Systems of Particles and Rotational Motion

Question : 23

Total: 33

A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg 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 90 cm to 20 cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6kg 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?

Solution:

Here, Mass in each hand =5kg
Moment of inertia of the man together with the platform, I=7.6kg m2
Distance of the weight from the axis, r1=90cm=0.9m
Distance of the weight from the axis, r2=20cm=0.2m
Initial moment of inertia of man, platform and weights
I1=I+Mr12 =7.6+2×5×(0.9)2 =7.6+8.1=15.7kgm2
Final moment of inertia of man, platform and weights
I2=7.6+2×Mr2 =7.6+2×5×(0.2)2 =8.0kg m2
According to Principle of conservation of angular momentum,
I2ω2=I1ω1
ω2=

ω1=

15.7×30

8.0

=58.9‌rpm=59‌rpm
(b) No, kinetic energy is not conserved in the process. Infact, as moment of inertia decreases kinetic energy of rotation increases. This change in K.E. is due to the work done by the man in decreasing the MI of the body.

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