Question

A ball on the end of a string is whirled around in a horizontal circle of radius 0.340 m. The plane of the circle is 1.50 m above the ground. The string breaks and the ball lands 1.50 m (horizontally) away from the point on the ground directly beneath the ball's location when the string breaks. Find the radial acceleration of the ball during its circular motion.

Magnitude
m/s²Direction

away from the center of curvature

toward the center of curvature

Answer 1

Given radius r = 0.340 m

height h = 1.5 m

horizontal distance x = 1.5 m

let 't' be the time to reach the ground the

when the wire breaks it has only horizontal velocity

the vertical velocity is zero

from the equation

s = (u*t) + (1/2 * a * t^2)

h = 0 + (1/2 * g * t^2

1.5 = 1/2 * 9.8 * t^2

t = 0.55 s

For horizontal motion

x = v * t

1.5 = v * 0.55

v = 2.72 m/s

This is the velocity of the ball just before it breaks so

radial acceleration is a = v^2 / r

a = (2.72)^2 / 0.340

a = 21.76 m/s^2 (toward the center of curvature)