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Rotational Motion Part 3

Section: Physics

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Question : 7 of 62

Marks: +1, -0

Two masses 400 g and 350 g are suspended from the ends of a light string passing over a heavy pulley of radius 2 cm . When released from rest the heavier mass is observed to fall 81 cm in 9 s . The rotational inertia of the pulley is

\_\_\_\_

\text{kg} \cdot \text{m}^2 \cdot (g=9.8 \text{ m/s}^2)

[24 jan 2026 Shift 1]

$8.3 \times 10^{-3}$
$4.75 \times 10^{-3}$
$1.86 \times 10^{-2}$
$9.5 \times 10^{-3}$

Solution:

Concept:

A heavy pulley has rotational inertia, so the tensions on both sides of the string are different.

Formula:

Kinematics:

h = \frac{1}{2} a t^2

(from rest).

Forces:

m_1 g - T_1 = m_1 a

,

T_2 - m_2 g = m_2 a

.

Torque:

(T_1 - T_2) R = I \alpha

, with

\alpha = a / R

(no slip).

Solution:

1. Find acceleration

a = \frac{2h}{t^2}

.

h = 81\ \text{cm} = 0.81\ \text{m}

,

t = 9\ \text{s}

.

a = \frac{2 \times 0.81}{81} = 0.02\ \text{m/s}^2

.

2. Write equations for masses:

m_1 = 0.4\ \text{kg}

,

m_2 = 0.35\ \text{kg}

,

g = 9.8\ \text{m/s}^2

.

T_1 = m_1(g - a)

and

T_2 = m_2(g + a)

.

Difference:

T_1 - T_2 = (m_1 - m_2)g - (m_1 + m_2)a

.

= (0.05)(9.8) - (0.75)(0.02) = 0.49 - 0.015 = 0.475\ \text{N}

.

3. Torque equation:

(T_1 - T_2) R = I \cdot \frac{a}{R}

.

So

I = \frac{(T_1 - T_2) R^2}{a}

.

Radius

R = 2\ \text{cm} = 0.02\ \text{m}

.

I = \frac{0.475 \times (0.02)^2}{0.02} = 0.475 \times 0.02 = 9.5 \times 10^{-3}\ \text{kg} \cdot \text{m}^2

.

Answer:

Option D:

9.5 \times 10^{-3}

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