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coldgirl [10]
3 years ago
13

A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.600 rev

/s. What is its angular velocity (in rev/s) after a 25.0 kg child gets onto it by grabbing its outer edge
Physics
1 answer:
natulia [17]3 years ago
3 0

Answer:

The final velocity \omega_f = 0.4235 \frac{rev}{s}

Explanation:

Given data

Mass of merry go round M_m = 120 kg

Radius = 1.8 m

Initial angular velocity \omega_i = 0.6 \frac{rev}{sec}

Mass of boy M_{boy} = 25 kg

We know that the final velocity is given by

\omega_f = \frac{\frac{1}{2}M_m \omega_i }{M_{boy} + \frac{1}{2} M_m }

Put all the values in above formula we get

\omega_f = \frac{\frac{1}{2}(120)  0.6}{25 + \frac{1}{2} (120) }

\omega_f = 0.4235 \frac{rev}{s}

This is the final velocity.

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