A2-7. Kinematics: Uniform Circular Motion

1. Speed, Velocity, and Period

Uniform Circular Motion (UCM) occurs when an object travels in a circular path at a constant speed. Even though the speed is constant, the velocity is constantly changing because the direction of motion is continuously altering. Because velocity is changing, the object must be accelerating.

Orbital Speed ($v$):

The speed of an object moving in a circle is the distance of one full circumference divided by the time it takes to complete one full revolution (the Period, $T$).

$$v = \frac{2\pi r}{T}$$
  • $v$: Tangential (linear) speed (m/s)
  • $r$: Radius of the circular path (m)
  • $T$: Period, the time for one complete revolution (s)

2. Centripetal Acceleration & Force

The acceleration that continuously turns the object into a circular path is called centripetal acceleration (meaning "center-seeking"). According to Newton's Second Law, this acceleration must be caused by a net force directed toward the center, known as the centripetal force.

Centripetal Acceleration ($a$):

Acceleration is directed exactly toward the center of the circle, perpendicular to the tangential velocity.

$$a = \frac{v^2}{r} = \frac{4\pi^2 r}{T^2}$$

Centripetal Force ($F$):

By substituting the centripetal acceleration equations into $F = ma$, we get the formulas for the net force required to maintain circular motion.

$$F = \frac{mv^2}{r} = \frac{m 4\pi^2 r}{T^2}$$
  • $F$: Centripetal force (N)
  • $m$: Mass of the orbiting object (kg)
  • Note: Centripetal force is not a new "type" of force. It is simply the net sum of real forces (like tension, gravity, or friction) that point toward the center of rotation.

Example 2: Tension in a Whirling String

Problem: A 0.20 kg ball is attached to a string and whirled in a horizontal circle of radius 0.80 m. It completes exactly one revolution every 1.5 s. Calculate the tension in the string.

Solution:

  • Step 1: Identify the given variables.
    $m = 0.20\text{ kg}$, $r = 0.80\text{ m}$, $T = 1.5\text{ s}$
  • Step 2: Recognize the source of the force.
    The tension in the string provides the necessary centripetal force ($F$).
  • Step 3: Select the appropriate formula and calculate.
    Since we are given the period ($T$), we use the period-based force formula:
    $$F = \frac{m 4\pi^2 r}{T^2} = \frac{(0.20)(4\pi^2)(0.80)}{(1.5)^2} = \frac{6.316}{2.25} \approx \mathbf{2.81\text{ N}}$$