Question

A solid metal ring and a disc of same radius and mass are rotating about their diameters with same angular frequency. The ratio of their respective rotational kinetic energy values is

• A. 1:1
• B. 1:2
• C. 2:1
• D. 1:4
• E. 4:1

Answer 1

The Correct Option is C

We are given a solid metal ring and a disc of the same radius and mass, rotating about their diameters with the same angular frequency. We need to find the ratio of their respective rotational kinetic energy values.

The formula for rotational kinetic energy is: \[ E_{\text{rot}} = \frac{1}{2} I \omega^2 \] Where: - \( I \) is the moment of inertia, - \( \omega \) is the angular velocity (or angular frequency). Since both the ring and the disc have the same mass and radius and are rotating with the same angular frequency, we need to compare their moments of inertia. 1. The moment of inertia of a solid disc about its diameter is: \[ I_{\text{disc}} = \frac{1}{2} m r^2 \] 2. The moment of inertia of a solid metal ring (thin ring) about its diameter is: \[ I_{\text{ring}} = m r^2 \] Here, \( m \) is the mass and \( r \) is the radius of both the ring and the disc. Now, the rotational kinetic energy of the disc is: \[ E_{\text{rot, disc}} = \frac{1}{2} I_{\text{disc}} \omega^2 = \frac{1}{2} \times \frac{1}{2} m r^2 \omega^2 = \frac{1}{4} m r^2 \omega^2 \] And the rotational kinetic energy of the ring is: \[ E_{\text{rot, ring}} = \frac{1}{2} I_{\text{ring}} \omega^2 = \frac{1}{2} \times m r^2 \omega^2 = \frac{1}{2} m r^2 \omega^2 \] Now, the ratio of their rotational kinetic energies is: \[ \frac{E_{\text{rot, ring}}}{E_{\text{rot, disc}}} = \frac{\frac{1}{2} m r^2 \omega^2}{\frac{1}{4} m r^2 \omega^2} = 2 \] So, the ratio of their rotational kinetic energies is 2:1.

Correct Answer:

Correct Answer: (C) 2:1