I_(CM) is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc. I_(AB) is it's moment of inertia about an axis AB perpendicular to plane and parallel to axis CM at a distance 2/3R from center. Where R is the radius of the disc. The ratio of I_(AB) and I_(CM) is x:9. The value of x is _____________.

Solution

Physics sir · 4 steps

Axis AB is just CM shifted sideways by 2R/3, so add Md² to the centre value and compare.

1. Centre value. For a uniform disc about the axis through its centre, perpendicular to its plane, I_(CM) = 1/2MR².
2. Shift the axis. AB is parallel to CM and d = 2/3R away from it. The parallel axis theorem applies because CM passes through the centre of mass.

   I_(AB) = I_(CM) + Md² = 1/2MR² + M(2R/3)² = 9/18MR² + 8/18MR² = 17/18MR²

3. Take the ratio. I_(AB)/I_(CM) = (17/18)/(1/2) = 17/9.
4. So. The ratio is 17:9, so x = 17.

The common slip: 21. Forgetting to square the distance gives I_(AB) = (1/2+2/3)MR² = 7/6MR². That makes the ratio 21:9, so x comes out as 21.

The idea to keep. The parallel axis theorem only works from an axis through the centre of mass. Start from the CM value, add Md² with the distance squared, then take ratios so M and R cancel.

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