Question

Mass per unit area of a circular disc of radius \( a \) depends on the distance r from its centre as \( \sigma \left( r \right) \) = A + Br . The moment of inertia of the disc about the axis, perpendicular to the plane and assing through its centre is:

• A. \( 2\pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} \right) \)
• B. \( \pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} \right) \)
• C. \( 2\pi {a^4}\left( {{{aA} \over 4} + {B \over 5}} \right) \)
• D. \( 2\pi {a^4}\left( {{A \over 4} + {B \over 5}} \right) \)

Hint:

Take a thin ring of radius r and width dr: \(dm=\sigma(r)\,2\pi r\,dr\) and \(dI=r^2dm\). Integrate from 0 to a.

Answer 1

The Correct Option is A

Step 1: Ring element: \(dm=(A+Br)(2\pi r\,dr)\).
Step 2: \(I=\int_0^a r^2\,dm=2\pi\int_0^a (A r^3+B r^4)\,dr\).
Step 3: \(I=2\pi\left[\dfrac{A a^4}{4}+\dfrac{B a^5}{5}\right]=2\pi a^4\left(\dfrac{A}{4}+\dfrac{aB}{5}\right)\).
Answer: (A) \(2\pi a^4\left(\dfrac{A}{4}+\dfrac{aB}{5}\right)\)