Question

A 60 g ball is tied to the end of a 50-cm-long string and swung in a vertical circle. The center of the circle, as shown in FIGURE P8.57, is 150 cm above the floor. The ball is swung at the
minimum speed necessary to make it over the top without the string going slack. If the string is released at the instant the ball is at the top of the loop, how far to the right does the ball hit the ground?

Short answer

The ball will hit the ground at a distance of 1.41 meters.

Step-by-step solution

Given information

mass = 60 gm=0.06kg

length = 50 cm=0.5 m

Height of center of circle =1.5 m = 150 cm

Explanation

Lets draw a free body diagram as below and equate forces

Lets consider the top point, we have

$T_{top}=\frac{m{v}^2}{r}-mg$

To find the minimum speed necessary to make it over the top we have, T=0,

$$\begin{gathered} 0=\frac{m{v}^2}{r}-mg \\ v=\sqrt{rg} \end{gathered}$$

Substitute values given, we get

$v=\sqrt{(0.5m)(9.8m/s^2)}=2.21m/s$

The height from which ball is falling =150 cm + 50cm=200 cm= 2m

Distance can be found by using the formula

$s=ut+\frac{1}{2}a{t}^2$

as u=0 and s= H , so

$$\begin{gathered} H=\frac{1}{2}g{t}^2 \\ t=\sqrt{\frac{2H}{g}} \end{gathered}$$

Substitute the values we get

$t=\sqrt{\frac{2\left(2m\right)}{(9.8m/s^2)}}=0.639s$

distance traveled in 0.639 sec with a velocity of 2.21 m/sec is

distance = (0.639 sec) x (2.21 m/sec) = 1.4121 meters.