JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
Seven identical circular planar disks, each of mass \(M\) and radius \(R\) are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point \(P\) is:
- A \(\frac{{55}}{2}M{R^2}\)
- B \(\frac{{73}}{2}M{R^2}\)
- C \(\frac{{181}}{2}M{R^2}\)
- D \(\;\frac{{19}}{2}M{R^2}\)
Answer & Solution
Correct Answer
(C) \(\frac{{181}}{2}M{R^2}\)
Step-by-step Solution
Detailed explanation
Using parallel axes theorem, moment of inertia about \('O'\) \(\begin{array}{l} {I_0} = {I_{cm}} + m{d^2}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{7M{R^2}}}{2} + 6\left( {M \times {{\left( {2R} \right)}^2}} \right) = \frac{{55M{R^2}}}{2} \end{array}\) Again, moment of inertia…
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