Question

A light rod of length l has two masses m1 and m2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod passing through the centre of mass will be what?

Answer 1

According to formula:

I = m₁r₁² + m₂r₂²...........mₓrₓ²

I = m₁r₁² + m₂r₂²

 = m₁(m₂l/m₁ + m₂)² + m₂(m₂l/m₁+m₂)²

 = m₁m₂(m₁+m₂)l² /(m₁ +m₂)²

 = m₁m₂l²/(m₁+m₂)

Moment of inertia is the tendency to change the angular acceleration.  It is given by the sum of the product of mass of each particle with the square of distance from axis of rotation.  

Answer 2

Hey dear,

◆ Answer -

I = m1.m2.r^2 / (m1+m2)

◆ Explaination-

Consider a light rod of length r with masses m1 & m2 at two ends.

Centre of mass is given by-

x = (m1x1+m2x2) / (m1+m2)

Here, x1=0 & x2=r, hence

x = m2.r / (m1+m2)

Now, let's calculate distance of C.M. from individual masses -

r1 = x = m2.r / (m1+m2)

r2 = r-x = m1.r / (m1+m2)

Moment of inertia is given by -

I = I1 + I2

I = m1.r1^2 + m2.r2^2

I = m1[m2.r/(m1+m2)]^2 + m2[m1.r/(m1+m2)]^2

I = (m1.m2^2.r^2 + m2.m1^2.r^2) / (m1+m2)^2

I = (m1+m2)(m1.m2.r^2) / (m1+m2)^2

I = m1.m2.r^2 / (m1+m2)

Therefore moment of inertia is m1.m2.r^2/(m1+m2).

Hope this helps...