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

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.

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

F_c = m r \omega ^2

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

40\ \text{rev/min} \times \frac{2\pi\ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{sec}} = 4.1888\ \text{rad/sec}

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

\begin{align*}
F_c & = m r \omega ^2 \\ \\
F_c & = \left( 22.0\ \text{kg}\right) \left( 1.25\ \text{m}\right) \left( 4.1888\ \text{rad/sec}\right)^2 \\ \\
F_c & = 482.5162\ \text{N} \\ \\
F_c & = 483\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

Let us convert the angular velocity to radians per second.

3.00\ \text{rev/min} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{sec}}=0.3142 \ \text{rad/sec}

Now, we can substitute the given values

\begin{align*}
F_c & = m r \omega ^2 \\ \\
F_c & = \left( 22.0\ \text{kg}\right) \left( 8.00\ \text{m}\right) \left( 0.3142\ \text{rad/sec}\right)^2 \\ \\
F_c & = 17.3750\ \text{N} \\ \\
F_c & = 17.4\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

For the first centripetal force we solved in Part A,

\frac{F_c}{w} = \frac{483\ \text{N}}{\left( 22\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)} = 2.24

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

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

\frac{F_c}{w} = \frac{17.4\ \text{N}}{\left( 22\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)} = 0.0806

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