Question

The moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane is 'I'. It is rotating with angular velocity 'ω'. Another identical ring is gently placed on it so that their centres coincide. If both the ring are rotating about the same axis, then loss in kinetic energy is ______.

Options

• `(Iomega^2)/3`

• `(Iomega^2)/4`

• Iω²

• `(Iomega^2)/2`

Answer 1

The moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane is 'I'. It is rotating with angular velocity 'ω'. Another identical ring is gently placed on it so that their centres coincide. If both the ring are rotating about the same axis, then loss in kinetic energy is `underline((Iomega^2)/4)`.

Explanation:

By law of conservation of angular momentum

I₁ω₁ = I₂ω₂

∴ Iω = 2Iω

`thereforeomega'=omega/2`

Initial kinetic energy `K_1=1/2Iomega^2`

Final kinetic energy `K_2=1/2(2I)omega'^2`

`=1/2xx2Ixxomega^2/4=(Iomega^2)/4`

`K_1-K_2=1/2Iomega^2-1/4Iomega^2=(Iomega^2)/4`