A disc has mass 9m. A hole ofradíus R/3 is cut from it as shown in th

1.	A disc has mass 9m. A hole ofradíus R/3 is cut from it as shown in the figure.The moment of inertia of remaining part about an axis passing through the centreO' of the disc and perpendicular to the plane of the disc is: [JEE'2005]4037(A) 8 mR2(B) 4 mR2
Answer» 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 mr^2 Moment of inertia of total disk of mass 9M and radius R = (1/2) 9M R^2 Iwhole= (1/2) 9M R^2 - Let us calculate the mass of the removed disk and the location of its centre of mass mass density = 9M / area of disk = 9M/ piR^2 mass of small disk of radius R/3 = area of small disk x mass density = pi(R/3)^2x 9M/ piR^2 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 = MR^2/18 + 4MR^2/9 Iremoved= MR^2/2 - hence the moment of inertia of remaining disk = Iwhole-Iremoved =(1/2) 9M R^2-MR^2/2 = 4MR2

A disc has mass 9m. A hole ofradíus R/3 is cut from it as shown in the figure.The moment of inertia of remaining part about an axis passing through the centreO' of the disc and perpendicular to the plane of the disc is: [JEE'2005]4037(A) 8 mR2(B) 4 mR2

Answer»

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 mr^2

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

Iwhole= (1/2) 9M R^2

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

mass density = 9M / area of disk = 9M/ piR^2

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

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

= MR^2/18 + 4MR^2/9

Iremoved= MR^2/2

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

=(1/2) 9M R^2-MR^2/2

= 4MR2