Question

A uniform disc of mass \(M\) and radius \(R\) is rotating about an axis through its center and perpendicular to its plane. What is its moment of inertia?

• A. \(MR^2\)
• B. \(\frac{1}{4}MR^2\)
• C. \(\frac{1}{2}MR^2\)
• D. \(2MR^2\)

Hint:

Important standard moments of inertia to remember: \[ \text{Ring about center} = MR^2 \] \[ \text{Solid disc about center} = \frac{1}{2}MR^2 \] \[ \text{Solid sphere about center} = \frac{2}{5}MR^2 \] Memorizing these results saves time in rotational dynamics problems.

Answer 1

The Correct Option is C

Concept: The moment of inertia measures how difficult it is to change the rotational motion of a body about a given axis. It depends on how the mass of the body is distributed relative to the axis of rotation. For a continuous body, \[ I = \int r^2 \, dm \] where:
• \(r\) is the perpendicular distance of the mass element from the axis,
• \(dm\) is the small mass element. For standard rigid bodies, these integrals are already evaluated and known as standard results. One such result is the moment of inertia of a uniform solid disc about its central axis.

Step 1: Identify the body and the axis of rotation. The object is a uniform solid disc of: \[ \text{Mass} = M, \qquad \text{Radius} = R \] The axis of rotation passes:
• Through the center of the disc
• Perpendicular to its plane This is the standard central axis for a disc.

Step 2: Use the standard formula for a solid disc. The moment of inertia of a uniform disc about this axis is: \[ I = \frac{1}{2}MR^2 \]

Step 3: Select the correct option. Thus, \[ I = \frac{1}{2}MR^2 \] Hence the correct answer is \[ \boxed{(C)\ \frac{1}{2}MR^2} \]