Question

Two small spheres of masses 10kg and 30kg are joined by a rod of length 0.5m and of negligible mass. The moment of inertia about the normal axis of the system is

Answer 1

Answer:

1.875 kgm^2

Explanation:

Take a look at the given picture

Hope you find it easy

Answer 2

I=13.125 units

Explanation:

We have two spheres connected at the ends of the rod.It is given that the rod has no mass so no need to involve the moment of inertia of the rod.

Also it is given that the spheres has negligible mass so no need to consider the moment of inertias of the spheres seperatly.

So to start with this problem we initially find out the centre of mass of the rod ,

which is given by the formula

$l1=\frac{m2*l}{m1+m2}$

where l is the total length of the rod and l1 is the distance from m1.

Let us calculate the centre of mass from the given data by substituting,

$l1=\frac{30*0.5}{40}\\ \\l1=\frac{3}{8}$

$l2=\frac{5}{8}$

So the moment of inertia of mass m at a distance of r is given by the formula,

$I=m*r^{2}$

By substituting the obtained data we get,

$I=10*(\frac{3}{8} )^{2}+30*(\frac{5}{8} )^{2}  \\\\I=13.125 units$

Hence this is the moment of inertia of the system.