Question

A disc of certain radius os cut from a disc of mass 9m and radius r. find its moment of inertia about an axis

Answer 1

The moment of inertia of remaining disk = Moment of inertia of whole disc - Moment of inertia of removed small disc

moment of inertia of any disk about its own central axis = 1/2 mr2

Moment of inertia of total disk of mass 9M and radius R = (1/2) 9M R2

Iwhole = (1/2) 9M R2

-

Let us calculate the mass of the removed disk and the location of its centre of mass

mass density = 9M / area of disk = 9M/ piR2

mass of small disk of radius R/3 = area of small disk x mass density = pi(R/3)2 x 9M/ piR2

We get the mass of the removed small disk = M

-

Its centre of mass is located at R - R/3 = 2R/3 from the main axis

its moment of inertia can be calculated using parallel axis theorem

moment of inertia of removed disk = moment of inertia of small disk about its own axis + moment of inerita of its centre of mass about the main axis

Iremoved = (1/2) M (R/3)2 + M (2R/3)2

= MR2 /18 + 4MR2 /9

Iremoved = MR2 /2

-

hence the moment of inertia of remaining disk = Iwhole - Iremoved

= (1/2) 9M R2 - MR2 /2

= 4MR2