Question

A boy whirls a stone in a horizontal circle of radius 1.1 m and at height 2.1 m above ground level. The string breaks, and the stone flies horizontally and strikes the ground after traveling a horizontal distance of 10 m. What is the magnitude of the centripetal acceleration of the stone while in circular motion? m/s2

Answer 1

Answer:

$212.8 m/s^{2}$

Explanation:

Time taken by stone to cover horizontal distance

$t=\sqrt{\frac {2h}{g}}$ where t is time, h is height of whirling the stone in horizontal circle, g is gravitational constant, Substituting h for 2.1 m and g for 9.81

$t=\sqrt{\frac {2*2.1}{9.81}}$= 0.654654 seconds

t=0.65 s

Velocity, v= distance/time

v=10/0.65= 15.27525 m/s

v=15.3 m/s

$a=\frac {v^{2}}{r}$ where r is radius of circle, substituting r with 1.1m

$a=\frac {15.3^{2}}{1.1}$

$a=212.8 m/s^{2}$

Therefore, centripetal acceleration is $212.8 m/s^{2}$