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

Step 1: Understanding the Question:
The question asks for the moment of inertia (\(I\)) of a standard geometric shape (a uniform disc) rotating about its most common axis: the transverse axis passing through the geometric center.
Step 2: Key Formula or Approach:
The moment of inertia for any rigid body is calculated using \(I = \int r^2 \, dm\).
For a uniform disc of mass \(M\) and radius \(R\), when the axis is perpendicular to its plane and passes through the center, the standard formula is derived by considering the disc as a collection of concentric rings.
Step 3: Detailed Explanation:
1. Consider a disc of mass \(M\) and radius \(R\).
2. The mass per unit area (surface density) is \(\sigma = \frac{M}{\pi R^2}\).
3. For an infinitesimal ring of radius \(r\) and thickness \(dr\), the mass is \(dm = \sigma(2\pi r dr)\).
4. Integrating the moment of inertia of these rings (\(r^2 dm\)) from \(0\) to \(R\):
\[ I = \int_{0}^{R} r^2 \cdot \left( \frac{M}{\pi R^2} \cdot 2\pi r \, dr \right) \]
\[ I = \frac{2M}{R^2} \int_{0}^{R} r^3 \, dr \]
\[ I = \frac{2M}{R^2} \left[ \frac{r^4}{4} \right]_{0}^{R} = \frac{1}{2}MR^2 \]
Step 4: Final Answer:
The moment of inertia of the disc about the central perpendicular axis is \(\frac{1}{2}MR^2\).