  Question

A kid of mass 50 kg stands at the edge of a platform of radius 1 m which can be freely rotated about its axis. The moment of inertia of the platform is 0.02 kg-m2. The system is at rest. The kid throws a ball of mass 2 kg horizontally to his friend (friend standing on the ground) in a direction tangential to the rim with a speed 10 m/s as seen by his friend. The angular velocity with which the platform will start rotating in rad/s is Solution

Given, Moment of inertia of platform (disc) =I=0.02 kg-m2 Radius of platform R=1 m Velocity of the ball v=10 m/s Mass of ball m=2 kg Mass of boy M=50 kg As we see, there is no external torque on the system. So the angular momentum of the system about the any axis will remain constant or conserved. Hence, →Li=−→Lf As we know, L=Iω MOI of the system (platform + boy) after throwing the ball is =I+MR2 Hence, Li=Lf ⇒0=mvR−(I+MR2)ω Because boy and platform will start moving in the opposite direction of the ball as shown in figure, ω=mvRI+MR2=2×10×10.02+50×12 ω=0.4 rad/sPhysics

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