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
Q. 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 ____ kg · m². (g = 9.8 m/s²)

(A) \(8.3 \times 10^{-3}\)

(B) \(4.75 \times 10^{-3}\)

(C) \(1.86 \times 10^{-2}\)

(D) \(9.5 \times 10^{-3}\)

Correct Answer: \(9.5 \times 10^{-3}\)

Explanation (Complete Step by Step Calculation)

Masses are:

\[ m_1 = 0.4\,\text{kg}, \quad m_2 = 0.35\,\text{kg} \]

Radius of pulley:

\[ R = 2\,\text{cm} = 0.02\,\text{m} \]

Distance fallen by heavier mass:

\[ s = 81\,\text{cm} = 0.81\,\text{m} \]

Time taken:

\[ t = 9\,\text{s} \]

Using equation of motion \( s = \frac{1}{2}at^2 \):

\[ 0.81 = \frac{1}{2} a (9)^2 \]
\[ 0.81 = 40.5a \]
\[ a = 0.02\,\text{m/s}^2 \]

For the two masses, equations of motion are:

\[ m_1 g - T_1 = m_1 a \]
\[ T_2 - m_2 g = m_2 a \]

Torque equation for the pulley:

\[ (T_1 - T_2)R = I \alpha \]

Angular acceleration:

\[ \alpha = \frac{a}{R} \]

Substituting tensions and simplifying:

\[ (m_1 - m_2)g - (m_1 + m_2)a = \frac{Ia}{R^2} \]

Substitute numerical values:

\[ (0.4 - 0.35)(9.8) - (0.75)(0.02) = \frac{I(0.02)}{(0.02)^2} \]
\[ 0.49 - 0.015 = \frac{I(0.02)}{0.0004} \]
\[ 0.475 = 50I \]
\[ I = 9.5 \times 10^{-3}\,\text{kg·m}^2 \]

Thus, the rotational inertia of the pulley is \(9.5 \times 10^{-3}\,\text{kg·m}^2\).

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