Key Characteristics

1. Speed is constant - the magnitude of velocity doesn't change
2. Velocity is NOT constant - the direction is continuously changing
3. There MUST be acceleration - changing direction means changing velocity
4. Acceleration points toward the center - called centripetal acceleration

💡 Critical Insight: Even though speed is constant, there IS acceleration because velocity is a vector (has direction). Changing direction = changing velocity = acceleration!

Centripetal Acceleration

The acceleration that points toward the center of the circular path is called centripetal acceleration (meaning "center-seeking").

Formula

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

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

(a) Find the period $T$:

The period is the time for one revolution:

$T = \frac{1}{f} = \frac{1}{2 \text{ Hz}} = 0.5 \text{ s}$

(b) Find the speed $v$:

Use the relationship between speed, radius, and period:

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

$v = \frac{2\pi (1.2 \text{ m})}{0.5 \text{ s}}$

$v = \frac{2.4\pi \text{ m}}{0.5 \text{ s}}$

$v = 4.8\pi \text{ m/s} \approx 15.1 \text{ m/s}$

(c) Find the centripetal acceleration $a_c$:

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

Using exact value $v = 4.8\pi$ m/s:

$a_c = \frac{(4.8\pi)^2}{1.2}$

$a_c = \frac{23.04\pi^2}{1.2}$

$a_c = 19.2\pi^2 \approx 189.5 \text{ m/s}^2$

Alternative method using $a_c = 4\pi^2 r f^2$:

$a_c = 4\pi^2 (1.2)(2)^2 = 4\pi^2 (1.2)(4) = 19.2\pi^2 \approx 189.5 \text{ m/s}^2$

Answers:

• (a) Period: 0.5 s
• (b) Speed: 15.1 m/s
• (c) Centripetal acceleration: 189.5 m/s² (toward center)

Note: This is about 19 times the acceleration due to gravity!

Find: Period $T$

Strategy: We need to work backwards from $a_c$ to find $v$, then use $v$ to find $T$.

Step 1: Find the required speed

From $a_c = \frac{v^2}{r}$, solve for $v$:

$v^2 = a_c \cdot r$

$v = \sqrt{a_c \cdot r}$

$v = \sqrt{(9.8 \text{ m/s}^2)(100 \text{ m})}$

$v = \sqrt{980 \text{ m}^2/\text{s}^2}$

$v = 31.3 \text{ m/s}$

Step 2: Find the period

From $v = \frac{2\pi r}{T}$, solve for $T$:

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

$T = \frac{2\pi (100 \text{ m})}{31.3 \text{ m/s}}$

$T = \frac{200\pi}{31.3} \text{ s}$

$T \approx 20.1 \text{ s}$

Alternative Method: Using $a_c = \frac{4\pi^2 r}{T^2}$

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

$T = 2\pi\sqrt{\frac{r}{a_c}}$

$T = 2\pi\sqrt{\frac{100}{9.8}}$

$T = 2\pi\sqrt{10.2}$

$T = 2\pi(3.20) \approx 20.1 \text{ s}$

Answer: The period of rotation should be approximately 20.1 seconds.

Interpretation: The station completes one rotation about every 20 seconds, creating a centripetal acceleration of $9.8$ m/s² at the outer edge, which would feel like Earth's gravity to people standing on the outer rim.