Rotational Kinematics

Angular displacement, velocity, acceleration, and rotational kinematic equations

🌀 Rotational Kinematics

Angular Quantities

Just as linear motion has position, velocity, and acceleration, rotational motion has corresponding angular quantities.

Angular Position θ\theta

Angle of rotation from reference line:

  • Units: radians (rad)
  • 2π2\pi rad = 360° = one full rotation
  • Conversion: θrad=θdeg×π180\theta_{rad} = \theta_{deg} \times \frac{\pi}{180}

Angular Displacement Δθ\Delta\theta

Change in angular position: Δθ=θfθi\Delta\theta = \theta_f - \theta_i

Units: radians (rad)

Positive: counterclockwise rotation Negative: clockwise rotation

Angular Velocity ω\omega

Rate of change of angular position:

Average: ωavg=ΔθΔt\omega_{avg} = \frac{\Delta\theta}{\Delta t}

Instantaneous: ω=dθdt\omega = \frac{d\theta}{dt}

Units: rad/s (radians per second)

Angular Acceleration α\alpha

Rate of change of angular velocity:

Average: αavg=ΔωΔt\alpha_{avg} = \frac{\Delta\omega}{\Delta t}

Instantaneous: α=dωdt=d2θdt2\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}

Units: rad/s² (radians per second squared)


Analogy with Linear Motion

| Linear Motion | Rotational Motion | |---------------|-------------------| | Position xx | Angular position θ\theta | | Velocity vv | Angular velocity ω\omega | | Acceleration aa | Angular acceleration α\alpha | | Mass mm | Moment of inertia II | | Force FF | Torque τ\tau |


Relationship Between Linear and Angular

For a point at distance rr from axis of rotation:

Arc Length

s=rθs = r\theta (where θ\theta is in radians)

Linear Velocity

v=rωv = r\omega

Tangential velocity - velocity tangent to circular path

Tangential Acceleration

at=rαa_t = r\alpha

Component of acceleration tangent to circle (changes speed)

Centripetal Acceleration

ac=v2r=ω2ra_c = \frac{v^2}{r} = \omega^2 r

Component of acceleration toward center (changes direction)


Rotational Kinematic Equations

For constant angular acceleration α\alpha:

The Big Four Equations

1. Angular velocity: ωf=ωi+αt\omega_f = \omega_i + \alpha t

2. Angular displacement: θ=θi+ωit+12αt2\theta = \theta_i + \omega_i t + \frac{1}{2}\alpha t^2

3. Velocity-displacement: ωf2=ωi2+2αΔθ\omega_f^2 = \omega_i^2 + 2\alpha\Delta\theta

4. Average velocity: θ=θi+12(ωi+ωf)t\theta = \theta_i + \frac{1}{2}(\omega_i + \omega_f)t

💡 These are EXACTLY analogous to linear kinematic equations! Just replace xθx \to \theta, vωv \to \omega, aαa \to \alpha.


Comparison: Linear vs. Rotational Kinematics

| Linear (constant aa) | Rotational (constant α\alpha) | |----------------------|--------------------------------| | vf=vi+atv_f = v_i + at | ωf=ωi+αt\omega_f = \omega_i + \alpha t | | x=xi+vit+12at2x = x_i + v_i t + \frac{1}{2}at^2 | θ=θi+ωit+12αt2\theta = \theta_i + \omega_i t + \frac{1}{2}\alpha t^2 | | vf2=vi2+2aΔxv_f^2 = v_i^2 + 2a\Delta x | ωf2=ωi2+2αΔθ\omega_f^2 = \omega_i^2 + 2\alpha\Delta\theta | | x=xi+12(vi+vf)tx = x_i + \frac{1}{2}(v_i + v_f)t | θ=θi+12(ωi+ωf)t\theta = \theta_i + \frac{1}{2}(\omega_i + \omega_f)t |


Period and Frequency

For uniform circular motion (constant ω\omega):

Period TT

Time for one complete rotation: T=2πωT = \frac{2\pi}{\omega}

Units: seconds

Frequency ff

Rotations per second: f=1T=ω2πf = \frac{1}{T} = \frac{\omega}{2\pi}

Units: Hz (hertz) = rev/s

Relationships

ω=2πf=2πT\omega = 2\pi f = \frac{2\pi}{T}


Rolling Motion

For an object rolling without slipping:

Constraint condition: vcm=rωv_{cm} = r\omega

where:

  • vcmv_{cm} = velocity of center of mass
  • rr = radius
  • ω\omega = angular velocity

No slipping means:

  • Point of contact is instantaneously at rest
  • Distance traveled = arc length: s=rθs = r\theta

Problem-Solving Strategy

For Rotational Kinematics:

  1. Identify knowns: θi\theta_i, ωi\omega_i, ωf\omega_f, α\alpha, tt, Δθ\Delta\theta
  2. Identify unknown: What are you solving for?
  3. Choose equation: Pick the one with known quantities and unknown
  4. Solve algebraically
  5. Check units: Should be rad, rad/s, or rad/s²
  6. Check reasonableness: Does answer make sense?

