Question

A disc of moment of inertia 5×10–4kgm2 is rotating freely about axis through its center at 40rpm. Calculate the new revolution per minute (rpm) if wax of mass 0.02kg is dropped gently on to the disc 0.08m from its axis.

Answer 1

Answer:this image may help you buddy.. make me brainilisthave a great day

Answer 2

Given info : A disc of moment of inertia 5 × 10⁻⁴ kgm²  is rotating freely about axis through its center at 40 rpm.

To find : he new revolution per minute (rpm) if wax of mass 0.02kg is dropped gently on to the disc 0.08m from its axis is ...

solution : initial moment of inertia, I₁ =   5 × 10⁻⁴ kgm²

final moment of inertia, I₂ = $I_1+I_{\text{wax}}$ = I₁ + mr²

=  5 × 10⁻⁴ kgm² + 0.02 kg × (0.08 m)² =  6.28 × 10⁻⁴ kgm²

as external torque acting on the disc is zero. so the angular momentum must be conserved.

⇒ I₁ω₁ = I₂ω₂

⇒  5 × 10⁻⁴ kgm² × 40 rpm =  6.28 × 10⁻⁴ kgm² × ω₂

⇒ ω₂ = 200/6.28 = 31.847 ≈ 32 rpm

therefore the new revolution per minute of the disc is 32 rpm.