__Question 7.23__:^{2}.

(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__:**(a)**Moment of inertia of the man-platform system = 7.6 kg m

^{2}

Moment of inertia when the man stretches his hands to a distance of 90 cm:

2 ×

*m*

*r*

^{2}

= 2 × 5 × (0.9)

^{2}

= 8.1 kg m

^{2}

Initial moment of inertia of the system, I

_{i}= 7.6 + 8.1 = 15.7 kg m

^{2}

Angular speed, ω

_{i}= 300 rev/min

Angular momentum, L

_{i}= I

_{i}ω

_{i}= 15.7 X 30 ....(i)

Moment of inertia when the man folds his hands to a distance of 20 cm:

2 ×

*mr*

^{2}

= 2 × 5 (0.2)

^{2}= 0.4 kg m

^{2}

Final moment of inertia, I

_{f}= 7.6 + 0.4 = 8 kg m

^{2}

Final angular speed = ω

_{f}

Final angular momentum, L

_{f}= I

_{f}ω

_{f}= 0.79 ω

_{f}...... (ii)

From the conservation of angular momentum, we have:

I

_{i}ω

_{i}= I

_{f}ω

_{f}

∴ ω

_{f}= 15.7 X 30 / 8 = 58.88 rev/min

**(b) No**

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.

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