Question 5.8: Acceleration of a revolving ball. A 150-g ball at the end of......
Acceleration of a revolving ball. A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m, as in Fig. 5-10 or 5-12. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?
APPROACH The centripetal acceleration is a_{\mathrm{R}}=v^2 / r. We are given r, and we can find the speed of the ball, \boldsymbol{v}, from the given radius and frequency.
If the ball makes two complete revolutions per second, then the ball travels in a complete circle in a time interval equal to 0.500 s, which is its period T. The distance traveled in this time is the circumference of the circle, 2 \pi r, where r is the radius of the circle. Therefore, the ball has speed
v=\frac{2 \pi r}{T}=\frac{2 \pi(0.600 \mathrm{~m})}{(0.500 \mathrm{~s})}=7.54 \mathrm{~m} / \mathrm{s}.
The centripetal acceleration { }^{\dagger} is
a_{\mathrm{R}}=\frac{v^2}{r}=\frac{(7.54 \mathrm{~m} / \mathrm{s})^2}{(0.600 \mathrm{~m})}=94.7 \mathrm{~m} / \mathrm{s}^2.
{ }^{\dagger} differences in the final digit can depend on whether you keep all digits in your calculator for v (which gives a_{\mathrm{R}} = 94.7 m/s²), or if you use v = 7.54 m/s in which case you get a_{\mathrm{R}} = 94.8 m/s². Both results are valid since our assumed accuracy is about + 0.1 m/s (see Section 1-3).