Question

(a) Derive an expression for the moment of inertia of a uniform thin circular ring about an axis passing through its centre and perpendicular to its plane.

(b) If the earth were to shrink to halfits radius what would be the duration of the day?​

Answer 1

(a)The moment of inertia of a thin uniform circular ring about an axis passing through its centre and perependicular to its plane is MR²

1.To calculate the moment of inertia we can take a small mass of dm

I(system)=ΣI(add up of all particles)

2.for ring mass is concentrated on its circumfurance

3.for small element dI=dm(R²)

4.for adding all particles we need to integrate

∫dI=∫dm(R²)

I=R²∫dm(taking constant outside)

=MR²

(b) If earth shrinks to half of its radius(without changing its mass) then the duration of the day will be 6 hours

1.the angular momentum of earth remains conserved

2.if radius changes from R to R/2 the

3.I(initial)=2/5MR²

I(final)=2/5M(R/2)²=MR²/10

therefore w₀=4 w₀

4.duration of day is rotation about its axis

therefore duration of day will be T/4 ie 6 hours