Two masses m and 2m are connected by a light string going over a pulley (disc) of mass 30m with radius r=0.1 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

Question

Two masses m and 2m are connected by a light string going over a pulley (disc) of mass 30m with radius r=0.1 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 m is __________ m/s. (Assume string does not slip and $g = 10 {m/s}^{2}$ ) [2026]

Smart Achievers · Accepted Answer

(2)

Using energy conservation

$\frac{1}{2} m v^{2} + \frac{1}{2} (2 m) v^{2} + \frac{1}{2} (\frac{30 m R^{2}}{2}) \frac{v^{2}}{R^{2}} = m g h$

$9 m v^{2} = m g h$

$v = \sqrt{\frac{g h}{9}} = \sqrt{\frac{10 \times 3.6}{9}}$

$v = \sqrt{4} = 2 m/s$

Solution

Ans.
(2)

Using energy conservation

$\frac{1}{2} m v^{2} + \frac{1}{2} (2 m) v^{2} + \frac{1}{2} (\frac{30 m R^{2}}{2}) \frac{v^{2}}{R^{2}} = m g h$

$9 m v^{2} = m g h$

$v = \sqrt{\frac{g h}{9}} = \sqrt{\frac{10 \times 3.6}{9}}$

$v = \sqrt{4} = 2 m/s$

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Solution

Ans. (2) Using energy conservation 12mv2+12(2m)v2+12(30mR22)v2R2=mgh 9mv2=mgh v=gh9=10×3.69 v=4=2 m/s

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Ans.
(2)

Using energy conservation

$\frac{1}{2} m v^{2} + \frac{1}{2} (2 m) v^{2} + \frac{1}{2} (\frac{30 m R^{2}}{2}) \frac{v^{2}}{R^{2}} = m g h$

$9 m v^{2} = m g h$

$v = \sqrt{\frac{g h}{9}} = \sqrt{\frac{10 \times 3.6}{9}}$

$v = \sqrt{4} = 2 m/s$