A man of mass M=50 kg standing on the edge of a platform (moment of inertia of the platform is I and radius R=1 m) is rotating in anticlockwise direction at an angular speed of 20 rad/s. The man starts walking along the rim with a speed 1 m/s relative to the platform, also in the anticlockwise direction. The new angular speed of the platform in (rad/s) is (take (I=0.02 kg-m2))

A man of mass M=50 kg standing on the edge of a platform (moment of in

1.	A man of mass M=50 kg standing on the edge of a platform (moment of inertia of the platform is I and radius R=1 m) is rotating in anticlockwise direction at an angular speed of 20 rad/s. The man starts walking along the rim with a speed 1 m/s relative to the platform, also in the anticlockwise direction. The new angular speed of the platform in (rad/s) is (take (I=0.02 kg-m2))
Answer» A man of mass $M = 50 kg$ standing on the edge of a platform (moment of inertia of the platform is $I$ and radius $R = 1 m$ ) is rotating in anticlockwise direction at an angular speed of $20 rad/s$ . The man starts walking along the rim with a speed $1 m/s$ relative to the platform, also in the anticlockwise direction. The new angular speed of the platform in $(rad/s)$ is (take $(I = 0.02 {kg-m}^{2})$ )

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