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A merry-go-round rotates at the rate of
0.28 rev/s with an 84 kg man standing at
a point 2.3 m from the axis of rotation.
What is the new angular speed when the
man walks to a point 0 m from the center?
Consider the merry-go-round is a solid 35 kg
cylinder of radius of 2.3 m.
Answer in units of rad/s.

Respuesta :

MathPhys

Explanation:

Angular momentum is conserved.

I₁ ω₁ = I₂ ω₂

(½ Mr² + md²) ω₁ = (½ Mr²) ω₂

(½ (35 kg) (2.3 m)² + (84 kg) (2.3 m)²) (0.28 rev/s) = (½ (35 kg) (2.3 m)²) ω

ω = 1.624 rev/s

ω = 10.2 rad/s

Round as needed.

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