Question

A ring of mass m and radius r rotates about an axis passing through its centre and perpendicular to its plane with angular velocity $\omega$. Its kinetic energy is

• A. $\frac{1}{2}mr^2\omega^2$
• B. mr$\omega^2$
• C. $mr^2\omega^2$
• D. $\frac{1}{2}mr\omega^2$

Answer 1

The Correct Option is A

To solve this problem, we need to find the kinetic energy of a rotating ring. The given parameters are the mass \(m\) of the ring, the radius \(r\), and the angular velocity \(\omega\).

The formula for the kinetic energy \(K\) of a rotating object is given by:

K = \frac{1}{2} I \omega^2

where:

• \(I\) is the moment of inertia of the object, and
• \(\omega\) is the angular velocity.

For a ring rotating about an axis through its center and perpendicular to its plane, the moment of inertia \(I\) is given by:

I = mr^2

Substituting \(I = mr^2\) into the kinetic energy formula, we have:

K = \frac{1}{2} (mr^2) \omega^2 = \frac{1}{2} mr^2 \omega^2

Thus, the kinetic energy of the ring is \frac{1}{2} mr^2 \omega^2, which matches the first option given.

Let's evaluate why the other options are incorrect:

• The option mr \omega^2 is incorrect because it does not take into account the moment of inertia of the ring.
• The option mr^2 \omega^2 omits the \(\frac{1}{2}\) factor, which arises naturally in the kinetic energy formula for rotation.
• The option \frac{1}{2} mr \omega^2 also incorrectly uses the term \(mr\) instead of \(mr^2\) for the moment of inertia.

Consequently, the correct answer is:

\frac{1}{2} mr^2 \omega^2