Question

A boy whirls a stone in a horizontal circle of radius 1.50m and at height 2.00m above level ground. The string breaks, and the stone flies off horizontally and strikes the ground after traveling a horizontal distance of 10.0m. What is the magnitude of the centripetal acceleration of the stone during the circular motion?

Answer 1

Answer:

$a_{cp}=162.2m/s^2$

Explanation:

The time the stone takes to fall can be calculated considering only the vertical component with the formula:

$y=y_0+v_{0y}t+\frac{a_yt^2}{2}$

Taking the inital height as 0m and downward direction positive, since it departs from (vertical) rest we have:

$y=\frac{gt^2}{2}$

Which gives us a time:

$t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2(2m)}{(9.8m/s^2)}}=0.64s$

Horizontally, on that time the stone travelled a distance x=10m, which means its horizontal speed was:

$v_x=\frac{x}{t}=\frac{10m}{0.64s}=15.6m/s$

Since this speed is the tangential velocity while whirling, the centripetal acceleration of the stone was:

$a_{cp}=\frac{v_x^2}{r}=\frac{(15.6m/s)^2}{(1.5m)}=162.2m/s^2$