Estimate the force a person must exert on a string attached to a 0.150-kg ball to make the ball revolve in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions per second. Ignore the string’s mass.

Solution:

First we need to draw the free-body diagram for the ball. The forces acting on the ball are the force of gravity (or weight), mg downward, and the tension force F_T that the string exerts toward the hand at the center (which occurs because the person exerts that same force on the string). The free-body diagram for the ball is shown in the figure below. The ball’s weight complicates matters and makes it impossible to revolve a ball with the cord perfectly horizontal. We estimate the force assuming the weight is small, and letting \phi = 0 from the figure. Then F_T will act nearly horizontally and, in any case, provides the force necessary to give the ball its centripetal acceleration.

Before, we can use the formula of the centripetal force, we need to solve for the value of the linear velocity first. The linear velocity of the ball can be computed by dividing the total arc length traveled by the total time of travel. That is, the ball traveled 2 revolutions (twice the circumference of the circle) for 1 second. Thus,

\begin{align*}
\text{v} &= \frac{2\cdot2 \pi \text{r}}{\text{t}} \\
\\
& = \frac{4\pi \text{r}}{\text{t}} \\
\\
& = \frac{4\pi\left( 0.600\ \text{m} \right)}{1 \ \text{s}} \\
\\
& = 7.54 \ \text{m/s}
\end{align*}

Using the formula for centripetal force, we have

\begin{align*}
\text{F}_\text{c} &=\text{ma}_\text{c} \\
\\
& = \text{m} \cdot \frac{\text{v}^{2}}{\text{r}} \\
\\
& = \left( 0.150\ \text{kg} \right) \cdot \frac{\left( 7.54\ \text{m/s} \right)^{2}}{0.600\ \text{m}}\\
\\
& = 14.2\ \text{N}
\end{align*}

Therefore, the force a person must exert on a string is about 14.2 N.