Four particles each of mass 'm' are lying symmetrically on the rim of the disc of mass M and radius R moment of inertia of this system about an axis passing through one of the particles and perpendicular to plane of disc is :
Answer & explanation
Correct answer: option 2
axis of rotation is passing through particle (3)
Moment of Inertia of disc about axis = (1/2)MR2 + MR2 = (3/2)MR2 by figure (2),
Moment of Inertia of particle (2) about given axis = m(√2R)2 = 2mR2
Moment of Inertia of particle (4) about given axis = m(√2R)2 = 2mR2
Moment of Inertia of particle (1) about given axis = m(2R)2 = 4mR2
Total moment of inertia = (3/2)MR2 + 2mR2 + 2mR2 + 4mR2
= (3/2)MR2 + 8mR2
= (R2/2) (3M + 16m)