- #1

Rongeet Banerjee

- 45

- 6

- Homework Statement
- Three particles each of mass m gram,are situated at the vertices of an equilateral triangle ABC of side l cm(as shown in the figure).The moment of inertia of the system about a line AX perpendicular to AB and in the plane of ABC in gcm² units will be:

1. (3/4)ml²

2. 2ml²

3. (5/4)ml²

- Relevant Equations
- Moment of Inertia I=summation

of(MR²)

If I take the three masses individually and try to calculate the moment of inertia of the system separately then

I=(m*0²)+(m*(l/2)²)+(m*l²)

=ml²/4 +ml²=(5/4)ml²

But If I try to calculate Moment of Inertia of the system using its Centre of mass then

As centre of mass is located at the the centroid,

I=3m*(l/2)²=(3/4)ml²

Acc to my book option 3 is correct.Why is my second method wrong?