Roller coasters, dips in the road, a bucket swung overhead: here the circle stands upright, and gravity takes part. At the top and bottom of the circle, gravity and the other force line up with the center, so Forces in Circular Motion works along one vertical axis.
Key idea
At the top or bottom, take "toward the center" as positive and write $\Sigma F_{toward} = \dfrac{mv^2}{r}$. At the bottom of a dip, $F_N = m\left(g + \dfrac{v^2}{r}\right)$; at the top of a hill, $F_N = m\left(g - \dfrac{v^2}{r}\right)$.
Why it works
At the bottom of a dip the center is above you, so the net force must point up: the seat pushes up harder than gravity pulls down, and you feel heavier. Over the top of a hill the center is below, so the net force points down: the seat pushes up less than gravity, and you feel lighter. If $v^2/r$ reaches $g$, the seat pushes nothing at all and you lift off. At the top of a loop, both gravity and the track push toward the center, so the smallest speed that keeps contact is $v = \sqrt{gr}$.
Worked examples
Worked example: Bottom of a dip
A $60\ \text{kg}$ rider passes through a dip of radius $20\ \text{m}$ at $12\ \text{m/s}$. Find the seat's normal force.
- Center above: $F_N - mg = \dfrac{mv^2}{r}$.
- $$F_N = 60\left(9.8 + \dfrac{144}{20}\right) = 60(9.8 + 7.2) = 1{,}020\ \text{N}.$$
- That is $1.7$ times the rider's weight of $588\ \text{N}$.
Answer: $1{,}020\ \text{N}$
Worked example: Top of a hill
The same rider goes over a hilltop of radius $20\ \text{m}$ at $10\ \text{m/s}$.
- Center below: $mg - F_N = \dfrac{mv^2}{r}$.
- $$F_N = 60\left(9.8 - \dfrac{100}{20}\right) = 60(4.8) = 288\ \text{N},$$ so the rider feels lighter.
Answer: $288\ \text{N}$
Worked example: Top of a loop
What is the slowest a car can go at the top of a loop of radius $10\ \text{m}$ and stay on the track?
- At the top, $mg + F_N = \dfrac{mv^2}{r}$. The slowest case has $F_N = 0$.
- $v = \sqrt{gr} = \sqrt{98} \approx 9.90\ \text{m/s}$.
Answer: about $9.90\ \text{m/s}$
Common mistake
Thinking the normal force at the bottom of a dip equals $mg$. It must be bigger, because the net force there points up toward the center. Check: you feel pressed into the seat at the bottom of a dip, and light at the top of a hill. Your answer should agree.