Question

A flywheel of mass 2 kg has radius of gyration 0.5 m. If it makes 10 r.p.s., then its rotational kinetic energy will be

• A. \( 100\pi^2 \, \text{J} \)
• B. \( 50\pi^2 \, \text{J} \)
• C. \( 100\pi^2 \, \text{erg} \)
• D. \( 50\pi^2 \, \text{erg} \)

Hint:

The rotational kinetic energy depends on the moment of inertia and the square of the angular velocity. Use \( \omega = 2\pi \times \text{r.p.s.} \) to calculate angular velocity.

Answer 1

The Correct Option is A

Step 1: Formula for rotational kinetic energy.
The rotational kinetic energy \( K_E \) of a rotating body is given by: \[ K_E = \frac{1}{2} I \omega^2 \] where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. The moment of inertia for a flywheel is given by: \[ I = m k^2 \] where \( m = 2 \, \text{kg} \) is the mass of the flywheel and \( k = 0.5 \, \text{m} \) is the radius of gyration.
Step 2: Calculating the energy.
The angular velocity \( \omega \) in terms of the number of revolutions per second is: \[ \omega = 2\pi \times 10 = 20\pi \, \text{rad/s} \] Substitute the values of \( I \) and \( \omega \) into the formula for \( K_E \): \[ K_E = \frac{1}{2} \times 2 \times (0.5)^2 \times (20\pi)^2 = 100\pi^2 \, \text{J} \] Step 3: Conclusion.
Thus, the rotational kinetic energy is \( 100\pi^2 \, \text{J} \), corresponding to option (A).