A 1.50-kg bucket of water is tied by a rope and whirled in a circle with a radius of 1.00 m. At the bottom of the circular loop, the speed of the bucket is 6.00 m/s. Determine the acceleration, the net force and the individual force values when the bucket is at the bottom of the circular loop.
Question
Answer: 36.0 m/s², 54.0 N, 14.7 N, 68.7 N
Explanation:
acceleration is centripetal
ac = v²/r = 6.00² /1.00 = 36.0 m/s²
The net force causes acceleration. In this case centripetal.
Fc = Fnet = 1.50(36.0) = 54.0 N
Fg = 1.50(9.81) = 14.7 N
The total force supplies the necessary centripetal acceleration and supports the weight of the bucket.
Ftot = mg + m(ac) = m(g + v²/r)
Ftot = 1.50(9.81 + 36.0) = 68.715 ≈ 68.7 N