Question

A thin uniform circular disc of mass \(\frac{10}{\pi^2}\) kg and radius 2 m is rotating about an axis passing through its centre and perpendicular to its plane. The work done to increase the angular speed of the disc from 90 rev/min to 120 rev/min is

• A. 35 J
• B. 70 J
• C. 140 J
• D. 210 J

Hint:

Always convert units to the standard SI system before calculation (e.g., rev/min to rad/s, g to kg). If your answer doesn't match any options, double-check for possible typos in the question's given values, especially with constants like \(\pi\).

Answer 1

The Correct Option is B

\textit{Note: The mass in the image appears to be \(\frac{10}{\pi}\), but this leads to an answer with \(\pi\). Assuming a typo and using \(M = \frac{10}{\pi^2}\) kg allows us to match the given options.}
Step 1: Understanding the Concept:
The work done on a rotating object is equal to the change in its rotational kinetic energy. This is the Work-Energy Theorem for rotation. We need to calculate the initial and final rotational kinetic energies.
Step 2: Key Formula or Approach:
Work Done, \(W = \Delta K_{rot} = K_{f} - K_{i} = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2 = \frac{1}{2}I(\omega_f^2 - \omega_i^2)\)
Moment of inertia of a disc about its central axis: \(I = \frac{1}{2}MR^2\)
Step 3: Calculations:
Convert Angular Speeds to SI units (rad/s):
The conversion factor is: 1 rev/min = \(\frac{2\pi \text{ rad}}{60 \text{ s}} = \frac{\pi}{30}\) rad/s.
Initial angular speed, \(\omega_i = 90 \text{ rev/min} = 90 \times \frac{2\pi}{60} = 3\pi\) rad/s.
Final angular speed, \(\omega_f = 120 \text{ rev/min} = 120 \times \frac{2\pi}{60} = 4\pi\) rad/s.
Calculate Moment of Inertia (I):
Given: \(M = \frac{10}{\pi^2}\) kg and \(R = 2\) m.
\[ I = \frac{1}{2}MR^2 = \frac{1}{2} \left( \frac{10}{\pi^2} \right) (2)^2 = \frac{1}{2} \times \frac{10}{\pi^2} \times 4 = \frac{20}{\pi^2} \text{ kg m}^2 \] Calculate Work Done (W):
Now, substitute the values of I, \(\omega_i\), and \(\omega_f\) into the work-energy theorem.
\[ W = \frac{1}{2}I(\omega_f^2 - \omega_i^2) = \frac{1}{2} \left( \frac{20}{\pi^2} \right) \left( (4\pi)^2 - (3\pi)^2 \right) \] \[ W = \frac{10}{\pi^2} (16\pi^2 - 9\pi^2) \] \[ W = \frac{10}{\pi^2} (7\pi^2) \] The \(\pi^2\) terms cancel out.
\[ W = 10 \times 7 = 70 \text{ J} \] Step 4: Final Answer:
The work done to increase the angular speed is 70 J. Therefore, option (B) is correct.