Question

A 200-kg, 2.0-m-radius, merry-go-round in the shape of a flat, uniform, circular disk parallel to level ground is rotating at 1.2 cycles/second about an axis through its center of mass and perpendicular to the ground. A 50-kg boy jumps onto the edge of the merry-go round and lands at a fixed point. What is the angular velocity of the merry-go-round after the boy lands on it

Answer 1

Step-by-step explanation:

Conservation of angular momentum.

Disk I = ½MR²

Point mass I = mR² (boy)

Initial angular momentum

L₀ = Iω = ½MR²ω₀

Final angular momentum

L₁ = Iω = (½MR² + mR²)ω₁

as momentum is conserved, these are equal

(½MR² + mR²)ω₁ = ½MR²ω₀

ω₁ = ω₀(½MR²/ (½MR² + mR²))

ω₁ = ω₀(½M/ (½M + m))

ω₁ = 1.2(½(200)/ (½(200) + 50))

ω₁ = 1.2(⅔)

ω₁ = 0.8 cycles/second or 0.8(2π) = 1.6π rad/s