Angular Position $\theta$

Angle of rotation from reference line:

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

Angular Displacement $\Delta\theta$

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

Units: radians (rad)

Positive: counterclockwise rotation Negative: clockwise rotation

Angular Velocity $\omega$

Rate of change of angular position:

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

Instantaneous: $\omega = \frac{d\theta}{dt}$

Units: rad/s (radians per second)

Angular Acceleration $\alpha$

Rate of change of angular velocity:

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

Instantaneous: $\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}$

Units: rad/s² (radians per second squared)

Analogy with Linear Motion

Relationship Between Linear and Angular

For a point at distance $r$ from axis of rotation:

Arc Length

$s = r\theta$ (where $\theta$ is in radians)

Linear Velocity

$v = r\omega$

Tangential velocity - velocity tangent to circular path

Tangential Acceleration

$a_t = r\alpha$

Component of acceleration tangent to circle (changes speed)

Centripetal Acceleration

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

Component of acceleration toward center (changes direction)

Rotational Kinematic Equations

For constant angular acceleration $\alpha$:

The Big Four Equations

1. Angular velocity: $\omega_f = \omega_i + \alpha t$

2. Angular displacement: $\theta = \theta_i + \omega_i t + \frac{1}{2}\alpha t^2$

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

4. Average velocity: $\theta = \theta_i + \frac{1}{2}(\omega_i + \omega_f)t$

💡 These are EXACTLY analogous to linear kinematic equations! Just replace $x \to \theta$, $v \to \omega$, $a \to \alpha$.

Comparison: Linear vs. Rotational Kinematics

Period and Frequency

For uniform circular motion (constant $\omega$):

Period $T$

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

Units: seconds

Frequency $f$

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

Units: Hz (hertz) = rev/s

Relationships

$\omega = 2\pi f = \frac{2\pi}{T}$

Rolling Motion

For an object rolling without slipping:

Constraint condition: $v_{cm} = r\omega$

where:

• $v_{cm}$ = velocity of center of mass
• $r$ = radius
• $\omega$ = angular velocity

No slipping means:

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

Problem-Solving Strategy

For Rotational Kinematics:

1. Identify knowns: $\theta_i$, $\omega_i$, $\omega_f$, $\alpha$, $t$, $\Delta\theta$
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 ($v$, $a_t$, $s$), convert using:

• $\omega = v/r$
• $\alpha = a_t/r$
• $\theta = s/r$

⚠️ Common Mistakes

Mistake 1: Degrees vs. Radians

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

Mistake 2: Confusing $a_t$ and $a_c$

• $a_t = r\alpha$: tangential (changes speed)
• $a_c = \omega^2 r$: centripetal (changes direction)
• Total: $a = \sqrt{a_t^2 + a_c^2}$

Mistake 3: Wrong Sign for $\alpha$

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

Mistake 4: Forgetting Initial Conditions

$\omega_i$ and $\theta_i$ are not always zero!

Special Cases

Starting from Rest

$\omega_i = 0$:

• $\omega_f = \alpha t$
• $\theta = \frac{1}{2}\alpha t^2$
• $\omega_f^2 = 2\alpha\theta$

Uniform Rotation

$\alpha = 0$ (constant $\omega$):

• $\omega_f = \omega_i = \omega$
• $\theta = \omega t$
• Period: $T = \frac{2\pi}{\omega}$

Coming to Rest

$\omega_f = 0$:

• $0 = \omega_i + \alpha t$ → $t = -\frac{\omega_i}{\alpha}$
• $\omega_i^2 = -2\alpha\Delta\theta$ → $\Delta\theta = -\frac{\omega_i^2}{2\alpha}$

Applications

Wheels and Gears

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

Rotating Machinery

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

Sports

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

Astronomy

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

Key Formulas Summary

Kinematic equations (constant $\alpha$):

1. $\omega_f = \omega_i + \alpha t$
2. $\theta = \theta_i + \omega_i t + \frac{1}{2}\alpha t^2$
3. $\omega_f^2 = \omega_i^2 + 2\alpha\Delta\theta$

Step 0: Convert units

$\omega_i = 10 \text{ rev/s} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} = 20\pi \text{ rad/s} \approx 62.8 \text{ rad/s}$

(a) Find angular acceleration

Step 1: Use first kinematic equation

$\omega_f = \omega_i + \alpha t$

$0 = 62.8 + \alpha(4)$

$4\alpha = -62.8$

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

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

(b) Find linear distance traveled

Step 2: Find angular displacement

$\theta = \omega_i t + \frac{1}{2}\alpha t^2$

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

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

$\theta = 251.2 - 125.6$

$\theta = 125.6 \text{ rad}$

Alternative: Use average velocity

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

Both methods agree! ✓

Step 3: Convert to linear distance

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

$s = 37.7 \text{ m}$

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 = $\frac{125.6}{2\pi} \approx 20$ revolutions during braking.

(a) Find angular acceleration

Step 1: Find final angular velocity

$v = r\omega$

$\omega_f = \frac{v_f}{r} = \frac{15}{0.5} = 30 \text{ rad/s}$

Step 2: Calculate angular acceleration

$\omega_f = \omega_i + \alpha t$

$30 = 0 + \alpha(10)$

$\alpha = 3 \text{ rad/s}^2$

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

(b) Find angular displacement

Step 3: Use displacement equation

$\theta = \omega_i t + \frac{1}{2}\alpha t^2$

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

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

$\theta = 150 \text{ rad}$

Alternative: Use average velocity

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

Answer (b): Angular displacement = 150 rad

(This is $\frac{150}{2\pi} \approx 23.9$ revolutions)

(c) Find total acceleration at t = 10 s

Step 4: Calculate tangential acceleration

$a_t = r\alpha = (0.5)(3) = 1.5 \text{ m/s}^2$

Step 5: Calculate centripetal acceleration

$a_c = \omega^2 r = (30)^2(0.5)$

$a_c = 900(0.5) = 450 \text{ m/s}^2$

Step 6: Find magnitude of total acceleration

Tangential and centripetal accelerations are perpendicular:

$a_{total} = \sqrt{a_t^2 + a_c^2}$

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

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

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

$a_{total} \approx 450 \text{ m/s}^2$

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).