Question 5.12: Revolving ball (vertical circle). A 0.150-kg ball on the end......
Revolving ball (vertical circle). 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).
APPROACH 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, Eq. 5-1 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 Fig. 5-18 for both positions.
a_R=\frac{v^2}{r}.\qquad\text{[centripetal (radial) acceleration]} (5-1)
(a) At the top (point 1), two forces act on the ball: m \overrightarrow{\mathbf{g}}, the force of gravity, and \overrightarrow{\mathbf{F}}_{\mathrm{T} 1}, 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_{\mathrm{R}}. We apply Newton’s second law, for the vertical direction, choosing downward as positive since the acceleration is downward (toward the center):
\begin{aligned}(\Sigma F)_{\mathrm{R}} & =m a_{\mathrm{R}} \\F_{\mathrm{T} 1}+m g & =m \frac{v_{1}^{2}}{r} .& \text{[at top]}\end{aligned}
From this equation we can see that the tension force F_{\mathrm{T} 1} at point 1 will get larger if v_{1} (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 v_{1} 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_{\mathrm{T} 1}=0, for which we have
m g=m \frac{v_{1}^{2}}{r} . \qquad \text{[minimum speed at top]}
We solve for v_{1}, keeping an extra digit for use in (b) :
v_{1}=\sqrt{g r}=\sqrt{\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)(1.10 \mathrm{~m})}=3.283 \mathrm{~m} / \mathrm{s} .
This is the minimum speed at the top of the circle if the ball is to continue moving in a circular path.
(b) When the ball is at the bottom of the circle (point 2 in Fig. 5-18), the cord exerts its tension force F_{\mathrm{T} 2} upward, whereas the force of gravity, m \overrightarrow{\mathbf{g}}, still acts downward. Choosing upward as positive, Newton’s second law gives:
\begin{aligned}(\Sigma F)_{\mathrm{R}} & =m a_{\mathrm{R}} \\F_{\mathrm{T} 2}-m g & =m \frac{v_{2}^{2}}{r} . & \text{[at bottom]}\end{aligned}
The speed v_{2} is given as twice that in (a), namely 6.566 \mathrm{~m} / \mathrm{s}. We solve for F_{\mathrm{T} 2} :
\begin{aligned}F_{\mathrm{T} 2} & =m \frac{v_{2}^{2}}{r}+m g \\& =(0.150 \mathrm{~kg}) \frac{(6.566 \mathrm{~m} / \mathrm{s})^{2}}{(1.10 \mathrm{~m})}+(0.150 \mathrm{~kg})\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)=7.35 \mathrm{~N}\end{aligned}