A bucket full of water is rotated in a vertical circle of radius 0.86 m (the approximate length of a person’s arm). What must be the minimum speed of the pail at the top of the circle if no water is to spill out? Answer in units of m/s. 011 10.0 point

Explanation:

At the top of the circular motion, we will have these forces:

Where:

N is the normal force

W is the weight

Fc is the centripretal force

Now, if we assume the normal force equal to zero, we could find the minimum value of the speed water to spill out, It means at any value lower than this value, water fallout of the pail.

So we have:

v(m) means the minimum value of speed.

I hope it helps you!

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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