Tip: If you know linear quantities (vv, ata_t, ss), convert using:

  • ω=v/r\omega = v/r
  • α=at/r\alpha = a_t/r
  • θ=s/r\theta = s/r

⚠️ Common Mistakes

Mistake 1: Degrees vs. Radians

MUST use radians in equations! Convert degrees to radians first.

Mistake 2: Confusing ata_t and aca_c

  • at=rαa_t = r\alpha: tangential (changes speed)
  • ac=ω2ra_c = \omega^2 r: centripetal (changes direction)
  • Total: a=at2+ac2a = \sqrt{a_t^2 + a_c^2}

Mistake 3: Wrong Sign for α\alpha

  • Speeding up in positive direction: α>0\alpha > 0
  • Slowing down in positive direction: α<0\alpha < 0

Mistake 4: Forgetting Initial Conditions

ωi\omega_i and θi\theta_i are not always zero!


Special Cases

Starting from Rest

ωi=0\omega_i = 0:

  • ωf=αt\omega_f = \alpha t
  • θ=12αt2\theta = \frac{1}{2}\alpha t^2
  • ωf2=2αθ\omega_f^2 = 2\alpha\theta

Uniform Rotation

α=0\alpha = 0 (constant ω\omega):

  • ωf=ωi=ω\omega_f = \omega_i = \omega
  • θ=ωt\theta = \omega t
  • Period: T=2πωT = \frac{2\pi}{\omega}

Coming to Rest

ωf=0\omega_f = 0:

  • 0=ωi+αt0 = \omega_i + \alpha tt=ωiαt = -\frac{\omega_i}{\alpha}
  • ωi2=2αΔθ\omega_i^2 = -2\alpha\Delta\thetaΔθ=ωi22α\Delta\theta = -\frac{\omega_i^2}{2\alpha}

Applications

Wheels and Gears

  • Angular velocity determines linear speed
  • Gear ratios change angular velocities
  • v=rωv = r\omega connects the two

Rotating Machinery

  • Turbines, engines, motors
  • Angular acceleration during startup
  • Constant ω\omega during normal operation

Sports

  • Figure skating spins (angular velocity)
  • Gymnastics rotations
  • Diving somersaults

Astronomy

  • Planetary rotation (Earth: T24T \approx 24 hr)
  • Orbital motion
  • Galaxy rotation

Key Formulas Summary

| Quantity | Formula | Units | |----------|---------|-------| | Angular velocity | ω=dθdt\omega = \frac{d\theta}{dt} | rad/s | | Angular acceleration | α=dωdt\alpha = \frac{d\omega}{dt} | rad/s² | | Linear velocity | v=rωv = r\omega | m/s | | Tangential acceleration | at=rαa_t = r\alpha | m/s² | | Centripetal acceleration | ac=ω2ra_c = \omega^2 r | m/s² | | Period | T=2πωT = \frac{2\pi}{\omega} | s | | Frequency | f=ω2πf = \frac{\omega}{2\pi} | Hz |

Kinematic equations (constant α\alpha):

  1. ωf=ωi+αt\omega_f = \omega_i + \alpha t
  2. θ=θi+ωit+12αt2\theta = \theta_i + \omega_i t + \frac{1}{2}\alpha t^2
  3. ωf2=ωi2+2αΔθ\omega_f^2 = \omega_i^2 + 2\alpha\Delta\theta

📚 Practice Problems

1Problem 1easy

Question:

A wheel starts from rest and accelerates uniformly at 2 rad/s² for 5 seconds. Find: (a) the final angular velocity, (b) the angular displacement during this time, and (c) the number of revolutions completed.

