Question

A bucket of mass 8 kg is supported by a light rope wound around a solid wooden cylinder of mass 16 kg and radius 20cm free to rotate about its Axis. initially the bucket and the cylinder are at rest and a man holds the free end of the rope .He now releases the rope and let the bucket for freely downward in a well 50m deep. neglecting friction, obtain the speed of the bucket and as the angular speed of the cylinder just before the bucket enters water .

[ take g = 10 m/s²]

Answer 1

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Answer 2

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mass of bucket ,m1 = 8 kg

mass of cylinder ,m2 = 16kg

radius of cylinder , r= 20 cm = 0.2 m

when the bucket falls down ,its potential energy is converted into the kinetic energy of the bucket and the rotational kinetic energy of the cylinder.

Loss of P.E= Translational k.E of bucket + rotational k.E

$m1gh = \frac{1}{2} m1v {}^{2} + \frac{1}{2} iw {}^{2}$

$m1gh = \frac{1}{2} m1v {}^{2} + \frac{1}{2} ( \frac{1}{2} m2r {}^{2}) \times \frac{v {}^{2} }{r {}^{2} }$

$m1gh = \frac{1}{2} v {}^{2} (m1 + \frac{1}{2} m2)$

put the given values;

$8 \times 10 \times 50 = \frac{1}{2} \times v {}^{2} (8 + \frac{1}{2} \times 16)$

$8 \times 10 \times 50 = \frac{1}{2} \times v {}^{2} \times 16$

$10 \times 50 = v {}^{2}$

$v = \sqrt{100 \times 5} \: \: ms {}^{ - 1}$

$v = 2.23 \times 10ms {}^{ - 1}$

speed of bucket just before the bucket touches the water

$= 22.3ms {}^{ - 1}$

the angular speed of the centre before the bucket touches the water, w = v/r

$= \frac{22.3}{0.2} = 111.5s {}^{ - 1}$