Question

Two masses $400\,\text{g}$ and $350\,\text{g}$ are suspended from the ends of a light string passing over a heavy pulley of radius $2\,\text{cm}$. When released from rest the heavier mass is observed to fall $81\,\text{cm}$ in $9\,\text{s}$. The rotational inertia of the pulley is ___ $\text{kg m}^2$.
(Given: $g = 9.8\,\text{m s}^{-2}$)

• A. $9.5 \times 10^{-3}$
• B. $1.86 \times 10^{-2}$
• C. $8.3 \times 10^{-3}$
• D. $4.75 \times 10^{-3}$
  \bigskip

Hint:

For pulley problems, always combine translational motion of masses with rotational motion of the pulley.

Answer 1

The Correct Option is B

Step 1: Finding acceleration using kinematics.
\[ s = \frac{1}{2}at^2 \Rightarrow 0.81 = \frac{1}{2}a(9)^2 \] \[ a = 0.02\,\text{m s}^{-2} \] Step 2: Writing equations of motion for masses.
\[ m_1 g - T_1 = m_1 a \] \[ T_2 - m_2 g = m_2 a \] Step 3: Torque equation for pulley.
\[ (T_1 - T_2)r = I\alpha \] \[ \alpha = \dfrac{a}{r} \] Step 4: Substituting values and solving.
After substituting all numerical values,
\[ I = 1.86 \times 10^{-2}\,\text{kg m}^2 \] Step 5: Final conclusion.
The rotational inertia of the pulley is $1.86 \times 10^{-2}\,\text{kg m}^2$.