Question

A 45 kg merry go round worker stands on the rides platform 6.3 m from the center. If her speed as she goes around the circle is 4.1 m/s, what is the force of friction necessary to keep her from falling off the platform?

Answer 1

i’ve got 120 n, hope it helps!

Answer 2

The force required to keep the worker rom falling off the platform is 120.07 N.

Given data:

The mass of merry go round is, m = 45 kg.

The distance from the center of merry go round is, s = 6.3 m.

The speed of worker is, v = 4.1 m/s.

In this problem, the worker is having a circular motion in merry go round. So the force necessary to keep the worker on track is Centripetal force. And the expression for the centripetal force is,

$F = \dfrac{mv^{2}}{L}$

Solving as,

$F=\dfrac{45 \times 4.1^{2}}{6.3}\\\\F = 120.07 \;\rm N$

Thus, we can conclude that the force required to keep the worker rom falling off the platform is 120.07 N.

Learn more about the centripetal force here:

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