Two masses m and 2 m are connected by a light string going over a pulley
Q. Two masses $m$ and $2m$ are connected by a light string going over a pulley (disc) of mass $30m$ with radius $r = 0.1\ \text{m}$. The pulley is mounted in a vertical plane and it is free to rotate about its axis. The $2m$ mass is released from rest and its speed when it has descended through a height of $3.6\ \text{m}$ is ____ m/s.

(Assume string does not slip and $g = 10\ \text{m/s}^2$)
Correct Answer: 2

Explanation

This question involves conservation of energy including translational and rotational kinetic energy, a key concept in JEE Main, JEE Advanced and IIT JEE.

Loss of gravitational potential energy of the system is

$$ \Delta U = (2m - m) g h = mgh $$

$$ \Delta U = m \times 10 \times 3.6 = 36m $$

This energy is converted into kinetic energy of both masses and rotational kinetic energy of the pulley.

Translational kinetic energy of the two masses is

$$ K_{\text{trans}} = \frac{1}{2} m v^2 + \frac{1}{2} (2m) v^2 = \frac{3}{2} m v^2 $$

Moment of inertia of the pulley (disc) is

$$ I = \frac{1}{2} (30m) r^2 = 15 m r^2 $$

Rotational kinetic energy of the pulley is

$$ K_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} (15 m r^2) \left(\frac{v}{r}\right)^2 = \frac{15}{2} m v^2 $$

Total kinetic energy is

$$ K_{\text{total}} = \frac{3}{2} m v^2 + \frac{15}{2} m v^2 = 9 m v^2 $$

Using energy conservation,

$$ 36m = 9 m v^2 $$

$$ v^2 = 4 $$

$$ v = 2\ \text{m/s} $$

Hence, the required speed is $2\ \text{m/s}$.

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