💡 Show Solution

Given Information:

  • Initial angular velocity: ωi=0\omega_i = 0 rad/s (starts from rest)
  • Angular acceleration: α=2\alpha = 2 rad/s²
  • Time: t=5t = 5 s

(a) Find final angular velocity


Step 1: Use first kinematic equation

ωf=ωi+αt\omega_f = \omega_i + \alpha t

ωf=0+(2)(5)\omega_f = 0 + (2)(5)

ωf=10 rad/s\omega_f = 10 \text{ rad/s}


Answer (a): Final angular velocity = 10 rad/s


(b) Find angular displacement


Step 2: Use displacement equation

θ=ωit+12αt2\theta = \omega_i t + \frac{1}{2}\alpha t^2

θ=0(5)+12(2)(5)2\theta = 0(5) + \frac{1}{2}(2)(5)^2

θ=0+12(2)(25)\theta = 0 + \frac{1}{2}(2)(25)

θ=25 rad\theta = 25 \text{ rad}


Alternative: Use average velocity

θ=12(ωi+ωf)t=12(0+10)(5)=25 rad\theta = \frac{1}{2}(\omega_i + \omega_f)t = \frac{1}{2}(0 + 10)(5) = 25 \text{ rad}

Both methods agree! ✓


Answer (b): Angular displacement = 25 rad


(c) Find number of revolutions


Step 3: Convert radians to revolutions

Revolutions=θ2π=252π\text{Revolutions} = \frac{\theta}{2\pi} = \frac{25}{2\pi}

Revolutions=256.28=3.98 rev\text{Revolutions} = \frac{25}{6.28} = 3.98 \text{ rev}


Answer (c): Number of revolutions ≈ 4.0 revolutions

Summary: The wheel accelerates from rest to 10 rad/s, turning through 25 radians (about 4 complete rotations) in 5 seconds.

2Problem 2medium

Question:

A wheel starts from rest and accelerates uniformly to 120 rpm in 8.0 seconds. (a) What is the angular acceleration in rad/s²? (b) How many revolutions does it make during this time? (c) What is the final angular velocity in rad/s?

💡 Show Solution

Solution:

Given: ω₀ = 0, ω_f = 120 rpm, t = 8.0 s

(a) Angular acceleration: Convert to rad/s: ω_f = 120 rev/min × (2π rad/rev) × (1 min/60 s) = 4π rad/s

α = (ω_f - ω₀)/t = (4π - 0)/8.0 = 1.57 rad/s² or π/2 rad/s²

(b) Number of revolutions: θ = ω₀t + ½αt² = 0 + ½(π/2)(8.0)² θ = ½(π/2)(64) = 16π rad

Convert to revolutions: 16π rad × (1 rev/2π rad) = 8.0 rev

(c) Final angular velocity: ω_f = 4π = 12.6 rad/s

Or 120 rpm as given.

3Problem 3medium

Question:

A car tire with radius 0.3 m is rotating at 10 rev/s. The car brakes, and the tire comes to rest in 4 seconds with constant angular acceleration. Find: (a) the angular acceleration, and (b) the linear distance traveled during braking.

💡 Show Solution

Given Information:

  • Radius: r=0.3r = 0.3 m
  • Initial angular velocity: ωi=10\omega_i = 10 rev/s
  • Final angular velocity: ωf=0\omega_f = 0 rad/s (comes to rest)
  • Time: t=4t = 4 s

Step 0: Convert units

ωi=10 rev/s×2π rad1 rev=20π rad/s62.8 rad/s\omega_i = 10 \text{ rev/s} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} = 20\pi \text{ rad/s} \approx 62.8 \text{ rad/s}


(a) Find angular acceleration


Step 1: Use first kinematic equation

ωf=ωi+αt\omega_f = \omega_i + \alpha t

0=62.8+α(4)0 = 62.8 + \alpha(4)

4α=62.84\alpha = -62.8

α=15.7 rad/s2\alpha = -15.7 \text{ rad/s}^2


Answer (a): Angular acceleration = −15.7 rad/s² (negative because it's slowing down)


(b) Find linear distance traveled


Step 2: Find angular displacement

θ=ωit+12αt2\theta = \omega_i t + \frac{1}{2}\alpha t^2

θ=(62.8)(4)+12(15.7)(4)2\theta = (62.8)(4) + \frac{1}{2}(-15.7)(4)^2

θ=251.2+12(15.7)(16)\theta = 251.2 + \frac{1}{2}(-15.7)(16)

θ=251.2125.6\theta = 251.2 - 125.6

θ=125.6 rad\theta = 125.6 \text{ rad}


Alternative: Use average velocity

θ=12(ωi+ωf)t=12(62.8+0)(4)=125.6 rad\theta = \frac{1}{2}(\omega_i + \omega_f)t = \frac{1}{2}(62.8 + 0)(4) = 125.6 \text{ rad}

Both methods agree! ✓


Step 3: Convert to linear distance

s=rθ=(0.3)(125.6)s = r\theta = (0.3)(125.6)

s=37.7 ms = 37.7 \text{ m}


Answer (b): Linear distance traveled = 37.7 m (about 38 meters)

Check: This is reasonable for a car braking from moderate speed over 4 seconds.

