The centripetal force of a child riding a merry-go-round

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Problem:

(a) A 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force must she exert to stay on if she is 1.25 m from its center?

(b) What centripetal force does she need to stay on an amusement park merry-go-round that rotates at 3.00 rev/min if she is 8.00 m from its center?

(c) Compare each force with her weight.

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Solution:

Part A

We are given the following values: $m=22.0\ \text{kg}$, $\omega = 40.0\ \text{rev/min}$, and $r=1.25\ \text{m}$. We are asked to solve for the centripetal force, $F_c$.

Centripetal force $F_c$ is any force causing uniform circular motion. It is a “center-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity $v$ and has magnitude $F_c=m a_c$, which can also be expressed as $F_c = m \frac{v^2}{r}$ or $F_c = m r \omega ^2$. Basing from the given values, we are going to solve the problem using the formula

First, we need to convert the angular velocity $\omega$ to $\text{rad/sec}$ for unit homogeneity.

Now, we can substitute the given values into our formula.

Part B

Let us convert the angular velocity to radians per second.

Now, we can substitute the given values

Part C

For the first centripetal force we solved in Part A,

The centripetal force is 2.24 times the weight of the child.

For the centripetal force we solved in Part B, we have

The centripetal force is only about 8% of the child’s weight.

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