Rotational MotionEasy
Question
A thin wire of length L and uniform linear mass density ρ is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XX′ is
Options
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Solution
Mass of the ring M = 
Let R be the radius of the ring, then
L = 2πR or R =
Moment of inertia anout an axis passing through O and parallel to XX′ will be
I0 =
MR2
Therefore, moment of inertia about XX′ (from parallel axis theorem) will be given by
IXX′ =MR2 + MR2 = 
MR2
Substituting values of M and R
IXX′ =(ρL)
= 
Let R be the radius of the ring, then
L = 2πR or R =
Moment of inertia anout an axis passing through O and parallel to XX′ will be
I0 =
MR2Therefore, moment of inertia about XX′ (from parallel axis theorem) will be given by
IXX′ =
MR2 + MR2 = MR2 Substituting values of M and R
IXX′ =
(ρL) = Create a free account to view solution
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