Note: Number of revolutions = 125.62π20\frac{125.6}{2\pi} \approx 20 revolutions during braking.

4Problem 4medium

Question:

A merry-go-round with radius 2.0 m rotates at 0.50 rev/s. A child stands at the outer edge. (a) What is the child's angular velocity? (b) What is the child's tangential (linear) speed? (c) What is the child's centripetal acceleration?

💡 Show Solution

Solution:

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

(a) Angular velocity: ω = 2πf = 2π(0.50) = π rad/s or 3.14 rad/s

(b) Tangential speed: v = rω = 2.0(π) = 2π m/s or 6.28 m/s

(c) Centripetal acceleration: a_c = v²/r = (2π)²/2.0 = 4π²/2.0 = 19.7 m/s²

Or: a_c = rω² = 2.0(π)² = 2π² = 19.7 m/s² ✓

5Problem 5hard

Question:

A disk of radius 0.5 m starts from rest and rotates with constant angular acceleration. After 10 seconds, a point on the rim of the disk has a tangential speed of 15 m/s. Find: (a) the angular acceleration, (b) the angular displacement in those 10 seconds, and (c) the magnitude of the total acceleration of a point on the rim at t = 10 s.

💡 Show Solution

Given Information:

  • Radius: r=0.5r = 0.5 m
  • Initial angular velocity: ωi=0\omega_i = 0 rad/s (starts from rest)
  • Time: t=10t = 10 s
  • Final tangential speed: vf=15v_f = 15 m/s

(a) Find angular acceleration


Step 1: Find final angular velocity

v=rωv = r\omega

ωf=vfr=150.5=30 rad/s\omega_f = \frac{v_f}{r} = \frac{15}{0.5} = 30 \text{ rad/s}


Step 2: Calculate angular acceleration

ωf=ωi+αt\omega_f = \omega_i + \alpha t

30=0+α(10)30 = 0 + \alpha(10)

α=3 rad/s2\alpha = 3 \text{ rad/s}^2


Answer (a): Angular acceleration = 3 rad/s²


(b) Find angular displacement


Step 3: Use displacement equation

θ=ωit+12αt2\theta = \omega_i t + \frac{1}{2}\alpha t^2

θ=0(10)+12(3)(10)2\theta = 0(10) + \frac{1}{2}(3)(10)^2

θ=12(3)(100)\theta = \frac{1}{2}(3)(100)

θ=150 rad\theta = 150 \text{ rad}


Alternative: Use average velocity

θ=12(ωi+ωf)t=12(0+30)(10)=150 rad\theta = \frac{1}{2}(\omega_i + \omega_f)t = \frac{1}{2}(0 + 30)(10) = 150 \text{ rad}


Answer (b): Angular displacement = 150 rad

(This is 1502π23.9\frac{150}{2\pi} \approx 23.9 revolutions)


(c) Find total acceleration at t = 10 s


Step 4: Calculate tangential acceleration

at=rα=(0.5)(3)=1.5 m/s2a_t = r\alpha = (0.5)(3) = 1.5 \text{ m/s}^2


Step 5: Calculate centripetal acceleration

ac=ω2r=(30)2(0.5)a_c = \omega^2 r = (30)^2(0.5)

ac=900(0.5)=450 m/s2a_c = 900(0.5) = 450 \text{ m/s}^2


Step 6: Find magnitude of total acceleration

Tangential and centripetal accelerations are perpendicular:

atotal=at2+ac2a_{total} = \sqrt{a_t^2 + a_c^2}

atotal=(1.5)2+(450)2a_{total} = \sqrt{(1.5)^2 + (450)^2}

atotal=2.25+202,500a_{total} = \sqrt{2.25 + 202,500}

atotal=202,502.25a_{total} = \sqrt{202,502.25}

atotal450 m/s2a_{total} \approx 450 \text{ m/s}^2


Answer (c): Total acceleration ≈ 450 m/s²

Note: The centripetal acceleration (450 m/s²) is MUCH larger than the tangential acceleration (1.5 m/s²), so the total acceleration is essentially just the centripetal acceleration. This makes sense at high rotational speeds!

Direction: The total acceleration points slightly inward from the purely radial direction (mostly toward center, with small tangential component).