Topic 2.9 Supplement: Vertical Circular Motion — The Water Bucket Problem
This page is optional enrichment for Topic 2.9 (Circular Motion) — it's not a new lesson covered in class. If you want to see the vertical-loop idea from our lesson applied to a different (and honestly more fun) example, this is for you.
The Demonstration
Swing a bucket of water in a vertical circle fast enough, and the water stays in — even when the bucket is completely upside down at the top. Why doesn't it just fall out?
Why the Water Stays In
It comes down to the same idea from our lesson: an object moving in a circle needs a net force pointing toward the center at all times, and that force can be supplied by whatever real forces are available — it's never a new, separate force of its own.
At the top of the swing, both gravity ($mg$, pointing down, which is toward the center at that point) and the normal force from the bucket bottom ($N$, also pushing down on the water, toward the center) point the same direction:
$$mg + N = \frac{mv^2}{r}$$
As long as the bucket is moving fast enough that $\dfrac{mv^2}{r}$ is at least as big as $mg$ alone, the normal force $N$ works out to zero or positive — meaning the bucket bottom is still pushing on the water (or just barely not), and the water has no reason to fall out. It's exactly the same reasoning as the top of the vertical loop from our lesson, just with water and a bucket standing in for a roller coaster car.
At the bottom of the swing, gravity points down (away from the center, which is now up) while the normal force points up (toward the center) — so this time they subtract:
$$N - mg = \frac{mv^2}{r} \implies N = mg + \frac{mv^2}{r}$$
Notice $N$ is bigger than $mg$ at the bottom — which is why the bucket (and the water) feels heaviest right at the bottom of the swing, and why the water feels like it's being pressed into the bucket rather than pulled out.
Minimum Speed
Just like the roller coaster car at the top of the loop, there's a minimum speed for the water to stay in at the top of the swing. At that minimum speed, the normal force from the bucket drops to exactly zero — gravity alone is providing all of the centripetal force needed:
$$mg = \frac{mv_{min}^2}{r} \implies v_{min} = \sqrt{gr}$$
Just like before, the mass cancels out completely — it doesn't matter how much water is in the bucket, only the radius of the swing.
Practice Problems
- A bucket is swung in a vertical circle of radius $0.75\text{ m}$. Find the minimum speed at the top of the swing needed to keep the water from falling out.
- A $2.0\text{ kg}$ bucket (water included) swings in a vertical circle of radius $0.60\text{ m}$ at a constant speed of $4.5\text{ m/s}$. Find the normal force the bucket bottom exerts on the water at (a) the top of the swing and (b) the bottom of the swing.
Try It: Swing the Bucket
Find the minimum speed and the string tension for a bucket swung in a vertical circle. Dynamics Dungeon: Swing the Bucket