AP Physics C Mechanics- 5.1 Rotational Kinematics- Study Notes- New Syllabus
AP Physics C Mechanics- 5.1 Rotational Kinematics – Study Notes
AP Physics C Mechanics- 5.1 Rotational Kinematics – Study Notes – per latest Syllabus.
Key Concepts:
- Angular Displacement
- Angular Velocity
- Angular Acceleration
- Kinematic Relationships for Constant Angular Acceleration
- Graphs of Angular Displacement, Angular Velocity, and Angular Acceleration as Functions of Time
Angular Displacement
Angular displacement is the measure of the angle, in radians, through which a point or line on a rigid system rotates about a specified axis.
\( \mathrm{\theta = \dfrac{s}{r}} \)
- \( \mathrm{\theta} \): angular displacement (in radians)
- \( \mathrm{s} \): arc length covered by the point
- \( \mathrm{r} \): distance of the point from the axis of rotation
Relevant Equation: \( \mathrm{\theta = \dfrac{s}{r}} \)
Rigid System
A rigid system is one that holds its shape but may have different points moving in different directions during rotation. A rigid system cannot be modeled as a single point object because not all points move identically.
- Each point maintains a fixed distance from every other point in the system.
- All points share the same angular displacement about the rotation axis, but their linear displacements differ.
Direction Convention
One direction of angular displacement (usually counterclockwise) is taken as mathematically positive, and the opposite direction (clockwise) as mathematically negative.
- This convention is consistent with the right-hand rule.
- Positive angular displacement → counterclockwise rotation.
- Negative angular displacement → clockwise rotation.
Simplifying Rotational Motion
When the rotational motion of a system can be effectively represented by the motion of its center of mass, the system may be treated as a single object.
For example, while Earth rotates about its own axis, the rotational motion is negligible compared to its revolution about the Sun. Therefore, in orbital motion analysis, Earth is treated as a single object located at its center of mass.
Example
A point on a rotating wheel moves through an arc length of \( \mathrm{0.50\,m} \) at a radius of \( \mathrm{0.25\,m} \) from the axis of rotation. Find the angular displacement of the point.
▶️ Answer / Explanation
Step 1: Use the formula for angular displacement:
\( \mathrm{\theta = \dfrac{s}{r}} \)
Step 2: Substitute the known values:
\( \mathrm{\theta = \dfrac{0.50}{0.25} = 2.0\,rad} \)
Result: The angular displacement of the point is \( \mathrm{2.0\,rad} \).
Angular Velocity
Angular velocity is the rate at which the angular position (or angular displacement) of a rotating body changes with respect to time.
\( \mathrm{\omega = \dfrac{d\theta}{dt}} \)
- \( \mathrm{\omega} \): angular velocity (in \( \mathrm{rad/s} \))
- \( \mathrm{\theta} \): angular displacement (in radians)
- \( \mathrm{t} \): time (in seconds)
Key Points:
- The direction of angular velocity is along the axis of rotation, determined by the right-hand rule.
- For uniform circular motion, angular velocity remains constant.
- Angular velocity is related to linear velocity by \( \mathrm{v = r\omega} \).
Relevant Equation: \( \mathrm{\omega = \dfrac{d\theta}{dt}} \)
Example
A wheel rotates such that its angular displacement changes from \( \mathrm{0} \) to \( \mathrm{10\,rad} \) in \( \mathrm{2.0\,s} \). Find the average angular velocity of the wheel.
▶️ Answer / Explanation
Step 1: Use the formula for average angular velocity:
\( \mathrm{\bar{\omega} = \dfrac{\Delta\theta}{\Delta t}} \)
Step 2: Substitute values:
\( \mathrm{\bar{\omega} = \dfrac{10 – 0}{2.0} = 5.0\,rad/s} \)
Result: The average angular velocity of the wheel is \( \mathrm{5.0\,rad/s} \).
Angular Acceleration
Angular acceleration is the rate at which angular velocity changes with respect to time. It describes how quickly an object speeds up or slows down its rotation.
\( \mathrm{\alpha = \dfrac{d\omega}{dt}} \)
- \( \mathrm{\alpha} \): angular acceleration (in \( \mathrm{rad/s^2} \))
- \( \mathrm{\omega} \): angular velocity (in \( \mathrm{rad/s} \))
- \( \mathrm{t} \): time (in seconds)
Key Points:
- Positive \( \mathrm{\alpha} \) indicates an increase in angular velocity (speeding up).
- Negative \( \mathrm{\alpha} \) indicates a decrease in angular velocity (slowing down).
- For constant angular acceleration, the rotational kinematic equations are analogous to linear motion equations.
Relevant Equation: \( \mathrm{\alpha = \dfrac{d\omega}{dt}} \)
Example
A rotating fan speeds up uniformly from \( \mathrm{10\,rad/s} \) to \( \mathrm{40\,rad/s} \) in \( \mathrm{6.0\,s} \). Find the angular acceleration.
