2 = \dfrac{1}{32}$
$x = \dfrac{1}{\sqrt{32}} \text{ m}$
Answer: $\dfrac{1}{\sqrt{32}}$
Hence, the correct answer is option: 'C'.