Question

A solid disc of radius \(40\ \text{cm}\) and the mass of \(20\ \text{kg}\) is rotating with an angular velocity of \(300\ \text{rpm}\), about an axis passing through its centre of mass and perpendicular to its plane. The magnitude of retarding torque required to bring the disc at rest in \(20\ \text{s}\) is \(x\pi \ \text{Nm}\). The value of \(x\) is

• A. \(0.8\)
• B. \(0.6\)
• C. \(0.4\)
• D. \(0.2\)

Hint:

Find the moment of inertia of the disc, convert rpm to rad/s, get the angular deceleration, and multiply.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The retarding torque equals the moment of inertia times the angular deceleration, \(\tau=I\alpha\).

Step 2: Moment of inertia
\(I=\dfrac{1}{2}MR^{2}=\dfrac{1}{2}\times20\times(0.4)^{2}=1.6\ \text{kg m}^{2}\)

Step 3: Angular quantities
\(\omega_0=300\ \text{rpm}=10\pi\ \text{rad/s}\). Coming to rest in \(20\ \text{s}\): \(\alpha=\dfrac{10\pi}{20}=\dfrac{\pi}{2}\ \text{rad/s}^{2}\).

Step 4: Torque
\(\tau=1.6\times\dfrac{\pi}{2}=0.8\pi\ \text{Nm}\), so \(x=0.8\).

Final Answer:
The value of x is 0.8. \[ \boxed{\text{(A) }0.8} \]