Rotational Kinematics
When an object is rotating, we measure its motion using radians
I. Radian = arc length that = radius of circle Video p. 1-3
360 degrees - One full rotation of a sphere is radians
90 degrees = radians
Ex 1) How many radians does the arc s represent?
Number of radians $\theta = $
Equation: $\theta = \frac{s}{r}$
And $d = r\theta$
Ex 2) A wheel with a radius of 2.0 m rolls 5.0 m along the floor. Find the number of radians rotated through. $\theta = $
Ex 1 Claim: The arc length $s$ is directly proportional to the angle $\theta$ in radians.
Line-by-Line explanation:
1. Fundamental Law: $\theta = \frac{\text{Arc Length } s}{\text{Radius } r}$
2. If $s = r$, then $\theta = r/r = 1\text{ radian}$.
3. If $s = 2r$, then $\theta = 2r/r = 2\text{ radians}$.
4. One full circle is $s = 2\pi r$, so $\theta = 2\pi r / r = 2\pi\text{ rad} \approx 6.28\text{ rad}$.
Ex 2 Claim: The wheel rotates through an angle of $2.5\text{ rad}$.
Evidence: Path length $s = 5.0\text{ m}$; Radius $r = 2.0\text{ m}$.
Line-by-Line explanation:
1. Select the formula: $\theta = s/r$
2. Substitution: $\theta = (5.0\text{ m}) / (2.0\text{ m})$
3. Mathematical evaluation: $5.0 / 2.0 = 2.5$
4. Result: $2.5\text{ radians}$.
II. Angular Velocity: $\omega$ = radians/sec
Relating angular velocity to linear velocity
Ex 3) A ball with a radius of .50 m rolls with a velocity of 5.0 m/s. Find its angular velocity.
Angular velocity = radians/sec
Formula: $\omega = \frac{v}{r}$ or $v = r\omega$
Ex 3 Claim: The ball rotates at an angular velocity of $10\text{ rad/s}$.
Evidence: Linear (tangential) velocity $v = 5.0\text{ m/s}$; Radius $r = 0.50\text{ m}$.
Line-by-Line explanation:
1. Relationship: $v = r\omega$
2. Algebra: $\omega = v/r$
3. Data Plug-in: $\omega = (5.0\text{ m/s}) / (0.50\text{ m})$
4. Solve: $10\text{ rad/s}$.
III. Period of a Rolling Object
Defined: Period – time for 1 full rotation
$$T = \frac{2\pi \text{ rad}}{\omega}$$
Ex 4) A ball with a radius of .50 m rolls with a velocity of 5.0 m/s. Find a period of rotation.
$T = \frac{2\pi \text{ rad}}{\omega}$
IV. Frequency - Rotations Per Second
$$f = \frac{1}{T} = \frac{\omega}{2\pi}$$
Ex 5) A ball with a radius of .50 m rolls with a velocity of 5.0 m/s. Find frequency of rotation.
Ex 4 Claim: Period $T \approx 0.63\text{ s}$.
Line-by-Line explanation:
1. Formula: $T = 2\pi / \omega$
2. Input: $\omega = 10\text{ rad/s}$ (from previous work)
3. Substitution: $T = (2 \times 3.14159) / (10\text{ rad/s})$
4. Result: $0.628\text{ seconds per rotation}$.
Ex 5 Claim: Frequency $f = 1.59\text{ Hz}$.
Line-by-Line explanation:
1. Inverse Relationship: $f = 1/T$
2. Substitution: $f = 1 / 0.628\text{ s}$
3. Result: $1.592\text{ Hertz (Rotations per second)}$.
IV. Angular Acceleration
Review: Relating Linear motion to Rotational motion
- $d = r\theta \quad (\theta = \text{distance rotated in radians})$
- $v = r\omega \quad (\omega - \text{angular velocity})$
- So .... $a = r\alpha$
Ex 6) A pool ball with a radius of 4.0 cm accelerates from rest to 5.0 m/s in .10 seconds. Find its angular acceleration.
Match the rotational variable with its linear kinematic variable Video
$\alpha$ _____________ $\omega$ _____________ $\theta$ ______________
$d = \_\_\_\_\_\_\_\_\_\_\_ \quad v = \_\_\_\_\_\_\_\_\_\_ \quad a = \_\_\_\_\_\_\_\_\_\_$
Linear Kinematics / Rotational Kinematics Conversion
| Linear Kinematics |
Rotational Kinematics |
| $a = \Delta V / \Delta t$ |
$\alpha = \Delta \omega / \Delta t$ |
| $V_f = V_i + at$ |
$\omega_f = \omega_i + \alpha t$ |
 $= \Delta d / \Delta t$ |
$\omega_{\text{avg}} = \Delta \theta / \Delta t$ |
| $V_f^2 = V_i^2 + 2a\Delta d$ |
$\omega_f^2 = \omega_i^2 + 2\alpha\Delta\theta$ |
| $\Delta d = V_it + \frac{1}{2}at^2$ |
$\Delta \theta = \omega_it + \frac{1}{2}\alpha t^2$ |
Ex 7) A ball with a radius of .10 m starts from rest and accelerates down a 10. m incline and attains a velocity of 8.0 m/s. Find its ANGULAR acceleration.
There are 2 ways to solve this. Explain.
Strategy 1 _________________________ Strategy 2 _________________________
Ex 6 Claim: Angular Acceleration $\alpha = 1250\text{ rad/s}^2$.
Line-by-Line explanation:
1. Linear acceleration $a = (V_f - V_i) / t = (5.0\text{ m/s} - 0) / 0.10\text{ s} = 50\text{ m/s}^2$.
2. Radius conversion: $4.0\text{ cm} = 0.040\text{ m}$.
3. Conversion Law: $a = r\alpha \implies \alpha = a/r$.
4. Solve: $\alpha = (50\text{ m/s}^2) / (0.040\text{ m}) = 1250\text{ rad/s}^2$.
Ex 7 Claim: $\alpha = 32\text{ rad/s}^2$.
Line-by-Line explanation:
1. Linear approach: $V_f^2 = V_i^2 + 2ad \rightarrow 8^2 = 2(a)(10) \rightarrow a = 3.2\text{ m/s}^2$.
2. Convert to angular: $\alpha = a/r = (3.2\text{ m/s}^2) / (0.10\text{ m}) = 32\text{ rad/s}^2$.
3. Angular approach: $\omega_f = v_f/r = 80\text{ rad/s}, \theta = d/r = 100\text{ rad}$.
4. Solve $\omega_f^2 = 2\alpha\theta$: $6400 = 2\alpha(100) \implies \alpha = 32\text{ rad/s}^2$.
