Circular motion

An object moving in a circle has a velocity that always points along the path: tangent to the circle. Even at constant speed, the direction of that velocity is changing every instant. A changing velocity means acceleration.

The acceleration that bends the path always points toward the center of the circle. It is called the centripetal acceleration, $\vec a_c$.

Changing direction is changing velocity, and changing velocity is acceleration. That is the whole story.

The size of the centripetal acceleration depends on speed and radius:

$$a_c = \dfrac{v^2}{r}.$$

A car going $v = 20$ m/s around a curve of radius $r = 50$ m has $a_c = (20)^2 / 50 = 8$ m/s$^2$, pointing inward.

Speed shows up squared, so doubling the speed at the same radius gives $4\times$ the acceleration. Doubling the radius at the same speed cuts $a_c$ in half. Tighter curves at higher speeds need more inward acceleration, which is why fast turns are wide or banked.

The most common FBD mistake on circular-motion problems is adding a separate arrow labelled "centripetal force" on top of tension, gravity, normal, or friction. Don't. Centripetal names a direction (toward the center), not a new force.

Only the real forces go on the FBD. Newton's second law in the radial direction reads

$$F_{\text{net, radial}} = \dfrac{m v^2}{r}.$$

The left side is the radial sum of the forces already on your FBD. That sum plays the role of the centripetal force; nothing new gets added.

Different setups, same idea. Friction does the job for a car on a flat curve. The horizontal piece of the tension does it for a ball on a string in a horizontal circle. The horizontal piece of the normal force does it for a car on a frictionless banked curve. Gravity does it for a satellite in orbit.

A car of mass $m$ rounds a flat circular curve of radius $r$ at constant speed $v$. Find the friction force.

Step 1: forces. Three forces act: gravity ($mg$ down), normal ($F_N$ up), and static friction ($f_s$ horizontal). The vertical pair cancels: $F_N = mg$. Friction is the only horizontal force, so it has to point toward the center.

Step 2: radial equation, with inward positive:

$$f_s = \dfrac{m v^2}{r}.$$

Step 3: solve. The friction force is exactly $m v^2 / r$. Static friction adjusts to whatever the situation requires, up to its cap of $\mu_s F_N = \mu_s m g$. As long as $m v^2 / r \le \mu_s m g$, the car holds the curve. Setting the two equal gives the maximum cornering speed: $v_{max} = \sqrt{\mu_s g r}$.

Note: friction here isn't a separate "centripetal force." It's the same static friction already on the FBD, playing the centripetal role.

A roller-coaster car moves on the inside of a vertical loop of radius $r$. At the very top, the car is upside down. Gravity pulls it straight down, which at this point is toward the center of the loop. The track pushes on the car from above, also down, also toward the center. Both forces point the same way.

Two traps live here. First trap: "the car is going slowest at the top, so the acceleration is small." Not how it works. The radial acceleration is $v^2 / r$, but it is also limited by which forces are available, and by the fact that the track can only push, never pull, so $F_N \ge 0$.

Second trap: "at minimum speed, the car is barely moving, so $\vec a \approx 0$." Speed and acceleration are independent. The velocity vector is still turning, and turning takes acceleration.

Cleanest way to see it: at minimum speed, $F_N = 0$, so gravity alone is bending the path. Gravity has not gone anywhere. The acceleration at the top equals $g$, exactly. Not zero.

A roller-coaster car rounds a vertical loop of radius $r = 5$ m. Find the minimum speed at the top such that the car stays in contact with the track.

Step 1: forces. At the top of the loop, two forces act: gravity ($mg$ down, toward the loop's center) and the normal force from the track ($F_N$, also down, since the track is above the inverted car). At minimum speed, the track is just barely in contact: $F_N = 0$. Only gravity remains.

Step 2: radial equation, with inward (down here) positive:

$$F_N + m g = \dfrac{m v^2}{r}.$$

Set $F_N = 0$:

$$m g = \dfrac{m v_{min}^2}{r} \quad\Rightarrow\quad v_{min}^2 = g r.$$

Step 3: solve. With $r = 5$ m and $g = 9.8$ m/s$^2$: $v_{min} = \sqrt{(9.8)(5)} \approx 7.0$ m/s.

The kicker: at this minimum-speed instant, $|\vec a| = g \approx 9.8$ m/s$^2$, not zero. The car is at its slowest, the track has stopped pushing, but the velocity vector is still bending around the curve. Gravity alone is exactly what it takes. Slowest speed and largest gravity-only acceleration, same instant.

Three failure modes show up over and over on circular-motion problems. Each one anchors a family of distractors on the AP exam.