Question

Two blocks of masses 400 g and 200 g are connected through a light string going over a pulley which is free to rotate about its axis. The pulley has a moment of inertia \[1 \cdot 6 \times  {10}^{- 4}   kg -  m^2\] and a radius 2⋅0 cm, Find (a) the kinetic energy of the system as the 400 g block falls through 50 cm, (b) the speed of the blocks at this instant.

Answer 1

From the free body diagram, we have

\[0 . 4g -  T_1  = 0 . 4a  ..........(1)\]

\[ T_2  - 0 . 2g = 0 . 2a  ..........(2)\]

\[\left( T_1 - T_2 \right)  r = \frac{la}{r}  ...........(3)\]

On solving the above equations, we get

\[a = \frac{\left( 0 . 4 - 0 . 2 \right)  g}{\left( 0 . 4 + 0 . 2 + \frac{1 . 6}{0 . 4} \right)} = \frac{g}{5}\]

On solving the (b) part of the question first, we have

Speed of the blocks = \[v = \sqrt{\left( 2ah \right)} = \sqrt{2 \times \frac{g}{5} \times 0 . 5} = \sqrt{\left( \frac{9 . 8}{5} \right)} = 1 . 4  m/s\]

(a) Total kinetic energy of the system

\[= \frac{1}{2} m_1  v^2  + \frac{1}{2} m_2  v^2  + \frac{1}{2}I \omega^2 \]

\[ = \frac{1}{2} m_1  v^2  + \frac{1}{2} m_2  v^2  + \frac{1}{2}I \left( \frac{v}{r} \right)^2 \]

\[ = \left( \frac{1}{2} \times 0 . 4 \times \left( 1 . 4 \right)^2 \right) + \left( \frac{1}{2} \times 0 . 2 \times \left( 1 . 4 \right)^2 \right) + \left( \frac{1}{2} \times 1 . 6 \times {10}^{- 4} \times \left( \frac{1 . 4}{2 \times {10}^{- 2}} \right)^2 \right)\]

\[ = \left( 0 . 2 + 0 . 1 + 0 . 2 \right) \left( 1 . 4 \right)^2 \]

\[ = 0 . 5 \times 1 . 96 = 0 . 98 \text{ joule}\]