MHT CET · Physics · Rotational Motion
The moment of inertia of a ring about an axis perpendicular to its plane and passing through its center is \(4 \mathrm{~kg} \mathrm{~m}^2\). Its moment of inertia about the tangent in the plane is
- A \(6 \mathrm{~kg} \mathrm{~m}^2\)
- B \(8 \mathrm{~kg} \mathrm{~m}^2\)
- C \(4 \mathrm{~kg} \mathrm{~m}^2\)
- D \(2 \mathrm{~kg} \mathrm{~m}^2\)
Answer & Solution
Correct Answer
(A) \(6 \mathrm{~kg} \mathrm{~m}^2\)
Step-by-step Solution
Detailed explanation
Ring has moment of inertia \(M R^2\) about the symmetric central axis. Using perpendicular axis theorem one can get moment of inertia about the planar diagonal as \(\frac{M R^2}{2}\)
Now, using to the parallel axis theorem, we shift by a distance R to be at the tangential position:
\(I_t=\frac{1}{2} M R^2+M R^2=\frac{3}{2} M R^2\)
Given \(M R^2=4 \mathrm{kgm}^2\) :
\(I_t=\frac{3}{2} \times 4 \mathrm{kgm}^2=6 \mathrm{kgm}^2\)
Now, using to the parallel axis theorem, we shift by a distance R to be at the tangential position:
\(I_t=\frac{1}{2} M R^2+M R^2=\frac{3}{2} M R^2\)
Given \(M R^2=4 \mathrm{kgm}^2\) :
\(I_t=\frac{3}{2} \times 4 \mathrm{kgm}^2=6 \mathrm{kgm}^2\)
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