Question:
A uniform disc of mass $4\ \text{kg}$ has radius $0.4\ \text{m}$. Its moment of inertia about an axis passing through a point on its circumference and perpendicular to its plane is
A uniform disc of mass $4\ \text{kg}$ has radius $0.4\ \text{m}$. Its moment of inertia about an axis passing through a point on its circumference and perpendicular to its plane is
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Use parallel axis theorem when axis does not pass through centre of mass.
Updated On: Feb 4, 2026
- $0.16\ \text{kg-m}^2$
- $0.32\ \text{kg-m}^2$
- $0.64\ \text{kg-m}^2$
- $0.96\ \text{kg-m}^2$
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The Correct Option is D
Solution and Explanation
Step 1: Moment of inertia of disc about central axis.
\[ I_{\text{centre}} = \dfrac{1}{2}MR^2 \] Step 2: Apply parallel axis theorem.
For an axis through circumference: \[ I = I_{\text{centre}} + MR^2 \] Step 3: Substitute given values.
\[ I = \dfrac{1}{2}(4)(0.4)^2 + 4(0.4)^2 \] Step 4: Simplify.
\[ I = 0.32 + 0.64 = 0.96\ \text{kg-m}^2 \]
\[ I_{\text{centre}} = \dfrac{1}{2}MR^2 \] Step 2: Apply parallel axis theorem.
For an axis through circumference: \[ I = I_{\text{centre}} + MR^2 \] Step 3: Substitute given values.
\[ I = \dfrac{1}{2}(4)(0.4)^2 + 4(0.4)^2 \] Step 4: Simplify.
\[ I = 0.32 + 0.64 = 0.96\ \text{kg-m}^2 \]
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