If an object has Centripetal Acceleration, the second law says a net force must point toward the center too. That force isn't a new kind of force. Some real push or pull, one you can name, does the job.
Key idea
For motion in a circle, the net inward force is $\Sigma F_{in} = \dfrac{mv^2}{r}$. Tension, friction, gravity or a normal force supplies it. Nothing pushes outward.
Why it works
Without an inward force, an object moves in a straight line (the first law). The inward force keeps bending its path. On a flat road, the only horizontal force on a car is static friction from the road, so friction must supply all of $mv^2/r$; that sets a top speed, $v = \sqrt{\mu_s g r}$. For the Moon, Earth's gravity does the job; for a ball on a string, the tension does.
Worked examples
Worked example: Tension in a string
A $0.5\ \text{kg}$ puck on frictionless ice circles a post at $4\ \text{m/s}$ on a $1.2\ \text{m}$ string. Find the tension.
- The string is the only horizontal force: $T = \dfrac{mv^2}{r} = \dfrac{(0.5)(16)}{1.2}$.
- $T \approx 6.67\ \text{N}$.
Answer: about $6.67\ \text{N}$
Worked example: Top speed on a flat curve
How fast can a car take a flat curve of radius $40\ \text{m}$ if $\mu_s = 0.7$?
- At the limit, $\mu_s mg = \dfrac{mv^2}{r}$. The mass cancels.
- $$v = \sqrt{\mu_s g r} = \sqrt{0.7 \times 9.8 \times 40} \approx 16.6\ \text{m/s}.$$
Answer: about $16.6\ \text{m/s}$
Common mistake
Adding an outward "centrifugal force" to the free-body diagram, or thinking a released ball flies straight outward. In a turning car you feel pushed outward because your body tends to keep going straight while the car turns under you. Check: every arrow on the diagram must be exerted by a real object, and a released object moves along the tangent.
Try it in the Science Lab: Circles and Gravity.