Question

A college student holds a pail full of water by the handle and whirls it around in a vertical circle at a constant speed. The radius of this circle is 1.2m. What is the minimum speed that the pail must have at the top of its circular motion if the water is not to spill out of the upside down pail?

Answer 1

The minimum speed of the water must be 3.4 m/s

Explanation:

There are two forces acting on the water in the pail when it is at the top of its circular motion:

• The force of gravity, mg, acting downward (where m is the mass of the water and g the acceleration of gravity)
• The normal reaction, N also acting downward

Since the water is in circular motion, the net force must be equal to the centripetal force, so:

$N+mg=m\frac{v^2}{r}$

Where:

$g=9.8 m/s^2$

v is the speed of the pail

r = 1.2 m is the radius of the circle

The water starts to spill out when the normal reaction of the pail becomes zero:

N = 0

When this occurs, the equation becomes:

$mg=m\frac{v^2}{r}\\v=\sqrt{gr}$

And substitutin the values of g and r, we find the minimum speed that the water must have in order not to spill out:

$v=\sqrt{(9.8)(1.2)}=3.4 m/s$

Learn more about circular motion:

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