Uniform Circular Motion

Centripetal acceleration and circular motion kinematics

🔄 Uniform Circular Motion

What is Uniform Circular Motion?

Uniform Circular Motion (UCM) occurs when an object moves in a circular path at constant speed.

Key Characteristics

Speed is constant - the magnitude of velocity doesn't change
Velocity is NOT constant - the direction is continuously changing
There MUST be acceleration - changing direction means changing velocity
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

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

Key Equations Summary

| Quantity | Formula | Units | |----------|---------|-------| | Centripetal acceleration | $a_c = \frac{v^2}{r} = \omega^2 r$ | m/s² | | Speed | $v = \frac{2\pi r}{T} = r\omega$ | m/s | | Period | $T = \frac{2\pi r}{v}$ | s | | Frequency | $f = \frac{1}{T}$ | Hz | | Angular velocity | $\omega = \frac{2\pi}{T} = 2\pi f = \frac{v}{r}$ | rad/s |

📝 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

📚 Practice Problems

1Problem 1medium

❓ Question:

A car travels at a constant speed of 20 m/s around a circular track with radius 50 m. (a) What is the car's centripetal acceleration? (b) What is the period of one complete lap? (c) What is the frequency of rotation?

💡 Show Solution

Solution:

Given: v = 20 m/s, r = 50 m

(a) Centripetal acceleration: a_c = v²/r = (20)²/50 = 400/50 = 8.0 m/s² (toward center)

(b) Period: Circumference = 2πr = 2π(50) = 100π m Period T = distance/speed = 100π/20 = 5π = 15.7 s

Note: Even though speed is constant, velocity changes direction, creating acceleration toward the center.

2Problem 2easy

❓ Question:

A car travels around a circular track with a radius of 50 m at a constant speed of 20 m/s. What is the magnitude of the car's centripetal acceleration?

💡 Show Solution

Given Information:

Radius: $r = 50$ m
Speed: $v = 20$ m/s
Motion: uniform circular motion (constant speed)

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.

3Problem 3medium

❓ Question:

💡 Show Solution

Solution:

Given: v = 20 m/s, r = 50 m

(a) Centripetal acceleration: a_c = v²/r = (20)²/50 = 400/50 = 8.0 m/s² (toward center)

(b) Period: Circumference = 2πr = 2π(50) = 100π m Period T = distance/speed = 100π/20 = 5π = 15.7 s

Note: Even though speed is constant, velocity changes direction, creating acceleration toward the center.

4Problem 4medium

❓ Question:

A 0.50 kg ball on a string is whirled in a horizontal circle of radius 1.2 m. The ball makes 2.0 revolutions per second. (a) What is the ball's speed? (b) What is the centripetal acceleration? (c) What is the tension in the string?

💡 Show Solution

Solution:

Given: m = 0.50 kg, r = 1.2 m, f = 2.0 rev/s

(a) Speed: Period T = 1/f = 1/2.0 = 0.50 s Circumference = 2πr = 2π(1.2) = 2.4π m v = 2πr/T = 2.4π/0.50 = 4.8π = 15.1 m/s

(b) Centripetal acceleration: a_c = v²/r = (15.1)²/1.2 = 228/1.2 = 190 m/s²

Or: a_c = 4π²r/T² = 4π²(1.2)/(0.50)² = 190 m/s² ✓

(c) Tension: For horizontal circular motion, tension provides centripetal force: F_c = ma_c T = 0.50 × 190 = 95 N

5Problem 5medium

❓ Question:

💡 Show Solution

Solution:

Given: m = 0.50 kg, r = 1.2 m, f = 2.0 rev/s

(a) Speed: Period T = 1/f = 1/2.0 = 0.50 s Circumference = 2πr = 2π(1.2) = 2.4π m v = 2πr/T = 2.4π/0.50 = 4.8π = 15.1 m/s

(b) Centripetal acceleration: a_c = v²/r = (15.1)²/1.2 = 228/1.2 = 190 m/s²

Or: a_c = 4π²r/T² = 4π²(1.2)/(0.50)² = 190 m/s² ✓

(c) Tension: For horizontal circular motion, tension provides centripetal force: F_c = ma_c T = 0.50 × 190 = 95 N

6Problem 6medium

❓ Question:

A 0.5 kg ball is attached to a 1.2 m string and swung in a horizontal circle, making 2 complete revolutions per second. Calculate: (a) the period, (b) the speed of the ball, and (c) the centripetal acceleration.

💡 Show Solution

Given Information:

Mass: $m = 0.5$ kg
Radius: $r = 1.2$ m
Frequency: $f = 2$ rev/s = 2 Hz

(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!

7Problem 7hard

❓ Question:

A space station rotates to create artificial gravity. If the station has a radius of 100 m and the centripetal acceleration at the outer edge is to equal Earth's gravity (9.8 m/s²), what should be the period of rotation?

💡 Show Solution

Given Information:

Radius: $r = 100$ m
Desired centripetal acceleration: $a_c = 9.8$ m/s² (to simulate Earth's 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.

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