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?

Answer 1

Answer:

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 2

Answer:

Moment of inertia of man+platform system =7.6kgm

2

Moment of inertia of the weights =2×mr

2

=2×5×0.9

2

=8.1kgm

2

total initial MOI =7.6+8.1=15.7kgm

2

Initial angular momentum of system =I

i

ω

i

=15.7×30 ......(i)

Moment of inertia of weights when man brings arms close=2×mr

′2

=2×5×(0.2)

2

=0.4kgm

2

Thus the final moment of inertia =7.6+0.4=8.0kgm

2

Let the final angular momentum be ω

′

Then from conservation of angular momentum,

I

i

ω

i

=I

f

ω

′

⟹ω

′

=

8

15.7×30

=58.88rev/min

(b) Kinetic energy is not conserved in the given process. In fact, with the decrease in the moment of inertia, kinetic energy increases. The additional kinetic energy comes from the work done by the man to fold his hands toward himself.