A wheel of moment of inertia 0.10 is rotating about a shaft at an angular speed of 160rev/min.A second wheel is set into rotation at 300 rev/min and is coupled to the same shaft so that both the wheels finally rotate with the common angular velocity of 200 rev/min.Find the moment of inertis of second wheel

For this purposeuse the law of conservation of angular momentum

I₁(omega)₁(Angular mom. of first wheel )+I₂(omega)₂ (Angular mom. of seconn d wheel )=(I₁+I₂)(omega)_f (Angular mom. of combined system wheel )

(0.1)(320pi rad/min) + I₂(600pi rad/min)=(0.1+I₂)(400pi rad/min)

Solve the above equation and u will get the value of I₂

I1 and I2 are moments of inertia

w1=160,w2=300,w3=200,

I1*w1 + I2*w2=(I1 + I2)*w3