a student holds a 3.0kg mass in each hand while sitting on a rotating stool. when his arms are extended horizontally, the masses
are 1.0m from the axis of rotation and he rotates with an angular speed of 0.75 rad/s. if the student pulls the masses horizontally to 0.30m from the axis of rotation, what is his new angular speed
Due to change in the position of 3 kg mass , the moment of inertia of the system changes , due to which angular speed changes . We shall apply conservation of angular momentum , because no external torque is acting .
Initial moment of inertia I₁ = M R² = 3 x 1 ² = 3 kg m²
Final moment of inertia I₂ = M R² = 3 x .3 ² = 0.27 kg m²
I think we will use the law of conservation of linear momentum; M1V1 = M2V2 M1 = 4 kg (mass of the water balloon launcher) V1=? M2= 0.5 kg ( mass of the balloon) V2 = 3 m/s
