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a 2.35 kg water bucket is swung in a full circle of radius 0.824 m just fast enough so that the water doesn't fall out at the top(meaning n = 0 there). what is the speed of the water at the top? velocity(m/s)

Respuesta :

the answer is 2.84 m/s

its ez lol

Answer:

2.84 m/s

Explanation:

Objects in circular motion experience centripetal acceleration, which is equal to the square of the tangential speed divided by the radius. According to Newton's second law, the net force on an object is equal to its mass times acceleration.

At the top of the loop, the tension force T and weight force mg are both pulling down towards the center. If the bucket is just moving fast enough that the water doesn't fall out, the tension force is zero.

Sum of forces in the centripetal direction:

∑F = ma

mg = m v²/r

v² = gr

v² = (9.8 m/s²) (0.824 m)

v = 2.84 m/s