A child of mass 30 kg stands at the rim of a merry-go-round of moment of inertia 200 kg m 2 and radius 2 m turning at 2 rad/s. The child walks to the centre. The new angular velocity is:

⚛️ Physics · System of Particles and Rotational Motion · NEET & JEE

A child of mass 30 kg stands at the rim of a merry-go-round of moment of inertia 200 kg·m<sup>2</sup> and radius 2 m turning at 2 rad/s. The child walks to the centre. The new angular velocity is:

Answer: 3.2 rad/s.

A 2.0 rad/s
B 3.2 rad/s
C 1.25 rad/s
D 4.0 rad/s

Correct answer: B. 3.2 rad/s

Explanation: Initial I = 200 + 30 × 2<sup>2</sup> = 320. At centre I = 200. Conserving L: 320 × 2 = 200ω, ω = 3.2 rad/s.

Torque about axis O equals the force F multiplied by the perpendicular distance r⊥ from the axis to the line of action of the force, producing rotation.

Concept context

Moment of inertia, torque, angular momentum, rolling motion. Very high JEE weightage.

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A child of mass 30 kg stands at the rim of a merry-go-round of moment of inertia 200 kg&middot;m<sup>2</sup> and radius 2 m turning at 2 rad/s. The child walks to the centre. The new angular velocity is:

Concept context

More System of Particles and Rotational Motion questions

A child of mass 30 kg stands at the rim of a merry-go-round of moment of inertia 200 kg·m<sup>2</sup> and radius 2 m turning at 2 rad/s. The child walks to the centre. The new angular velocity is: