Topic 2.10 Circular Motion: relate centripetal acceleration to speed and radius, identify the real force that supplies the centripetal force, and apply Newton's second law to circular motion including vertical circles.

What this topic is asking

The College Board (Topic 2.10) wants you to relate the centripetal acceleration to speed and radius, to identify the real force that supplies the centripetal force, and to apply Newton's second law along the radial direction. The crucial conceptual point, and the source of the classic trap, is that centripetal force is not a new force but the net inward force provided by tension, gravity, friction or the normal force, depending on the setup.

Centripetal acceleration

Even at constant speed, circular motion is accelerated motion, because the direction of the velocity is always changing, and velocity is a vector. The acceleration points toward the center, perpendicular to the velocity, which is why it changes the direction of motion without changing the speed. The $v^2/r$ dependence means doubling the speed quadruples the required acceleration, while a tighter turn (smaller $r$) also raises it. If the speed is also changing, there is an additional tangential acceleration along the velocity, but uniform circular motion has only the centripetal part.

The centripetal force is a real force

The centripetal acceleration requires a net inward force by Newton's second law:

directed toward the center. The essential point is that "centripetal force" is a role, not a new kind of force. Some real force, tension in a string, gravity on a satellite, friction on a car's tires, the normal force on a passenger in a loop, plays that role by pointing inward. On a free-body diagram you draw only the real forces; the centripetal force is their net inward sum, not a separate arrow. There is no outward centrifugal force acting on the object in an inertial frame: the sensation of being "thrown outward" is just inertia carrying you straight while the seat pushes you inward.

Applying Newton's second law radially

The method for any circular-motion problem is to draw the free-body diagram, pick the radial direction (inward positive), and set the net radial force equal to $mv^2/r$. On a flat curve, static friction supplies the force: $\mu_s mg = \dfrac{mv^2}{r}$ gives the maximum cornering speed. For a conical pendulum or banked turn, a component of the tension or normal force points inward. The tangential direction usually has no net force in uniform circular motion (constant speed), so the radial equation does the work.

Vertical circles

A vertical circle is the richest case because gravity's role changes around the loop. At the top, both gravity and the tension (or normal force) point downward, toward the center, so $T + mg = \dfrac{mv^2}{r}$. The string goes slack when $T = 0$, giving the minimum speed at the top, $v_{min} = \sqrt{gr}$, at which gravity alone provides the centripetal force. At the bottom, the tension points up (inward) while gravity points down (outward), so $T - mg = \dfrac{mv^2}{r}$ and the tension is largest there. Combining these radial equations with energy conservation between top and bottom is a standard multi-part FRQ.

Try this

Q1. A $0.30$ kg ball on a $1.2$ m string moves in a horizontal circle at $4.0$ m/s. Calculate the tension in the string (ignore gravity). [2 points]

• Cue. $T = \dfrac{mv^2}{r} = \dfrac{(0.30)(4.0)^2}{1.2} = \dfrac{4.8}{1.2} = 4.0$ N.

Q2. State and explain the minimum speed at the top of a vertical circle of radius $r$ for an object on a string. [2 points]

• Cue. $v_{min} = \sqrt{gr}$: when the tension drops to zero, gravity alone provides $mv^2/r$, giving $mg = mv^2/r$.