Question

A uniform circular disc of mass $\dfrac{\pi}{40}$ kg is rotating about an axis passing through its center and perpendicular to its plane with an angular speed of 150 rev/min. If the angular momentum of the disc is 6.25 Js, then its radius is:

• A. 25 cm
• B. 50 cm
• C. 12.5 cm
• D. 100 cm

Hint:

Always convert rotational speed from rev/min to rad/s using $1$ rev = $2\pi$ radians. Angular momentum depends directly on both the moment of inertia and angular velocity. Ensure unit consistency between $L$, $I$, and $\omega$. For discs, $I = \frac{1}{2}MR^2$, while for rings, $I = MR^2$.

Answer 1

The Correct Option is B

• Given: $M = \dfrac{\pi}{40}$ kg, $\omega = 150$ rev/min $= \dfrac{150 \times 2\pi}{60} = 5\pi$ rad/s.
• Moment of inertia of a disc about its center is $I = \dfrac{1}{2}MR^2$.
• Angular momentum $L = I\omega = \dfrac{1}{2}MR^2\omega$.
Substituting the values: \[ 6.25 = \frac{1}{2} \times \frac{\pi}{40} \times R^2 \times 5\pi \] \[ R^2 = \frac{6.25 \times 80}{5\pi^2} = \frac{100}{\pi^2} \] \[ R = \frac{10}{\pi} \approx 3.18 \text{ m} = 50 \text{ cm} \] • Hence, the radius of the disc is 50 cm.