Question

A thin circular plate of mass M and radius R has its density varying as p (r) = pₒ
r with Pₒ as constant and r is the distance from its center. The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is I = aMR² .
The value of the coefficient a is:
(A) 8/5
(B) 1/2
(C) 3/5
(D) 3/2

Answer 1

The value of the coefficient a is equal to 3/5.

• Consider a ring of thickness dr at a distance r from the centre.
• So, the mass of this thin ring dm = p₀.r.2πr.dr
• Hence the mass of the disc M = ∫dm = ∫p₀.2πr²dr = 2πp₀R³/3
• The moment of inertia is defined as the product of the mass and the square of the distance from the axis.
• I = ∫p₀.r.2πr.dr.r² = ∫p₀.2πr⁴dr = 2πp₀R⁵/5
• The ratio of the moment of inertia to the mass of the disc is equal to 3R²/5.
• Hence, the moment of inertia of the disc is equal to 3MR²/5.
• Therefore, the value of the constant a is equal to 3/5.