If you swing a bucket of water over your head quickly, the water stays in, but you’ll get a shower if you swing too slowly. Why? We’ll answer this question by starting with an equivalent situation, a roller coaster doing a loop-the-loop.

FIGURE 8.14 shows a roller-coaster car going around a vertical loop-the-loop of radius $r$. Why doesn’t the car fall off at the top of the circle? Now, motion in a vertical circle is not uniform circular motion; the car slows down as it goes up one side and speeds up as it comes back down the other. But at the very top and very bottom points, only the car’s direction is changing, not its speed, so at those points the acceleration is purely centripetal. Thus there must be a net force toward the center of the circle.

FIGURE 8.14 A roller-coaster car going around a loop-the-loop.

First consider the very bottom of the loop. To have a net force toward the center—upward at this point—requires $n>F_{\text{G}}$. The normal force has to exceed the gravitational force to provide the net force needed to “turn the corner” at the bottom of the circle. This is why you “feel heavy” at the bottom of the circle or at the bottom of a valley on a roller coaster.

We can analyze the situation quantitatively by writing the r-component of Newton’s second law. At the bottom of the circle, with the r-axis pointing upward, we have

From Equation 8.16 we find

The normal force at the bottom is larger than mg.

Things are a little trickier as the roller-coaster car crosses the top of the loop. Whereas the normal force of the track pushes up when the car is at the bottom of the circle, it presses down when the car is at the top and the track is above the car. Think about the free-body diagram to make sure you agree.

The car is still moving in a circle, so there must be a net force toward the center of the circle. The r-axis, which points toward the center of the circle, now points downward. Consequently, both forces have positive components. Newton’s second law at the top of the circle is

Thus at the top the normal force of the track on the car is

The normal force at the top can exceed $mg$ if $v_{\text{top}}$ is large enough. Our interest, however, is in what happens as the car goes slower and slower. As $v_{\text{top}}$ decreases, there comes a point when $n$ reaches zero. “No normal force” means “no contact,” so at that speed, the track is not pushing against the car. Instead, the car is able to complete the circle because gravity alone provides sufficient centripetal acceleration.

The speed at which $n=0$ is called the critical speed $v_{\text{c}}\text{:}$

The critical speed is the slowest speed at which the car can complete the circle. Equation 8.19 would give a negative value for $n$ if $v<v_{\text{c}}$, but that is physically impossible. The track can push against the wheels of the car $(n>0)$, but it can’t pull on them. If $v<v_{\text{c}}$, the car cannot turn the full loop but, instead, comes off the track and becomes a projectile! FIGURE 8.15 summarizes our reasoning.

FIGURE 8.15 A roller-coaster car at the top of the loop.

Water stays in a bucket swung over your head for the same reason: Circular motion requires a net force toward the center of the circle. At the top of the circle—if you swing the bucket fast enough—the bucket adds to the force of gravity by pushing down on the water, just like the downward normal force of the track on the roller-coaster car. As long as the bucket is pushing against the water, the bucket and the water are in contact and thus the water is “in” the bucket. As you swing slower and slower, requiring the water to have less and less centripetal acceleration, the bucket-on-water normal force decreases until it becomes zero at the critical speed. At the critical speed, gravity alone provides sufficient centripetal acceleration. Below the critical speed, gravity provides too much downward force for circular motion, so the water leaves the bucket and becomes a projectile following a parabolic trajectory toward your head!