Question

A boy whirls a stone in a horizontal circle of radius 1.5 m and at height 2.0 m above ground. The string breaks, and the stone flies off and strikes the ground 15 m away. What is the magnitude of the centripetal acceleration of the stone during its time under uniform circular motion?

(Please explain your answers.)

Answer 1

after string breaks the stone flis off tangential to the

cicular path with a speed v

the stones flies horizontal to the ground and travels a parabolic path

along vertical

initial velocity voy = 0

acceleration = -g = -9.8 m/s^2

displacement y = -2 m

y = voy*t + (1/2)*ay*t^2

-2 = 0 - (1/2)*9.8*t^2

time taken to land on ground t = 0.64 s

along horizontal

initial velocity vox = v

displacement = 15 m

acceleration ax = 0

x = vox*t + (1/2)*ax*t^2

15 = v*0.64

v = 23.5 m/s

for the circular motion

centripetal acceleration a = v^2/r = 23.5^2/1.5 = 368.2 m/s^2