Easy MCQ +4 / -1

The moment of inertia of a thin ring of mass $M$ and radius $R$ about an axis through its centre and perpendicular to its plane is:

  1. A $\frac{1}{2}MR^2$
  2. B $MR^2$ Correct answer
  3. C $\frac{2}{3}MR^2$
  4. D $\frac{2}{5}MR^2$

Solution

All the mass lies at distance $R$ from the axis, so $I = MR^2$.

About this question

Subject: Physics · Chapter: Rotational Motion · Topic: Moment of Inertia

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