Question

M.I of a solid sphere of mass 10 kg and radius 0.5 m rotating about an axis passing through a point at a distance of 0.2 m from the centre of the sphere

Answer 1

Answer:

Explanation:

I=MR^2.

I=10×(0.5)^2

I=10×0.25

Thus I=2.5kg-m^2..

Hope this helps u!

Answer 2

Given:

Mass of the solid sphere = 10 kg

The radius of the solid sphere = 0.5m

Distance of axis of rotation from the center of sphere = 0.2 m

To find:

M.I of the solid sphere rotating about an axis passing through a point at a distance of 0.2 m from the center of the sphere.

Solution:

It is given that

the mass of the solid sphere (M) = 10 kg

The radius of the solid sphere (R) = 0.5 m

We know that the M.I about an axis parallel to the diameter is given by

M.I = $\frac{2}{5}MR^2 + Mx^2$

where x is the distance of the axis from the diameter, therefore

M.I = 0.4 x 10 x 0.25 + 10 x 0.04

M.I = 1 + 0.4

M.I = 1.4 kgm²

Therefore the M.I of the solid sphere will be 1.4 kgm².