Question

(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2MR2/5, where M is the mass of the sphere and R is the radius of the sphere. (b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be MR2/4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Answer 1

(1) Given,
Moment of inertia of the sphere about its diametre = (2/5)mR²

Use, parallel axis theorem ,
Moment of inertia of the sphere about tangent = I + mR²
= (2/5)mR² + mR²
= (7/5)mR²

(B) given,
Moment of inertia of disc of mass m and radius R about any of its diametre = mR²/4

See the figure ,
Moment of inertia about diametre = Ix = Iy= (1/4)mR²

Use , perpendicular axis theorem ,
We know, Iz = Ix + Iy
Where Iz is moment of inertia about perpendicular axis of plane of disc .

Iz = (1/4)mR² + (1/4)mR²
= (1/2)mR²

Now, moment of inertia of disc about passing through a point of its edge
______________________________
Use , parallel axis theorem ,
I = Iz + mR²
= (1/2) mR² + mR²
= (3/2)mR²

Answer 2

Explanation:

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