Question 5.11: Force on revolving ball (horizontal). Estimate the force a p......
Force on revolving ball (horizontal). 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 (T = 0.500 s), as in Example 5-8. Ignore the string’s mass.
APPROACH First we need to draw the free-body diagram for the ball. The forces acting on the ball are the force of gravity, m \overrightarrow{ g } downward, and the tension force \overrightarrow{ 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 as shown in Fig. 5-17. The ball’s weight complicates matters and makes it impossible to revolve a ball with the cord perfectly horizontal. We assume the weight is small, and put \phi \approx 0 in Fig. 5-17. Thus \overrightarrow{ F }_{ T } will act nearly horizontally and, in any case, provides the force necessary to give the ball its centripetal acceleration.
We apply Newton’s second law to the radial direction, which we assume is horizontal:
(\Sigma F)_{ R }=m a_{ R },
where a_{ R }=v^2 / r \text { and } v=2 \pi r / T=2 \pi(0.600 m)/(0.500 s) = 7.54 m/s. Thus
F_{ T }=m \frac{v^2}{r}=(0.150 \ kg ) \frac{(7.54 \ m / s )^2}{(0.600 \ m )} \approx 14 \ N.
NOTE We keep only two significant figures in the answer because we ignored the ball’s weight; it is mg = (0.150 kg)(9.80m/s²) = 1.5 N, about \frac{1}{10} of our result, which is small but not so small as to justify stating a more precise answer for F _{ T }.
NOTE To include the effect of m \overrightarrow{ g }, resolve \overrightarrow{ F }_{ T } in Fig. 5-17 into components, and set the horizontal component of \overrightarrow{ F }_{ T } equal to m v^2 / r and its vertical component equal to mg.