Suresh Sonkar
Last Activity: 5 Years ago
Given circular disc of radius R and mass M, rotating with angular velocity ω about its centre.
Moment of inertia (I) about an axis passing through the centre (centre of mass) of the disc and perpendicular to the plane of the disc
Icm = MR2/2
Now using parallel axis theorem:
Moment of inertia about a point P (Ip):
Ip = Icm + Md2 = MR2/2 + M (R/2)2 = 3/4MR2
(Here d is given R/2 and Icm = MR2/2)
Angular momentum (L) about an axis passing through point P and perpendicular to the disc
= Ipω = 3/4MR2 ω