A 22.0-kg child is riding a playground merrygo-round that is rotating at 40.0 rev/min. What centripetal force is exerted if he is 1.25 m from its center?

Respuesta :

Answer:

F=480.491 N

Explanation:

Given that

mass ,m = 22 kg

Angular speed ω = 40 rev/min

[tex]\omega=\dfrac{2\pi \times 40}{60}\ rad/s[/tex]

ω =4.18 rad/s

The radius  r= 1.25 m

We know that centripetal force is given as

F=m ω² r

Now by putting the values in the above equation we get

[tex]F=22\times 4.18^2\times 1.25\ N[/tex]

F=480.491 N

Therefore the centripetal force on the child will be 480.491 N.

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