A uniformly thick wheel with moment of inertia $I$

$\frac{1}{R} \sqrt{\frac{2 (m_{1} - m_{2}) g h}{(m_{1} + m_{2}) + \frac{I}{R^{2}}}}$

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$\frac{g h}{R} \sqrt{\frac{(m_{1} + m_{2})}{(m_{1} + m_{2}) + \frac{I}{R^{4}}}}$

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$\frac{g h}{R} \sqrt{\frac{(m_{1} - m_{2})}{(m_{1} + m_{2}) + \frac{I}{R}}}$

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$\frac{1}{R} \sqrt{\frac{2 (m_{1} + m_{2}) g h}{(m_{1} + m_{2}) + \frac{I}{R^{2}}}}$

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Solution

The correct option is A
$\frac{1}{R} \sqrt{\frac{2 (m_{1} - m_{2}) g h}{(m_{1} + m_{2}) + \frac{I}{R^{2}}}}$

Step 1. Assume, since the blocks are at rest initial potential energy and initial kinetic energy will be zero.
Given,
$I =$ Inertia
$R =$ Radius
$m_{1} =$ mass of block $1$
$m_{2} =$ mass of block $2 (w h e r e, m_{1} > m_{2})$
$h =$ height
Step 2. Calculate the angular speed of the wheel,
Final P.E. $= m_{2} g h - m_{1} g h$
If the final speed of the blocks is $v$ and the angular velocity of the pulley is $ω$ then,
Final K.E $= \frac{1}{2} (m_{1} + m_{2}) v^{2} + \frac{1}{2} I ω^{2}$
The total energy is conserved. Hence,
$E = P . E . + K . E . 0 = m_{2} g h - m_{1} g h + \frac{1}{2} (m_{1} + m_{2}) v^{2} + \frac{1}{2} I ω^{2}$
$v = ω R$
Step 3. Due to no-slip condition
$\Rightarrow \frac{1}{2} (m_{1} + m_{2}) ω^{2} R^{2} + \frac{1}{2} I ω^{2} = (m_{1} - m_{2}) g h \Rightarrow ω^{2} [\frac{1}{2} (m_{1} + m_{2}) R^{2} + \frac{1}{2} I] = (m_{1} - m_{2}) g h \Rightarrow ω^{2} = \frac{[2 (m_{1} - m_{2}) g h]}{R^{2} [(m_{1} + m_{2}) + \frac{I}{R^{2}}]} \Rightarrow ω = \frac{1}{R} \sqrt{\frac{[2 (m_{1} - m_{2}) g h]}{[(m_{1} + m_{2}) + \frac{I}{R^{2}}]}}$
Hence, option A is correct.

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A uniformly thick wheel with moment of inertia I

A uniformly thick wheel with moment of inertia $I$