Question

A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90cm to 20cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m2. (a) What is his new angular speed? (Neglect friction.) (b) Is kinetic energy conserved in the process? If not, from where does the change come about? Question 7.23 System Of Particles And Rotational Motion

Answer 1

Concept : - according to law of conservation of angular momentum, " when total external torque acting on system is zero. Then, angular momentum remains conserved.
e.g Iw = constant
Given, mass of each weight (m) = 5kg
Initial angular speed (w1) = 30rpm
r1 = 90 cm = 0.9 m
r2 = 20cm = 0.2 m
Moment of inertia of man and platform = 7.6 kgm²
Moment of inertia of two given, weight (I1) = mr1² + mr1²
= 2mr1²
= 2× 5 × (0.9)² = 8.1 kgm² .

Moment of inertia of this two weight when man close his arms
(I2) = mr2² + mr2²
= 2mr2²
= 2×5 × (0.2)² = 0.4 kgm²
By law of conservation of angular momentum,
I'1w1 = I'2w2
(I + I1)w1 = (I+I2)w2
(7.6+ 8.1)×30 = (7.6+0.4)×w2
w2 = 15.7×30/8 = 58.88 rpm

Here, I'1w1 = I'2w2
Take square both sides,
I'1²w1² = I'2²w2²
Multiply with (1/2) both sides,
1/2×I'1(I'w1²) = 1/2×I'2(I'2w²)
1/2×I'1w1²/1/2×I'2w2² = I2/I1
KE1/KE2 = I2/I1
From above , it is clear that
I'1 > I'2
So, I1/I2 > 1
KE1/KE2 <1
Hence, kinetic energy is not conserved in this process.as when moment of inertia decreases , the rotational KE increases . the change in KE is obtained from the workdone by man in bringing his arms closer to his body