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Question
The descending pulley shown in the following figure has a radius 20 cm and moment of inertia 0⋅20 kg-m2. The fixed pulley is light and the horizontal plane frictionless. Find the acceleration of the block if its mass is 1⋅0 kg.
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Solution
Let the mass of block be m1 and mass of pulley be m2.
Acceleration of the massive pulley will be half of that of the block.
From the free body diagram, we have
\[T_1 = m_1 a..........(1)\]
\[\left( T_2 - T_1 \right) r = I\alpha\]
\[ T_2 - T_1 = \frac{Ia}{2 r^2} = \frac{5a}{2}..........(2)\]
\[ m_2 g - m_2 \frac{a}{2} = T_1 + T_2............(3)\]
Putting the value of mass in equation (1) and using equation (1) in equation (2), we get
\[T_1 = a\text{ and }T_2 = \frac{7}{2}a\]
\[m_2 g = m_2 \frac{a}{2} + \frac{7}{2}a + a\]
On replacing the value of \[m_2 using\frac{1}{2}m r^2 = I,\] we get
\[\frac{2I}{r^2}g = \frac{2I}{r^2}\frac{a}{2} + \frac{9}{2}a\]
\[ \Rightarrow 98 = 5a + 4 . 5a\]
\[ \Rightarrow a = \frac{98}{9 . 5} = 10 . 3 m/ s^2\]
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