You swing a 4.5-kg bucket of water in a vertical circle of radius 1.1 m. a) What speed must the bucket have if it is to complete the circle without spilling any water? b) How does your answer depend on the mass of the bucket?
2007-02-28 13:33:58 · 3 answers · asked by 123haha 1 in Science & Mathematics ➔ Physics
Ac = Ag
w^2r = 9.81 m/s^2
w = sqrt(9.81/1.1) = 2.98 rad/s = 0.475 rev/s = 3.28 m/s
2007-02-28 13:57:56 · answer #1 · answered by catarthur 6 · 0⤊ 0⤋
When g = v^2 / R the bucket is accelerating at the same rate as the water in free fall, so they will not become separated, and the water will stay in the bucket.
9.8 = v^2 / 1.1 and solve for v to get your answer.
The mass of the bucket will not change the speed that you must move it to keep the water from spilling out at the top of the circle. (Increasing the mass of the bucket DOES increase the force you must apply to give the bucket the acceleration needed to keep the water in it, though.)
2007-02-28 21:45:57 · answer #2 · answered by Dennis H 4 · 0⤊ 0⤋
The acceleration of the bucket must be greater than the acceleration of gravity. The mass has nothing to do with this answer.
acceleration of gravity = 9.81 m/s/s
to not spill a drop accel must be 9.82 m/s/s
Acceleration = w X radius = 9.82 m/s/s
9.82 = w X 1.1
w = 9.82/1.1 = 8.93 rad/sec of about 1.42 rev/sec
2007-02-28 21:49:03 · answer #3 · answered by eric l 6 · 0⤊ 1⤋