(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 $$\dfrac{ 2MR^2}{5}$$, 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 $$\dfrac{ MR^2}{4}$$, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Asked by Pragya Singh | 1 year ago |  84

##### Solution :-

Given,

(a) The moment of inertia (M.I.) of a sphere about its diameter =$$\dfrac{ 2MR^2}{5}$$

According to the theorem of parallel axes, M.I of a sphere about a tangent to the sphere

= $$\dfrac{ 2MR^2}{5}$$+ MR =$$\dfrac{ 7MR^2}{5}$$

(b) Given, moment of inertia of a disc about its diameter =$$\dfrac{ ( MR^2 )1}{4}$$

(i) According to the theorem of perpendicular axis, the moment of inertia of a planar body (lamina) about an axis passing through its center and perpendicular to the disc

= 2 x ($$\dfrac{1}{4}$$)MR2 = $$\dfrac{ ( MR^2 )}{2}$$

The situation is shown in the given figure.

(ii) Using the theorem of parallel axes:

Moment of inertia about an axis normal to the disc and  going through a point on its circumference

$$\dfrac{ ( MR^2 )}{2}$$+ MR2

$$\dfrac{ (3 MR^2 )}{4}$$

Answered by Abhisek | 1 year ago

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