Circular motion

In uniform circular motion an object goes around a circle at a steady speed. The speed is constant, but the velocity is not: velocity is a vector, and its direction keeps changing all the way around. A changing velocity is an acceleration, so a circling object is accelerating even when its speed never changes.

That acceleration points toward the center of the circle. It is the centripetal acceleration, $a_c = v^2/r$, or $a_c = \omega^2 r$ with $\omega = v/r$ the angular speed. The faster the object or the tighter the circle, the larger the inward acceleration.

By Newton’s second law a centripetal acceleration needs an inward net force, the centripetal force $F_{net} = mv^2/r$. The rest of this lesson is about what that force really is, what it is not, and the three habits that lose points: reading constant speed as zero acceleration (§2), drawing centripetal force as a separate arrow (§3), and inventing an outward force (§4).

We keep the speed constant, so the acceleration is purely inward. If the speed also changed there would be a forward or backward piece too, but the inward $v^2/r$ is always there. Throughout, $g = 10$ m/s$^2$.

The most common circular-motion mistake is to say the acceleration is zero because the speed is constant. Speed is a scalar; velocity is a vector. The velocity changes whenever its direction changes, and in a circle the direction is always changing, so the acceleration is never zero.

A satellite in a circular orbit at steady speed is accelerating the whole time. That inward acceleration is what bends its straight-line path into an orbit. Zero acceleration would mean a constant velocity, which is straight-line motion at steady speed, not a circle at all.

The size of the acceleration is $a_c = v^2/r$, directed toward the center. Constant speed only means the size of $a_c$ is steady; its direction keeps turning to stay pointed inward. So “the speed is constant” does not settle whether the object accelerates: it does.

A quick tell: any claim that an object “is not accelerating because its speed is constant,” or that it “accelerates only when it speeds up or slows down,” is the speed-for-velocity slip.

“Centripetal” means toward the center. The centripetal force is the name for the net inward force, $F_{net} = mv^2/r$. It is not a new kind of force and not an extra arrow on the free-body diagram. It is the job done by whatever real force already acts.

On a puck whirled on a string, the string tension is the centripetal force. For a satellite, gravity is. For a car on a flat curve, static friction is. On a banked curve, the inward component of the normal force is. In every case you draw the real forces, and their inward sum is the centripetal force.

The mistake is to list the real forces and then add one more arrow labeled “centripetal force.” That double-counts. Either a real force is the centripetal force, or there is no separate centripetal arrow. Treating centripetal force as a fundamental force, on the same list as gravity and tension, is the same error.

To apply Newton’s second law, add up the real forces and set their inward sum equal to $mv^2/r$: $\sum F_{\text{in}} = mv^2/r$. The centripetal force is the right-hand side, not an added term on the left.

In the ground (inertial) frame, the frame this course uses, the net force on an object going around a curve points inward, toward the center, and there is no outward force. The skill the College Board names here is to describe the motion of an object in uniform circular motion, and the inward net force is the whole story.

The strong feeling of being thrown outward in a turning car is inertia, not a force. Your body tends to keep moving in a straight line while the car turns inward beneath you, so the door pushes you inward. The inward push from the door is the real force; the outward feeling is your inertia, and it does not belong on the diagram.

A diagram that pairs the correct inward force with a matching outward “centrifugal force” is tempting because the arrows look balanced. But balanced forces give zero net force, and zero net force means straight-line motion, not a circle. The fact that the object turns is proof the net force is inward.

(A spinning frame adds a centrifugal term, but that is a different frame from the one this course uses. Keep the diagram in the ground frame: on a flat curve the only horizontal force is the inward one.)

The centripetal acceleration is $a_c = v^2/r$, or $\omega^2 r$, and the matching force is $F_{net} = mv^2/r$. Two standard setups show real forces playing that role.

Vertical loop. At the top of a vertical loop, both gravity and the normal force point downward, toward the center, so $N + mg = mv^2/r$. The slowest speed that keeps the object on the track is where $N \to 0$: then $mg = mv_{\text{top}}^2/r$, giving $v_{\text{top}} = \sqrt{gr}$. At the bottom the normal force points up (inward) and gravity down (outward), so $N - mg = mv^2/r$.

Banked curve. On a frictionless curve banked at angle $\theta$, the inward component of the normal force supplies the centripetal force; balancing the vertical direction gives $\tan\theta = v^2/(rg)$. A larger speed or a tighter radius needs a steeper bank.

The recipe never changes: identify the real forces, take their components toward the center, and set the inward sum equal to $mv^2/r$. The centripetal force is always that sum, not a separate arrow, and the acceleration it produces is inward and nonzero even at constant speed.

Setup. A $0.5$ kg ball is swung in a vertical circle of radius $r = 0.4$ m on a string. Take $g = 10$ m/s$^2$. (a) Find the minimum speed at the top that keeps the string taut. (b) At that speed, find the string tension and the ball’s acceleration at the top. (c) Identify the centripetal force, and say whether the ball is accelerating.

(a) Minimum speed at the top. At the top, gravity and tension both point toward the center: $T + mg = mv^2/r$. The string goes slack when $T \to 0$, so the minimum is $mg = mv_{\text{min}}^2/r$, giving $v_{\text{min}} = \sqrt{gr} = \sqrt{10 \cdot 0.4} = 2$ m/s.

(b) Tension and acceleration there. At $v = v_{\text{min}} = 2$ m/s the tension is $T = mv^2/r - mg = (0.5)(4)/0.4 - (0.5)(10) = 5 - 5 = 0$ N, exactly the slack point. The acceleration is $a = v^2/r = 4/0.4 = 10$ m/s$^2$ toward the center; here it equals $g$, because gravity alone is supplying the whole centripetal force.

(c) What the centripetal force is. The centripetal force at the top is the net inward force $mv^2/r = 5$ N, supplied by gravity (and by tension as well once the ball moves faster), not by a separate arrow. The ball is accelerating because its velocity direction is turning. In a vertical circle the speed is not even constant: it is slowest here at the top and fastest at the bottom, so this is not uniform circular motion, but the inward acceleration is present at every point. It stays on the circle because the inward force provides exactly $mv^2/r$; nothing pushes it outward.

The traps to avoid. Do not add a separate centripetal arrow to the gravity-and-tension diagram, do not invent an outward force to “balance” gravity at the top, and do not call the acceleration zero just because the speed is momentarily unchanging at the top.