A 0.150-kg ball on the end of a 1.10-m-long cord (negligible mass) is swung in a vertical circle. (a) Determine the minimum speed the ball must have at the top of its arc so that the ball continues moving in a circle. (b) Calculate the tension in the cord at the bottom of the arc, assuming the ball is moving at twice the speed of part (a).

Solution:

The ball moves in a vertical circle and is not undergoing uniform circular motion. The radius is assumed constant, but the speed v changes because of gravity. Nonetheless, the equation for centripetal acceleration \text{a}_\text{c} = \frac{\text{v}^2}{\text{r}} is valid at each point along the circle, and we use it at the top and bottom points. The free body diagram is shown in the figure below for both positions.

Part A

At the top (point 1), two forces act on the ball: mg, the force of gravity (or weight), and F_T1, the tension force the cord exerts at point 1. Both act downward, and their vector sum acts to give the ball its centripetal acceleration a_c. We apply Newton’s second law, for the vertical direction, choosing downward as positive since the acceleration is downward (toward the center):

\begin{align*}
\sum_{}^{}\text{F}_\text{v}& =\text{ma}_\text{c}\\
\\
\text{F}_\text{T1}\ +\ \text{mg}&= \text{m} \cdot \frac{\text{v}_1^2}{\text{r}}
\end{align*}

From this equation we can see that the tension force F_T1at point 1 will get larger if v₁ (ball’s speed at top of circle) is made larger, as expected. But we are asked for the minimum speed to keep the ball moving in a circle. The cord will remain taut as long as there is tension in it. But if the tension disappears (because v₁ is too small) the cord can go limp, and the ball will fall out of its circular path. Thus, the minimum speed will occur if F_T1 = 0 (the ball at the topmost point), for which the equation above becomes

\text{mg}=\text{m}\cdot \frac{\left( \text{v}_1 \right)^2}{\text{r}}

We solve for v₁, we have

\begin{align*}
\text{v}_1&=\sqrt{\text{gr}} \\
\\
&=\sqrt{\left( 9.81\  \text{m/s}^2 \right)\left( 1.10\ \text{m} \right)} \\
\\
&=3.285 \ \text{m/s}
\end{align*}

Therefore, the minimum speed at the top of the circle if the ball is to continue moving in a circular path is about 3.285 m/s.

Part B

When the ball is at the bottom of the circle, the cord exerts its tension force F_T2 upward, whereas the force of gravity, mg still acts downward. Choosing upward as positive, Newton’s second law gives:

\begin{align*}
\sum_{}^{}\text{F}_\text{v}& =\text{ma}_\text{c}\\
\\
\text{F}_\text{T2}\ -\ \text{mg}&= \text{m} \cdot \frac{\text{v}_2 ^2}{\text{r}}
\end{align*}

The speed v₂ is given as twice that in (a). We solve for F_T2

\begin{align*}
F_\text{T2} & = m\cdot \frac{v^2}{r}+mg\\
\\
& = \left( 0.150\ \text{kg} \right)\cdot \frac{\left( 2\times 3.285 \ \text{m/s}\right)^2}{1.10\ \text{m}}+\left( 0.150\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)\\
\\
&=7.358 \ \text{N}
\end{align*}