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6.2 Uniform Circular Motion

6.2 Uniform Circular Motion

Written by the Fiveable Content Team • Last updated August 2025
Written by the Fiveable Content Team • Last updated August 2025
⚾️Honors Physics
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Uniform Circular Motion

Uniform circular motion describes what happens when an object moves along a circular path at constant speed. Even though the speed stays the same, the velocity is always changing because the direction of motion is constantly shifting. That continuous change in direction requires a net force pointed toward the center of the circle, and understanding that force is the core of this topic.

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Centripetal Acceleration and Velocity

In uniform circular motion, the object's velocity vector is always tangent to the circular path. Think of a ball on a string: at any instant, if you cut the string, the ball would fly off in a straight line tangent to the circle. The velocity's magnitude (speed) doesn't change, but its direction does, and any change in velocity means there's an acceleration.

That acceleration is centripetal acceleration (aca_c), and it always points toward the center of the circular path. Because it's perpendicular to the velocity at every moment, it only redirects the motion without speeding the object up or slowing it down.

The magnitude of centripetal acceleration depends on two things:

ac=v2ra_c = \frac{v^2}{r}

  • Higher speed → larger aca_c (and it scales with the square of speed, so doubling your speed quadruples the acceleration)
  • Smaller radius → larger aca_c (a tighter curve demands more acceleration to keep bending the path)
Centripetal acceleration and velocity, File:Centripetal force diagram.svg - Wikipedia

Forces in Uniform Circular Motion

Newton's second law still applies here. The net force on an object in uniform circular motion is called the centripetal force (FcF_c), and it points toward the center of the circle:

Fc=mac=mv2rF_c = ma_c = \frac{mv^2}{r}

A critical point: centripetal force is not a new type of force. It's just the label for whatever real force (or combination of forces) happens to be directed toward the center. Without a continuous inward force, the object's inertia would carry it off in a straight line.

Here are the most common forces that can act as the centripetal force:

  • Tension — A ball on a string swung in a horizontal circle. The string's tension pulls the ball inward. Examples: tetherball, a yo-yo spun overhead.
  • Static friction — A car rounding a curve on a flat road. Static friction between the tires and the road surface pushes the car toward the center of the turn. If friction isn't strong enough (icy road, too much speed), the car slides outward.
  • Gravity — A satellite orbiting Earth. The gravitational pull between the satellite and Earth provides the centripetal force that keeps the satellite in orbit. GPS and weather satellites stay in circular orbits this way.

In many real situations, more than one force contributes. A car on a banked curve, for instance, gets centripetal force from components of both friction and the normal force.

Centripetal acceleration and velocity, Uniform Circular Motion – University Physics Volume 1

Applying the Centripetal Equations

When solving circular motion problems, follow these steps:

  1. Identify the force(s) providing the centripetal acceleration. Draw a free-body diagram and determine which force (tension, friction, gravity, normal force, or a combination) points toward the center of the circle.
  2. List your known variables: mass (mm), speed (vv), and radius (rr).
  3. Set the centripetal force equal to the real force(s) directed toward the center, then solve for the unknown.

The two key equations you'll use:

  • ac=v2ra_c = \frac{v^2}{r}
  • Fc=mv2rF_c = \frac{mv^2}{r}

Example Problem: A 1000 kg car makes a turn with a radius of 50 m at a speed of 15 m/s. Calculate the centripetal force and identify the force responsible.

  1. Known: m=1000 kgm = 1000 \text{ kg}, v=15 m/sv = 15 \text{ m/s}, r=50 mr = 50 \text{ m}
  2. Apply the formula: Fc=mv2r=(1000 kg)(15 m/s)250 m=1000×22550=4500 NF_c = \frac{mv^2}{r} = \frac{(1000 \text{ kg})(15 \text{ m/s})^2}{50 \text{ m}} = \frac{1000 \times 225}{50} = 4500 \text{ N}
  3. On a flat road, the force providing this 4500 N toward the center is static friction between the tires and the road.

Notice that if the car doubled its speed to 30 m/s, the required force would jump to 18,000 N (four times as much), because force scales with v2v^2. That's why sharp turns at high speed are dangerous.

Special Cases of Circular Motion

Banked curves are roads or tracks tilted at an angle so that a component of the normal force points toward the center of the curve. This reduces (or even eliminates) the reliance on friction. At the "ideal" banking angle, a car can make the turn with no friction at all. You find this angle by setting the inward component of the normal force equal to the required centripetal force.

Conical pendulum describes a mass on a string that swings in a horizontal circle while the string traces out a cone shape. The horizontal component of the string's tension provides the centripetal force, while the vertical component balances gravity. Analyzing it requires breaking the tension into components and applying Newton's second law in both the horizontal and vertical directions.