Question

A ball on the end of a string is whirled around in a horizontal circle of radius 0.478 m. The plane of the circle is 1.02 m above the ground. The string breaks and the ball lands 2.5 m away from the point on the ground directly beneath the ball’s location when the string breaks. The acceleration of gravity is 9.8 m/s 2 . Find the centripetal acceleration of the ball during its circular motion. Answer in units of m/s 2 .

Answer 1

Answer:

Explanation:

We shall find first the velocity of ball at the time when string breaks. Let it be v . During its fall on the ground , 1.02 m below, we use the formula

h = 1/2 gt² where t is time of fall .

1.02 = 1/2 x 9.8 x t²

t²= .2081

t = .456

During this time it travels horizontally at distance of 2.5 m with uniform velocity of v

v x .456 = 2.5

v  = 5.48 m /s

centripetal acceleration

= v² / r where r is radius of the circular path

= 5.48² / .478

= 62.82 m /s²