Question

A 4.6 kg bucket is held by a string and whirled in a vertical circe. The radius of the circle is 9.4 m. The speed of the bucket is 8.0 m/s at the top of the loop and 9,9 m/s at
the bottom of the loop. What is the acceleration the bucket will experience at the top of the loop?

Answer 1

Answer:

Let T + m g = m v^2 / R      

At the top of the loop T can be zero and the force of gravity provides the necessary centripetal force

v^2 / R = a = g

a = 8^2 / 9.4 = 6.81 m/s^2