Question

A thin circular ring of mas 'M' and radius 'R' is rotating about a transverse axis passing through its centre with constant angular velocity 'ω'. Two objects each of mass 'm' are attached gently to the opposite ends of a diameter of the ring. What is the new angular velocity?

Options

• `("M"omega)/("M"+"m")`

• `(("M"+"2m")omega)/"M"`

• `(("M"-"2m")omega)/("M"+"2m")`

• `("M"omega)/("M"+"2m")`

Answer 1

`("M"omega)/("M"+"2m")`

Explanation:

Initial angular momentum of the ring = Iω = MR²ω

If the new angular velocity is ω' then the final angular momentum = I'ω'

where I'= MR² + 2mR² = (M + 2m)R²

By law of conservation of momentum

I'ω' = Iω

(M + 2m)R² ω' = MR² ω

`thereforeomega^'=("M"omega)/("M"+"2m")`