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Question
A string is wrapped on a wheel of moment of inertia 0⋅20 kg-m2 and radius 10 cm and goes through a light pulley to support a block of mass 2⋅0 kg as shown in the following figure. Find the acceleration of the block.
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Solution
Moment of inertia of the bigger pulley, I = 0.20 kg-m2
r = 10 cm = 0.1 m,
Smaller pulley is light. Therefore, on neglecting its moment of inertia, we have
Mass of the block, m = 2 kg
From the free body diagram, we get
\[mg - T = ma........(1)\]
\[Tr = I\alpha\] And
\[a = \alpha r\]
\[\Rightarrow T = \frac{Ia}{r^2} ........(2)\]
Using equations (1) and (2), we get
\[mg = \left( m + \frac{I}{r^2} \right) a\]
\[ \Rightarrow a = \frac{mg}{m + \frac{I}{r^2}}\]
\[= \frac{2 \times 9 . 8}{2 + \left( \frac{0 . 2}{0 . 01} \right)}\]
\[= \frac{19 . 6}{22} = 0 . 89 m/s^2\]
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