Question

After release, the blocks moves 81 cm in 9 seconds. Find moment of inertia of the pulley :
(Given $m_{1} = 400$ gm, $m_{2} = 350$ gm, R = 2 cm, g = 10 m/s$^2$)

• A. $97 \times 10^{-4}$ Kg-m$^2$
• B. $100 \times 10^{-4}$ Kg-m$^2$
• C. $21 \times 10^{-4}$ Kg-m$^2$
• D. $87 \times 10^{-4}$ Kg-m$^2$

Hint:

The equivalent mass of a pulley in translation is $\frac{I}{R^2}$. So, the standard Atwood's acceleration formula $a = \frac{Net Force}{Total Mass}$ easily modifies to $a = \frac{(m_1 - m_2)g}{m_1 + m_2 + I/R^2}$. Remembering this modified form saves time on drawing FBDs for all three components.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This is an Atwood's machine problem where the pulley has mass and moment of inertia. We are given the kinematics of the system (distance and time) from which we can find the acceleration. Then, using Newton's laws for rotation and translation, we can find the moment of inertia ($I$) of the pulley.
Step 2: Key Formula or Approach:
1. Kinematic equation to find acceleration: $S = ut + \frac{1}{2}at^2$.
2. Acceleration of Atwood's machine with a massive pulley: $a = \frac{(m_{1} - m_{2})g}{m_{1} + m_{2} + \frac{I}{R^2}}$.
Step 3: Detailed Explanation:
Given values:
$m_{1} = 400 \text{ gm} = 0.4 \text{ kg}$
$m_{2} = 350 \text{ gm} = 0.35 \text{ kg}$
Radius of pulley, $R = 2 \text{ cm} = 0.02 \text{ m}$
Distance travelled, $S = 81 \text{ cm} = 0.81 \text{ m}$
Time taken, $t = 9 \text{ s}$
Initial velocity, $u = 0$
Part 1: Find the acceleration (a)
Using the second equation of motion:
$S = ut + \frac{1}{2}at^2$
$0.81 = 0 + \frac{1}{2} \times a \times (9)^2$
$0.81 = \frac{81a}{2}$
$a = \frac{2 \times 0.81}{81} = 0.02 \text{ m/s}^2$
Part 2: Find the Moment of Inertia (I)
Using the dynamic equation for the system:
$a = \frac{(m_{1} - m_{2})g}{m_{1} + m_{2} + \frac{I}{R^2}}$
Substitute the known values (using $g = 10 \text{ m/s}^2$):
$0.02 = \frac{(0.4 - 0.35) \times 10}{0.4 + 0.35 + \frac{I}{R^2}}$
$0.02 = \frac{0.05 \times 10}{0.75 + \frac{I}{R^2}}$
$0.02 = \frac{0.5}{0.75 + \frac{I}{R^2}}$
$0.75 + \frac{I}{R^2} = \frac{0.5}{0.02} = 25$
$\frac{I}{R^2} = 25 - 0.75 = 24.25 \text{ kg}$
Now, calculate $I$:
$I = 24.25 \times R^2$
$I = 24.25 \times (0.02)^2$
$I = 24.25 \times 0.0004$
$I = 0.0097 \text{ Kg-m}^2 = 97 \times 10^{-4} \text{ Kg-m}^2$
Step 4: Final Answer:
The moment of inertia of the pulley is $97 \times 10^{-4}$ Kg-m$^2$.