where:

• $a_c$ = centripetal acceleration (m/s²)
• $v$ = speed (m/s)
• $r$ = radius of circular path (m)

Direction

• Always points toward the center of the circle
• Perpendicular to the velocity vector
• Changes direction as the object moves around the circle

Alternative Form

Using $v = \frac{2\pi r}{T}$ (where $T$ is period):

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

Also, using angular velocity $\omega = \frac{v}{r}$:

$a_c = \omega^2 r$

Period and Frequency

Period (T)

The period is the time for one complete revolution:

$T = \frac{2\pi r}{v}$

Units: seconds (s)

Frequency (f)

The frequency is the number of revolutions per second:

$f = \frac{1}{T}$

Units: hertz (Hz) or revolutions per second (rev/s)

Relationship Between v, r, T, and f

$v = \frac{2\pi r}{T} = 2\pi r f$

Angular Velocity

Angular velocity ($\omega$) measures how fast the angle changes:

$\omega = \frac{\Delta \theta}{\Delta t} = \frac{2\pi}{T} = 2\pi f$

Units: radians per second (rad/s)

Relationship to Linear Velocity

$v = r\omega$

This connects the speed along the circular path ($v$) to the angular velocity ($\omega$).

Common Scenarios

Scenario 1: Object on a String

A ball swung in a horizontal circle:

• Tension provides centripetal force
• $a_c = \frac{v^2}{r}$ points toward center
• If string breaks, object flies off tangent to circle (not radially outward!)

Scenario 2: Car Rounding a Curve

A car turning on a flat road:

• Friction provides centripetal force
• Maximum safe speed: $v_{max} = \sqrt{\mu_s g r}$
• Radius affects speed: tighter curves require slower speeds

Scenario 3: Satellite in Orbit

A satellite orbiting Earth:

• Gravity provides centripetal force
• $a_c = g$ at Earth's surface
• Orbital period depends on altitude

⚠️ Common Misconceptions

Misconception 1: "Centrifugal Force"

❌ Wrong: There's an outward "centrifugal force" on the object ✅ Right: There's no outward force. The object wants to move in a straight line (Newton's 1st Law), but centripetal force pulls it inward toward the center. "Centrifugal force" is a fictitious force felt in the rotating reference frame.

Misconception 2: Constant Velocity

❌ Wrong: Uniform circular motion has constant velocity ✅ Right: UCM has constant speed but changing velocity (because direction changes)

Misconception 3: Acceleration and Speed

❌ Wrong: If speed is constant, acceleration must be zero ✅ Right: Acceleration can be perpendicular to velocity, changing direction without changing speed

Misconception 4: Direction After Release

❌ Wrong: If the string breaks, the object flies radially outward ✅ Right: The object flies off tangent to the circle (in the direction of instantaneous velocity)

Problem-Solving Strategy

1. Draw a diagram showing the circular path and center
2. Identify the radius of the circular path
3. Find or calculate the speed (may need to use $v = \frac{2\pi r}{T}$ or $v = r\omega$)
4. Calculate centripetal acceleration: $a_c = \frac{v^2}{r}$
5. Direction: Always toward the center

Key Equations Summary

📝 Important Notes

• Centripetal acceleration exists even though speed is constant
• The acceleration changes direction continuously (always pointing toward center)
• Period and frequency are inversely related: $f = \frac{1}{T}$
• For a given radius, higher speed requires greater centripetal acceleration
• All points on a rigid rotating object have the same angular velocity but different linear velocities

Find: Centripetal acceleration $a_c$

Solution:

Use the centripetal acceleration formula:

$a_c = \frac{v^2}{r}$

Substitute the values:

$a_c = \frac{(20 \text{ m/s})^2}{50 \text{ m}}$

$a_c = \frac{400 \text{ m}^2/\text{s}^2}{50 \text{ m}}$

$a_c = 8 \text{ m/s}^2$

Answer: The centripetal acceleration is 8 m/s² directed toward the center of the circular track.

Note: This acceleration is less than $g$ (9.8 m/s²), so friction alone could provide this if the coefficient is sufficient.