Question

Four point masses each of value M are placed at the corners of square of side l.The moment of inertia of the system about Axis passing through A and parallel to BD is

a)3ml^2. b)ml^2
c)2ml^2. c)√3ml^2

Answer 1

a should be the ans.

Answer 2

Answer:

(C) 2ml^2

Explanation:

Given,

• Mass of each four particles = M
• Length of the side of the square = l

Moment of inertia of the system of particles rotating about the same axis of rotation is equal to the algebraic sum of the individual moment of inertia of each particle.  

The rotational axis is passing through A and parallel to BD, this means the axis is about to AD. therefore the moment of inertia of particles at the place A and D will be zero.

Let I be the total moment of inertia of the system of particle.

$\therefore I\ =\ I_a\ +\ I_b\ +\ I_c\ +\ I_d\\\Rightarrow I\ =\ 0\ +\ ml^2\ +\ ml^2\ +\ 0\\\Rightarrow I\ =\ 2ml^2$

Hence the moment of inertial of the system about the axis passing through A and parallel to BD is $2ml^2.$