Question

Find the moment of inertia of a sphere about a tangent to the sphere, given the M.I. of the sphere about any of its diameter to be 2 MR²/5, Where m is the mass of the sphere and R is the radius of the sphere.

(b) Given the M.I of disc of mass M and radius R about any of its diameters to the MR²/4, find its M.I. about an axis normal to the disc and passing through a point on its edge.

Answer 1

(A) Let I_AB be the M.I. of the given sphere about the diameter AB of radius R and mass m.

∴ I_AB = 2/5 MR²    ....(i)

It means the sphere is solid

Let CD be a tangent of the sphere parallel to the diameter AB of the sphere.

∴ Distance between the two parallel axes is R. If I_CD be is MI about CD axis, then according to the theorem of parallel axes,

I_CD = I_AB + MR² = 2/5 MR² + MR² = 7/5 MR²

(b) Here AB and CD are the diameter sof the disc of radius R and mass M. Let EF be an axis ⊥ ar to the plane of disc and passing through a point D on its edge.

Clearly the axis DG is parallel to the axis EF

∴ If I_EF be the M.I. of the disc about EF axis,

Then according to theorem of ⊥ ar axes.

I_EF = I_AB + I_CD = {MR²}/{4} + {MR²}/{4} = 1/2 MR²

Here ⊥ ar distance between EF and DG axes = R

∴ If I_DG be the M.I. of the disc about the required axes, then according to theorem of parallel axes.

I_DG = I_EF + MR² = 1/2 MR² + MR² = 3/2 MR²