Question

A thin circular right of mass M and radius R is rotating about its axis with an angular speed omega_(0) two particles each of mass m are now attached at diametrically opposite points. Find new angular speed of the ring.

Answer 1

The new angular velocity is   $\frac{M}{M+2m}$ × ω₀

A thin circular ring of mass M and radius R is rotating about its axis with an angular speed ω₀

Now, two masses are attached to diametrically opposite ends of the ring.

• Here we use the principle of conservation of momentum.

        Initial angular momentum = final angular momentum

•         I₁ ω₁ = I₂ ω₂

Moment of inertia of ring about its axis is given by :

• I = M.R²
• So, I₁ = M.R²

Moment of inertia of particle with mass m about the axis will be I = m.R²

• I₂ = Moment of inertia of ring + 2(moment of inertia of particle)
• I₂ = M.R² + 2m.R²

Substituting the values of I₁ and I₂

• M.R². ω₀ = ( M.R² + 2.m.R ²)ω

• M.ω₀ = (M + 2m )ω
• ω = $\frac{M}{M+2m}$ × ω₀

Answer 2

In a thin circular right of mass M and radius R is rotating about its axis with an angular speed omega-

Since, no external force is acting on the system so the angular momentum will be conserved.

According to law of conservation of angular momentum:-

Initial angular momentum = final angular momentum

 Let the final angular velocity be ωfinal

Angular momentum is conserved about center of the ring.

Li=Lf

Iintialω0=Ifinalωfinal

Iintial=MR2

Ifinal=MR2+mR2+mR2                       ( here M= mass of disc)

by substituting above in equation,    ( m= mass of added particle)

MR2ω0=(MR2+mR2+mR2)ωfinal       ( R= radius)

ωfinal=M ω0 / M+2m