Circular Motion

Question

Kerel is swinging a bucket of water above his head to show off in front of a girl he likes. If the distance from Kerel's shoulder to the end of the bucket is 67 cm, what is the minimum speed required to keep the water in the bucket and keep the girl from laughing at him? (Hint: for the minimum speed, you want gravity, and only gravity, to provide the centripetal force; your hand should be providing no "central facing" force - gravity, and only gravity, becomes the centripital force needed to keep the water going in a circle)

Answer 1

Let

r = 67 cm = 0.67 m be the radius of rotation,

m (unknown) be the mass of the bucket and water,

v (to be computed) be speed of the bucket,

g = 9.81 m/s² be gravitational acceleration.

The bucket of water experiences the force of gravity

mg

For the bucket of water to remain in circular motion it requires centripetal force

mv²/r

As the hint suggests, the minimal speed is reached when gravity alone provides the centripetal force.

(This is in contrast to higher speeds, when tension in the arm would provide additional force.)

So at minimal speed

mv²/r = mg

Hence

v = √(gr)

Substituting actual numbers

v = √(9.81×0.67) = 2.56 m/s