Question

The moment of inertia of a solid cylinder of mass M, length 2R and radius R about an axis passing through the centre of mass and perpendicular to the axis of the cylinder is I_(1) and about an axis passing through one end of the cylinder and perpendicular to the axis of cylinder is I_(2)

Answer 1

Answer:

If the length of the cylinder is L then the moment of inertia about an axis passing through centre of mass and perpendicular to the axis of cylinder is

Ic = MR^2/4 + ML^2/12

But the length of the cylinder is 2R therefore L = R

Ic = MR^2/4 + MR^2/8

Ic = 3MR^2/8

And the moment of inertia about an axis passing through one end of the cylinder and the perpendicular to the axis of cylinder is

Now, L = 2R

Therefore Io = Ic + Mh^2

Io = 3MR^2/8 + M(2R)^2

= 3MR^2/8 + 4MR^2

= 35MR^2/8

Hope this helps you!!!!