Centripetal Force in Uniform Circular Motion

An object moving in a uniform circle at constant speed is constantly changing direction — and a change in velocity (even just direction) means there must be a net force. That force is called centripetal force, and it always points toward the center of the circle. It's not a "type" of force like gravity or friction — it's whatever combination of real forces keeps the object on its circular path.

The Formula

For an object of mass $m$ moving in a circle of radius $r$ at speed $v$:

$F_c = \dfrac{m\,v^{2}}{r}$

This is the magnitude of the net inward force required to maintain the circular motion. The direction is always perpendicular to the velocity, pointing toward the center.

Step-by-Step Solution

Given: $m = 2\;\text{kg}$, $r = 1\;\text{m}$, $v = 5\;\text{m/s}$.

Find: The centripetal force $F_c$.

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Step 1 — Plug into the formula.

$F_c = \dfrac{m\,v^{2}}{r}$

Step 2 — Compute $v^{2}$.

$v^{2} = (5)^{2} = 25\;\text{m}^{2}/\text{s}^{2}$

Step 3 — Multiply by mass.

$m\,v^{2} = 2 \times 25 = 50\;\text{kg}\cdot\text{m}^{2}/\text{s}^{2}$

Step 4 — Divide by radius.

$F_c = \dfrac{50}{1} = 50\;\text{N}$

Step 5 — Compute the centripetal acceleration.

$a_c = \dfrac{v^{2}}{r} = \dfrac{25}{1} = 25\;\text{m/s}^{2}$

That's about 2.5 g of acceleration — like riding a hard-banking turn. The ball is constantly changing direction at this rate, even though its speed is constant.

Step 6 — Compare to a slower motion.

If you slowed the ball to half the speed ($v = 2.5\;\text{m/s}$):

$F_c = \dfrac{2(6.25)}{1} = 12.5\;\text{N}$

The force drops by a factor of 4 when you halve the speed — because $v$ enters the formula squared. To double the speed of the ball requires four times the force.

Step 7 — How does it scale with $r$?

If you doubled the radius (keeping speed and mass the same):

$F_c = \dfrac{2(25)}{2} = 25\;\text{N}$

Doubling the radius halves the centripetal force — gentler turning radius, less force needed to bend the path.

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Answer:

$\boxed{\;F_c = \dfrac{mv^{2}}{r} = 50\;\text{N}\;}$

The centripetal acceleration is $a_c = 25\;\text{m/s}^{2}$ — about 2.5 times the acceleration of gravity. Without this constant inward force (e.g., the tension in a string, gravity for an orbit, friction for a car turning), the ball would shoot off in a straight line tangent to the circle.

Try It

• Adjust the mass, radius, and speed sliders.
• Watch the centripetal force vector rotate around to always point toward the center.
• Try doubling the speed — the force quadruples.
• Try doubling the radius (with constant speed) — the force halves.