Question

Find the moment of inertia of a dise of mass 3Kg and radius 50 cm

1) About the axis passing through the center and perpendicular to the plane of the disc.

2) Axis passing through touching the edge & axis tou plane of the disc.

3) Axis passing through the center and lying on the plane of the disc

on the plane of the disc.​

Answer 1

Question:

Find the rotational kinetic energy of a ring of mass 9 kg and radius 3 m rotating with 240 rpm about an axis passing through its centre and perependicualr to its plane.

Explanation:

The mass M=3K g radius, R=50cm=50×10−2m=0.5m

(i) The moment of inertia (ii) about an axis passing through the center and perpendicular to the plame of the disc is

I=12MR2

I=12×3×(0.5)2=0.5×3×0.5×0.5

I=0.375kgm2

(ii) The moment of inertia (i) about an axis touching the edge and perpendicular to the plane of the disc by parallel axis theorem is,

I=IC+Md2

where IC=12MR2andd=R

I=32MR2+MR2=32MR2

I=32×3×(0.5)2=1.5x3×0.5×0.5

I=1.125kgm2

(iii) The moment of inertia (i) about an axis passing through the center and lying on the

IZ=IX+IY

Where IX=IY=IandIZ=1)2MR2

IZ=2I,I=12Iz

I=12×12MR2=14MR2

I=14×3×(0.5)2=0.25×3×0.5×0.5

I=0.1875kgm2