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Q7.23(a):
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 $7.6 kg m^2$. (a) What is his new angular speed? (Neglect friction.)

Solution :

Step 1: Identifying Given Parameters and Constants

To determine the new angular speed, we define the initial and final states of the system based on the Principle of Conservation of Angular Momentum [Per Newton's Second Law for Rotation, $\tau_{ext} = \frac{dL}{dt}$; since friction is neglected, $\tau_{ext} = 0$, implying $L_i = L_f$].

  • Initial angular speed ($\omega_i$): $30 \text{ rpm} = 30 \times \frac{2\pi}{60} \text{ rad/s} = \pi \text{ rad/s} \approx 3.14159 \text{ rad/s}$.
  • Mass of each weight ($m$): $5 \text{ kg}$. Total mass of weights $= 10 \text{ kg}$.
  • Initial distance ($r_1$): $90 \text{ cm} = 0.9 \text{ m}$.
  • Final distance ($r_2$): $20 \text{ cm} = 0.2 \text{ m}$.
  • Moment of inertia of man + platform ($I_p$): $7.6 \text{ kg m}^2$.

Step 2: Calculating Moments of Inertia

The total moment of inertia ($I$) is the sum of the moment of inertia of the man/platform and the point-mass weights ($I = I_p + \sum mr^2$).

Initial Moment of Inertia ($I_i$):

$I_i = I_p + 2(mr_1^2)$ $I_i = 7.6 + 2(5 \times 0.9^2)$ $I_i = 7.6 + 2(5 \times 0.81) = 7.6 + 8.1 = 15.7 \text{ kg m}^2$

Final Moment of Inertia ($I_f$):

$I_f = I_p + 2(mr_2^2)$ $I_f = 7.6 + 2(5 \times 0.2^2)$ $I_f = 7.6 + 2(5 \times 0.04) = 7.6 + 0.4 = 8.0 \text{ kg m}^2$

Step 3: Applying Conservation of Angular Momentum

Since the net external torque is zero, the angular momentum remains constant [Law of Conservation of Angular Momentum, $L = I\omega$]:

$I_i \omega_i = I_f \omega_f$ $15.7 \times \pi = 8.0 \times \omega_f$

Step 4: Solving for New Angular Speed ($\omega_f$)

$\omega_f = \frac{15.7 \times \pi}{8.0}$ $\omega_f = 1.9625 \times \pi \text{ rad/s}$

Converting back to revolutions per minute (rpm):

$\text{New speed in rpm} = \frac{\omega_f \times 60}{2\pi}$ $\text{New speed in rpm} = \frac{1.9625 \times \pi \times 60}{2\pi} = 1.9625 \times 30 = 58.875 \text{ rpm}$ Axis r=0.9m (Initial) r=0.2m (Final) Rotation ω

Final Solution: The new angular speed is $58.875 \text{ rpm}$ (or approximately $6.16 \text{ rad/s}$).


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