Physics, asked by gopal4515, 1 year ago

A thin rod of length l and mass m is bent to form a semicircular ring. The moment of inertia of ring about the diameter is

Answers

Answered by khushirjobanputra

2

Answer:

m(l2/2π2)

Explanation:

l =πr

r=l/π

Symmetry moment of inertia = (2m) r2/2

Moment of inertia of half Ring = m (r2/2)

= m(l2/2π2)

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