Saturday, May 16, 2009

IIT JEE Level Revision Questions - 10. Rotational mechanics

1. A circular disc of radius r/3 is removed from the outer-edge of a bigger circular disc of mass 9M and radius r. the moment of inertia of the remaining portion of the disc about the centre O of the disc perpendicular to the plane of the disc is

a. 4Mr²
b. (37/9)Mr²
c. (40/9)Mr²
d. 9Mr²

JEE 2005

Solution

Mass of the disc removed

Total mass – M
Total area = π r²
Area of piece removed π (r/3) ²

Hence mass of the disc removed = M[π (r/3) ²]/( π r²) = M

Moment of inertia of removed about its own centre is = ½ (M)*(r/3) ²

Parallel axis theorem

Moment of inertia of a body about an axis parallel to its centre of mass is equal to
Moment of the body about the axis passing through its centre of mass plus Mr² (where M is the mass of body and r is the perpendicular distance between two axes.

According to the parallel axis theorem

Moment of inertia of the removed piece about an axis passing through the centre of the big disc = ½ (M)*(r/3) ² + M(2r/3) ²
=1/2(M) (r²/9) + M(4r²/9) = (½)(M) (r²/9 + 8r²/9) = (½)Mr²

Moment of inertia of bigger disc before cutting the piece

= ½ (9M)(r²)

Hence moment of inertia after removal of the piece
=( ½) (9M)(r²) - (½)Mr² = 4Mr²