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.

Answer 1

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

\[\frac{2I}{r^2}g = \frac{2I}{r^2}\frac{a}{2} + \frac{9}{2}a\]

\[ \Rightarrow 98 = 5a + 4 . 5a\]

\[ \Rightarrow a = \frac{98}{9 . 5} = 10 . 3  m/ s^2\]