The moment of inertia of a semicircular ring about an axis, passing through the center and perpendicular to the plane of ring, is 1/x MR², where R is the radius and M is the mass of the semicircular ring. The value of x will be __________.

Solution

Physics sir · 3 steps

Every bit of this semicircular ring sits at distance R from the axis, so I is just MR².

1. Look at distances. The axis passes through the centre and is perpendicular to the plane of the ring. Every small mass element dm on the semicircle is at the same distance R from this axis.

   I = ∫ r² dm = R² ∫ dm = MR²

2. Mass already halved. A full ring of mass M also has I = MR². Cutting it to a semicircle halves the mass, but here M is already the mass of the semicircle, and every element is still at distance R.
3. So. I = MR² = 1/1MR², so x = 1.

The common slip: 2. Halving a full ring's MR² gives 1/2MR², that is 2. But M is already the semicircle's own mass, so nothing should be halved.

The idea to keep. For an axis through the centre perpendicular to the plane, any arc of a ring has I = MR², with M its own mass. Only the distance of the mass from the axis matters.

Answer from the NTA final answer key. Solution by PrepWiser.