Uniform Circular Motion

Centripetal acceleration and circular motion kinematics

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šŸ”„ 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

  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

šŸ’” 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

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

where:

  • aca_c = centripetal acceleration (m/s²)
  • vv = speed (m/s)
  • rr = 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=2Ļ€rTv = \frac{2\pi r}{T} (where TT is period):

ac=4Ļ€2rT2a_c = \frac{4\pi^2 r}{T^2}

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

ac=ω2ra_c = \omega^2 r


Period and Frequency

Period (T)

The period is the time for one complete revolution:

T=2Ļ€rvT = \frac{2\pi r}{v}

Units: seconds (s)

Frequency (f)

The frequency is the number of revolutions per second:

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

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

Relationship Between v, r, T, and f

v=2Ļ€rT=2Ļ€rfv = \frac{2\pi r}{T} = 2\pi r f


Angular Velocity

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

ω=ΔθΔt=2Ļ€T=2Ļ€f\omega = \frac{\Delta \theta}{\Delta t} = \frac{2\pi}{T} = 2\pi f

Units: radians per second (rad/s)

Relationship to Linear Velocity

v=rωv = r\omega

This connects the speed along the circular path (vv) to the angular velocity (ω\omega).


Common Scenarios

Scenario 1: Object on a String

A ball swung in a horizontal circle:

  • Tension provides centripetal force
  • ac=v2ra_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: vmax=μsgrv_{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
  • ac=ga_c = g at Earth's surface
  • Orbital period depends on altitude

āš ļø Common Misconceptions

Misconception 1: "Centrifugal Force"

āŒ Wrong: There's an outward "centrifugal force" on the object āœ… 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

āŒ Wrong: Uniform circular motion has constant velocity āœ… Right: UCM has constant speed but changing velocity (because direction changes)

Misconception 3: Acceleration and Speed

āŒ Wrong: If speed is constant, acceleration must be zero āœ… Right: Acceleration can be perpendicular to velocity, changing direction without changing speed

Misconception 4: Direction After Release

āŒ Wrong: If the string breaks, the object flies radially outward āœ… 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=2Ļ€rTv = \frac{2\pi r}{T} or v=rωv = r\omega)
  4. Calculate centripetal acceleration: ac=v2ra_c = \frac{v^2}{r}
  5. Direction: Always toward the center

Key Equations Summary

| Quantity | Formula | Units | |----------|---------|-------| | Centripetal acceleration | ac=v2r=ω2ra_c = \frac{v^2}{r} = \omega^2 r | m/s² | | Speed | v=2Ļ€rT=rωv = \frac{2\pi r}{T} = r\omega | m/s | | Period | T=2Ļ€rvT = \frac{2\pi r}{v} | s | | Frequency | f=1Tf = \frac{1}{T} | Hz | | Angular velocity | ω=2Ļ€T=2Ļ€f=vr\omega = \frac{2\pi}{T} = 2\pi f = \frac{v}{r} | rad/s |


šŸ“ 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=1Tf = \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

šŸ“š Practice Problems

1Problem 1easy

ā“ 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?

šŸ’” Show Solution

Given Information:

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

Find: Centripetal acceleration aca_c


Solution:

Use the centripetal acceleration formula:

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

Substitute the values:

ac=(20Ā m/s)250Ā ma_c = \frac{(20 \text{ m/s})^2}{50 \text{ m}}

ac=400Ā m2/s250Ā ma_c = \frac{400 \text{ m}^2/\text{s}^2}{50 \text{ m}}

ac=8Ā m/s2a_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 gg (9.8 m/s²), so friction alone could provide this if the coefficient is sufficient.

2Problem 2medium

ā“ 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.

šŸ’” Show Solution

Given Information:

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

(a) Find the period TT:

The period is the time for one revolution:

T=1f=12Ā Hz=0.5Ā sT = \frac{1}{f} = \frac{1}{2 \text{ Hz}} = 0.5 \text{ s}


(b) Find the speed vv:

Use the relationship between speed, radius, and period:

v=2Ļ€rTv = \frac{2\pi r}{T}

v=2Ļ€(1.2Ā m)0.5Ā sv = \frac{2\pi (1.2 \text{ m})}{0.5 \text{ s}}

v=2.4π m0.5Ā sv = \frac{2.4\pi \text{ m}}{0.5 \text{ s}}

v=4.8π m/sā‰ˆ15.1Ā m/sv = 4.8\pi \text{ m/s} \approx 15.1 \text{ m/s}


(c) Find the centripetal acceleration aca_c:

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

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

ac=(4.8Ļ€)21.2a_c = \frac{(4.8\pi)^2}{1.2}

ac=23.04Ļ€21.2a_c = \frac{23.04\pi^2}{1.2}

ac=19.2Ļ€2ā‰ˆ189.5Ā m/s2a_c = 19.2\pi^2 \approx 189.5 \text{ m/s}^2

Alternative method using ac=4Ļ€2rf2a_c = 4\pi^2 r f^2:

ac=4Ļ€2(1.2)(2)2=4Ļ€2(1.2)(4)=19.2Ļ€2ā‰ˆ189.5Ā m/s2a_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!

3Problem 3hard

ā“ 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?

šŸ’” Show Solution

Given Information:

  • Radius: r=100r = 100 m
  • Desired centripetal acceleration: ac=9.8a_c = 9.8 m/s² (to simulate Earth's gravity)

Find: Period TT


Strategy: We need to work backwards from aca_c to find vv, then use vv to find TT.


Step 1: Find the required speed

From ac=v2ra_c = \frac{v^2}{r}, solve for vv:

v2=acā‹…rv^2 = a_c \cdot r

v=acā‹…rv = \sqrt{a_c \cdot r}

v=(9.8Ā m/s2)(100Ā m)v = \sqrt{(9.8 \text{ m/s}^2)(100 \text{ m})}

v=980Ā m2/s2v = \sqrt{980 \text{ m}^2/\text{s}^2}

v=31.3Ā m/sv = 31.3 \text{ m/s}


Step 2: Find the period

From v=2Ļ€rTv = \frac{2\pi r}{T}, solve for TT:

T=2Ļ€rvT = \frac{2\pi r}{v}

T=2Ļ€(100Ā m)31.3Ā m/sT = \frac{2\pi (100 \text{ m})}{31.3 \text{ m/s}}

T=200Ļ€31.3Ā sT = \frac{200\pi}{31.3} \text{ s}

Tā‰ˆ20.1Ā sT \approx 20.1 \text{ s}


Alternative Method: Using ac=4Ļ€2rT2a_c = \frac{4\pi^2 r}{T^2}

T2=4Ļ€2racT^2 = \frac{4\pi^2 r}{a_c}

T=2Ļ€racT = 2\pi\sqrt{\frac{r}{a_c}}

T=2Ļ€1009.8T = 2\pi\sqrt{\frac{100}{9.8}}

T=2Ļ€10.2T = 2\pi\sqrt{10.2}

T=2Ļ€(3.20)ā‰ˆ20.1Ā sT = 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 9.89.8 m/s² at the outer edge, which would feel like Earth's gravity to people standing on the outer rim.