▶️ Answer / Explanation
Step 1: Use the formula for angular acceleration:
\( \mathrm{\alpha = \dfrac{\Delta\omega}{\Delta t}} \)
Step 2: Substitute values:
\( \mathrm{\alpha = \dfrac{40 – 10}{6.0} = 5.0\,rad/s^2} \)
Result: The angular acceleration of the fan is \( \mathrm{5.0\,rad/s^2} \).
Kinematic Relationships for Constant Angular Acceleration
For rotation about a fixed axis with constant angular acceleration (\( \mathrm{\alpha = const.} \)), the angular quantities obey the same mathematical relationships as linear kinematics:
\( \mathrm{\omega = \omega_0 + \alpha t} \)
\( \mathrm{\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2} \)
\( \mathrm{\omega^2 = \omega_0^2 + 2\alpha(\theta – \theta_0)} \)
- \( \mathrm{\omega_0} \): initial angular velocity
- \( \mathrm{\omega} \): angular velocity at time \( \mathrm{t} \)
- \( \mathrm{\alpha} \): constant angular acceleration
- \( \mathrm{\theta_0} \): initial angular position
- \( \mathrm{\theta} \): angular position after time \( \mathrm{t} \)
Example
A wheel starts from rest and rotates with a constant angular acceleration of \( \mathrm{3.0\,rad/s^2} \). Find the angular velocity after \( \mathrm{4.0\,s} \) and the angular displacement in that time.
▶️ Answer / Explanation
Step 1: Use \( \mathrm{\omega = \omega_0 + \alpha t} \) to find angular velocity.
\( \mathrm{\omega = 0 + (3.0)(4.0) = 12\,rad/s} \)
Step 2: Use \( \mathrm{\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2} \) to find angular displacement.
\( \mathrm{\theta = 0 + \tfrac{1}{2}(3.0)(4.0)^2 = 24\,rad} \)
Result: After \( \mathrm{4.0\,s} \), the wheel’s angular velocity is \( \mathrm{12\,rad/s} \), and it has rotated through an angular displacement of \( \mathrm{24\,rad} \).
Graphs of Angular Displacement, Angular Velocity, and Angular Acceleration as Functions of Time
Graphs of rotational quantities—angular displacement \( \mathrm{(\theta)} \), angular velocity \( \mathrm{(\omega)} \), and angular acceleration \( \mathrm{(\alpha)} \)—as functions of time provide a visual way to understand their relationships. These graphs are analogous to the linear motion graphs of displacement, velocity, and acceleration.
Key Relationships from Graphs:
- The slope of the \( \mathrm{\theta\text{-}t} \) graph gives the angular velocity \( \mathrm{\omega} \).
- The slope of the \( \mathrm{\omega\text{-}t} \) graph gives the angular acceleration \( \mathrm{\alpha} \).
- The area under the \( \mathrm{\omega\text{-}t} \) graph gives the angular displacement \( \mathrm{\theta} \).
- The area under the \( \mathrm{\alpha\text{-}t} \) graph gives the change in angular velocity \( \mathrm{\Delta\omega} \).
Interpreting the Graphs:
| Graph Type | Slope Represents | Area Represents |
|---|---|---|
| \( \mathrm{\theta\text{-}t} \) | Angular velocity \( \mathrm{\omega} \) | — |
| \( \mathrm{\omega\text{-}t} \) | Angular acceleration \( \mathrm{\alpha} \) | Angular displacement \( \mathrm{\theta} \) |
| \( \mathrm{\alpha\text{-}t} \) | — | Change in angular velocity \( \mathrm{\Delta\omega} \) |
Key Idea: The relationships between \( \mathrm{\theta} \), \( \mathrm{\omega} \), and \( \mathrm{\alpha} \) can be derived visually: • The slope gives the rate of change, and the area gives the total change in the quantity below.
Example
An object starts from rest and experiences a constant angular acceleration of \( \mathrm{2\,rad/s^2} \). Using the graphical method, determine:
- The angular velocity after \( \mathrm{3\,s} \).
- The angular displacement during this time.
▶️ Answer / Explanation
Step 1: The \( \mathrm{\alpha\text{-}t} \) graph is a horizontal line at \( \mathrm{2\,rad/s^2} \). The area under this graph gives the change in angular velocity:
\( \mathrm{\Delta\omega = \alpha t = (2)(3) = 6\,rad/s} \)
Thus, \( \mathrm{\omega = 6\,rad/s} \) after \( \mathrm{3\,s} \).
Step 2: The \( \mathrm{\omega\text{-}t} \) graph is a straight line from \( \mathrm{0} \) to \( \mathrm{6\,rad/s} \). The area under this graph (a triangle) gives the angular displacement:
\( \mathrm{\theta = \tfrac{1}{2}(base)(height) = \tfrac{1}{2}(3)(6) = 9\,rad} \)
Result: After \( \mathrm{3\,s} \), the angular velocity is \( \mathrm{6\,rad/s} \) and the angular displacement is \( \mathrm{9\,rad} \).