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Q7.23(b):
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 $7.6 kg m^2$. (b) Is kinetic energy conserved in the process? If not, from where does the change come about?

Solution :

1. Given Variables and Initial Parameters

  • Mass of each weight ($m$): $5 \text{ kg}$
  • Number of weights: $2$
  • Initial distance from axis ($r_1$): $90 \text{ cm} = 0.9 \text{ m}$
  • Final distance from axis ($r_2$): $20 \text{ cm} = 0.2 \text{ m}$
  • Moment of inertia of man + platform ($I_p$): $7.6 \text{ kg m}^2$
  • Initial angular speed ($\omega_1$): $30 \text{ rpm} = 30 \times \frac{2\pi}{60} \text{ rad/s} = \pi \text{ rad/s}$

2. Calculating Moments of Inertia

The moment of inertia of the two weights is calculated as $I_w = 2 \times (m \times r^2)$.

Initial Moment of Inertia ($I_1$):

$I_1 = I_p + 2(m \times r_1^2)$ $I_1 = 7.6 + 2(5 \times 0.9^2) = 7.6 + 10(0.81) = 7.6 + 8.1 = 15.7 \text{ kg m}^2$ [By Parallel Axis Theorem / Definition of Point Mass MI]

Final Moment of Inertia ($I_2$):

$I_2 = I_p + 2(m \times r_2^2)$ $I_2 = 7.6 + 2(5 \times 0.2^2) = 7.6 + 10(0.04) = 7.6 + 0.4 = 8.0 \text{ kg m}^2$

3. Calculating Final Angular Velocity

Since there are no external torques acting on the system, the angular momentum is conserved [Per Law of Conservation of Angular Momentum].

$L_1 = L_2 \implies I_1\omega_1 = I_2\omega_2$ $\omega_2 = \frac{I_1 \omega_1}{I_2} = \frac{15.7 \times \pi}{8.0} = 1.9625\pi \text{ rad/s}$

4. Analyzing Kinetic Energy (K.E.)

The rotational kinetic energy is defined as $K = \frac{1}{2}I\omega^2$.

Initial Kinetic Energy ($K_1$):

$K_1 = \frac{1}{2} \times 15.7 \times (\pi)^2 \approx 0.5 \times 15.7 \times 9.87 \approx 77.48 \text{ J}$

Final Kinetic Energy ($K_2$):

$K_2 = \frac{1}{2} \times 8.0 \times (1.9625\pi)^2 = 4 \times (3.8514 \times 9.87) \approx 152.09 \text{ J}$

5. Comparative Analysis

State Moment of Inertia ($I$) Angular Velocity ($\omega$) Kinetic Energy ($K$)
Initial $15.7 \text{ kg m}^2$ $\pi \text{ rad/s}$ $77.48 \text{ J}$
Final $8.0 \text{ kg m}^2$ $1.96\pi \text{ rad/s}$ $152.09 \text{ J}$

Since $K_2 > K_1$, the kinetic energy is not conserved.

6. Source of Kinetic Energy Increase

The increase in kinetic energy ($\Delta K \approx 74.61 \text{ J}$) arises from the work done by the man in pulling his arms inward towards the axis of rotation against the centrifugal force. This internal work done by the human muscular system is converted into the rotational kinetic energy of the system.

Final Solution: Kinetic energy is not conserved; it increases because the man performs work on the system to move the weights against the centrifugal force, effectively transferring chemical energy into rotational kinetic energy.


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