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A 19.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force (in N) must she exert to stay on if she is 2.00 m from its center

Respuesta :

Answer:

The centripetal force is 663.95 N.

Explanation:

Given that,

Mass of the child, m = 19 kg

Angular velocity of the merry-go-round, [tex]\omega=40\ rev/min=4.18\ rad/s[/tex]

Distance from the center, r = 2 m

We need to find the centripetal force acting on the child. The centripetal force is given by the formula as :

[tex]F=m\omega^2r[/tex]

[tex]F=19\times (4.18)^2\times 2[/tex]

F = 663.95 N

So, the centripetal force is 663.95 N. Hence, this is the required solution.