Question

The torque required to increase the angular speed of a uniform solid disc of mass 10 kg and diameter 0.5 m from zero to 120 rotations per minute in 5 sec is:

• A. \( \frac{\pi}{4} \, {Nm} \)
• B. \( \pi \, {Nm} \)
• C. \( \frac{\pi}{2} \, {Nm} \)
• D. \( \frac{\pi}{3} \, {Nm} \)
• E. \( \frac{3\pi}{4} \, {Nm} \)

Hint:

To calculate torque, use \( \tau = I \alpha \) where \( I \) is the moment of inertia and \( \alpha \) is the angular acceleration.

Answer 1

The Correct Option is A

Step 1: Moment of Inertia of a Uniform Solid Disc The moment of inertia \( I \) of a uniform solid disc about its central axis is given by: \[ I = \frac{1}{2} M R^2 \] where \( M = 10 \) kg (mass of the disc), \( R = \frac{d}{2} = \frac{0.5}{2} = 0.25 \) m (radius of the disc). Substituting the values, \[ I = \frac{1}{2} \times 10 \times (0.25)^2 \] \[ I = \frac{1}{2} \times 10 \times 0.0625 \] \[ I = \frac{10 \times 0.0625}{2} = 0.3125 { kg m}^2 \] Step 2: Angular Acceleration Calculation The angular acceleration \( \alpha \) is given by: \[ \alpha = \frac{\omega_f - \omega_i}{t} \] Given: Initial angular velocity, \( \omega_i = 0 \), Final angular velocity, \( \omega_f = 120 \) rpm \( = 120 \times \frac{2\pi}{60} = 4\pi \) rad/s, Time, \( t = 5 \) s. Thus, \[ \alpha = \frac{4\pi - 0}{5} \] \[ \alpha = \frac{4\pi}{5} { rad/s}^2 \] Step 3: Torque Calculation Torque \( \tau \) is given by: \[ \tau = I \alpha \] Substituting the values, \[ \tau = (0.3125) \times \left(\frac{4\pi}{5}\right) \] \[ \tau = \frac{0.3125 \times 4\pi}{5} \] \[ \tau = \frac{1.25\pi}{5} = \frac{\pi}{4} { Nm} \] Thus, the required torque is \( \frac{\pi}{4} \) Nm.