CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
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.

Open in App
Solution

(a)
The moment of inertia (M.I.) of a sphere about its diameter=2MR2/5
According to the theorem of parallel axes, the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.
The M.I. about a tangent of the sphere =2MR2/5+MR2=7MR2/5

(b)
The moment of inertia of a disc about its diameter =MR2/4
According to the theorem of perpendicular axis, the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.
The M.I. of the disc about an axis normal to the disc through the center =MR2/4+MR2/4=MR2/2
Now, Applying the theorem of parallel axes:
The moment of inertia about an axis normal to the disc and passing through a point on its edge
=MR2/2+MR2=3MR2/2

476766_458310_ans.png

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Moment of Inertia of Solid Bodies
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon