​ The moment of inertia of a thin ring of radius R about an axis passing through any diameter is 1/2MR^2

Question

The moment of inertia of a thin ring of radius R about an axis passing through any diameter is \(\frac{1}{2}\)MR²

1. To find the moment of inertia of the same ring about an axis passing through its center of mass and perpendicular to its plane, which of the following theorem is used and state the theorem.

• Perpendicular axis theorem.
• Parallel axis theorem

2. What is the radius of gyration of the ring about an axis passing through its centre of mass and perpendicular to its plane?

3. A thin metal ring has a diameter 0.20 cm and mass 1 kg. Calculate its moment of inertia about an axis passing through any tangent.

Answer 1

1. perpendicular axis theorem.

2. Moment of inertia of ring,

I = mr² ____(1)

Moment of inertia of ring in terms of radius of gyration,

I = mk² ____(2)

From eq(1) and eq(2), we get

mk² = mr²

radius of gyration, k = r.

3. I = I₀ + ma²