Question

J
57. Two small spheres of mass 5kg and 15 kg are
joined by a rod of length 0.5m and of
negligible mass. The M.I. of the system about
an axis passing through centre of rod and
normal to it is​

Answer 1

Explanation:

at first find the distance where centre of mass lie by x=m1x+m2(0.5-x)/m1+m2

then MI passing through the centre is ml2/12

Answer 2

The moment of inertia of the system about an axis passing through center of rod and  normal to it is​ $1.25kg.m^2$

Explanation:

Given masses of two sphere, M = 5 kg and m =  15 kg.

The length of rod is  0.5 m

So the distance of each mass from center of the rod,

$r_{1} = r_{2} = \frac{0.5}{2} =0.25 m$

The moment of inertia of the system about a normal axis through center of rod,

$I =M\times (r_{1})^2+ m\times (r_{2})^2$

Substitute the given values, we get

$I=5kg \times(0.25 m)^2+15kg \times(0.25 m)^2$

$I=1.25kg.m^2$

Thus, the moment of inertia of the system about

an axis passing through center of rod and  normal to it is​ $1.25kg.m^2$

Topic: Moment of inertia

Learn More:

https://brainly.in/question/12906628