Question

A bucket of mass 2.00 kg is whirled in a vertical circle of radius 1.20 meters. At the lowest point of its motion the tension in the rope supporting the bucket is 20.0 N.

A) Find the speed of the bucket at its lowest point.

B) Find the minimum speed of the bucket at the top of the circle so that the rope will not slack?

Answer 1

Answer:

(A) Applying the Newton's second law of force, T-mg = F_c

where T is the tension in the rope, mg is the weight of the bucket and F_c is the centripetal force and is equal to mv²/r

Therefore, T - mg = mv²/r or 20 N - (2.00 kg)(9.8 m/s²) = (2.00 kg) v² / (1.20 m)

then    0.4 N = 1.67 kg/m v²   or v = [ (0.4 N) / (1.67 kg/m) ]^1/2 = 0.48 m/s.

(B) When T = 0 N, then mg = mv²/r

therefore, v² = g/r or v = (g x r)^1/2 = [(9.8 ms^-2) (1.20 m)]^1/2 = 3.42 m/s.