Question

A thin ring has mass of 0.25 Kg and radius 0.5m, calculate its moment of inertia about:

a) An axis passing through its centre and perpendicular to its plane

b) An axis passing through a point on its circumference and perpendicular to its plane.
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Answer 1

Answer:

◆ Answer -

(A) 0.0625 kg/m^2

● Explanation -

# Given -

m = 0.25 kg

r = 0.5 m

# Solution -

Moment of inertia about an axis passing through center of the ring and perpendicular to its plane is given by formula -

I = mr^2

I = 0.25 × 0.5^2

I = 0.0625 kg/m^2

Hence, moment of inertia about an axis passing through center of the ring and perpendicular to its plane is 0.0625 kg/m^2.