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