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

`a_(cp)=162.2m/s^2`

Step-by-step 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+(a_yt^2)/(2)`

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

`y=(gt^2)/(2)`

Which gives us a time:

`t=\sqrt{(2y)/(g)}=\sqrt{(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=(x)/(t)=(10m)/(0.64s)=15.6m/s`

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

`a_(cp)=(v_x^2)/(r)=((15.6m/s)^2)/((1.5m))=162.2m/s^2`