System of Particles and Rotational Motion - Result Question 53

57. From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?

[2016]

(a) 15MR2/32

(c) 11MR2/32

(b) 13MR2/32

(d) 9MR2/32

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Answer:

Correct Answer: 57. (b)

Solution:

  1. (b) Moment of inertia of complete disc about point ’ O ‘.

ITotal disc =MR22

Mass of removed disc

MRemoved =M4 (Mass area )

Moment of inertia of removed disc about point ‘O’.

IRemoved  (about same perpendicular axis)

=Icm+mx2

=M4(R/2)22+M4(R2)2=3MR232

Therefore the moment of inertia of the remaining part of the disc about a perpendicular axis passing through the centre,

IRemaing disc =ITotal IRemoved 

=MR22332MR2=1332MR2

Moment of inertia of a part of a rigid body (Symmetrically cut from the whose mass) is the same as that of the whole body when whole mass is replaced by mass of that part.



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