Question
A thin wire of length 'L' and linear mass density 'm' is bent into a circular ring (in x-y plane) with centre 'C' as shown in figure. The moment of inertia of the ring about an axis yy' will be :
A thin wire of length 'L' and linear mass density 'm' is bent into a circular ring (in x-y plane) with centre 'C' as shown in figure. The moment of inertia of the ring about an axis yy' will be :
B. 3\ mL^ 3 8 ^ 2
Total mass of the ring, M = m L. Since the wire of length L is bent into a circular ring of radius R, its circumference is L. 2 R = L R = L 2 The axis yy' is a tangent to the ring in its plane. The moment of inertia of a ring about its diametric axis is I_ d = 1 2 MR^2. Using the parallel axis theorem, the moment of inertia of the ring about the tangent yy' is: I_ yy' = I_ d + MR^2 I_ yy' = 1 2 MR^2 + MR^2 = 3 2 MR^2 Substituting the values of M and R: I_ yy' = 3 2 (mL) ( L 2 )^2 I_ yy' = 3 2 mL ( L^2 4 ^2 ) I_ yy' = 3mL^3 8 ^2 Answer: 3\ mL^ 3 8 ^ 2
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