MHT CET · Physics · Rotational Motion
Two discs A and B each of radius \(r\) and mass \(m\) are attached as shown to form a system. The moment of inertia of this system about an axis perpendicular to the plane of the discs and passing through the center of disc \(\mathrm{A}\) is
- A \(\frac{9}{2} m r^2\)
- B \(m r^2\)
- C \(2 m r^2\)
- D \(5 m r^2\)
Answer & Solution
Correct Answer
(D) \(5 m r^2\)
Step-by-step Solution
Detailed explanation
The moment of inertia of the disc about the centroidal axis is \(\frac{m r^2}{2}\)
The moment of inertia of the disc at A about the axis passing through B is given using the parallel axis theorem as \(\frac{m r^2}{2}+m(2 r)^2\)
Thus, the total moment of inertia of the whole system is given as \(\frac{m r^2}{2}+\frac{m r^2}{2}+m(2 r)^2=5 m r^2\)
The moment of inertia of the disc at A about the axis passing through B is given using the parallel axis theorem as \(\frac{m r^2}{2}+m(2 r)^2\)
Thus, the total moment of inertia of the whole system is given as \(\frac{m r^2}{2}+\frac{m r^2}{2}+m(2 r)^2=5 m r^2\)
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