AP Physics C Mechanics- 5.2 Connecting Linear and Rotational Motion- Study Notes- New Syllabus
AP Physics C Mechanics- 5.2 Connecting Linear and Rotational Motion – Study Notes
AP Physics C Mechanics- 5.2 Connecting Linear and Rotational Motion – Study Notes – per latest Syllabus.
Key Concepts:
- Relationship Between Arc Length and Angular Displacement
- Relationships Between Linear and Angular Velocity and Acceleration
- Common Angular Velocity and Acceleration in a Rigid System
Relationship Between Arc Length and Angular Displacement
For a point at a distance \( \mathrm{r} \) from a fixed axis of rotation, the linear distance \( \mathrm{s} \) traveled by the point as the system rotates through an angle \( \mathrm{\Delta\theta} \) is given by:
\( \mathrm{s = r\Delta\theta} \)
- \( \mathrm{s} \): linear distance traveled by the point (in meters)
- \( \mathrm{r} \): perpendicular distance from the axis of rotation (in meters)
- \( \mathrm{\Delta\theta} \): angular displacement (in radians)
Key Idea: Each point on a rotating rigid system travels a different linear distance depending on its radius from the axis, but they all rotate through the same angular displacement.
Example
The tip of a fan blade is \( \mathrm{0.30\,m} \) from the center. The blade rotates through an angle of \( \mathrm{60^\circ} \). Find the linear distance the tip travels.
▶️ Answer / Explanation
Step 1: Convert the angular displacement to radians:
\( \mathrm{\Delta\theta = 60^\circ = \dfrac{\pi}{3}\,rad} \)
Step 2: Use the relation \( \mathrm{s = r\Delta\theta} \):
\( \mathrm{s = (0.30)\left(\dfrac{\pi}{3}\right) = 0.314\,m} \)
Result: The tip travels \( \mathrm{0.314\,m} \).
Relationships Between Linear and Angular Velocity and Acceleration
The linear velocity and tangential acceleration of a point in circular motion are directly related to its angular quantities.
\( \mathrm{v = r\omega} \)
\( \mathrm{a_t = r\alpha} \)
- \( \mathrm{v} \): linear (tangential) velocity (\( \mathrm{m/s} \))
- \( \mathrm{a_t} \): tangential acceleration (\( \mathrm{m/s^2} \))
- \( \mathrm{r} \): radius from the axis (\( \mathrm{m} \))
- \( \mathrm{\omega} \): angular velocity (\( \mathrm{rad/s} \))
- \( \mathrm{\alpha} \): angular acceleration (\( \mathrm{rad/s^2} \))
| Linear Quantity | Rotational Analogue | Relationship |
|---|---|---|
| Displacement (\( \mathrm{x} \)) | Angular Displacement (\( \mathrm{\theta} \)) | \( \mathrm{x = r\theta} \) |
| Velocity (\( \mathrm{v} \)) | Angular Velocity (\( \mathrm{\omega} \)) | \( \mathrm{v = r\omega} \) |
| Acceleration (\( \mathrm{a} \)) | Angular Acceleration (\( \mathrm{\alpha} \)) | \( \mathrm{a = r\alpha} \) |
Key Idea: Each point on a rotating rigid system has the same angular velocity and angular acceleration, but its linear velocity and acceleration increase linearly with radius.
Example
A disk rotates with an angular velocity of \( \mathrm{10\,rad/s} \) and an angular acceleration of \( \mathrm{2.0\,rad/s^2} \). Find the linear velocity and tangential acceleration of a point \( \mathrm{0.15\,m} \) from the center.
▶️ Answer / Explanation
Step 1: Use \( \mathrm{v = r\omega} \) to find linear velocity:
\( \mathrm{v = (0.15)(10) = 1.5\,m/s} \)
Step 2: Use \( \mathrm{a_t = r\alpha} \) to find tangential acceleration:
\( \mathrm{a_t = (0.15)(2.0) = 0.30\,m/s^2} \)
Result: The point has a linear velocity of \( \mathrm{1.5\,m/s} \) and a tangential acceleration of \( \mathrm{0.30\,m/s^2} \).
Common Angular Velocity and Acceleration in a Rigid System
In a rigid system (a body that maintains its shape during rotation), all points rotate through the same angular displacement in the same time. Therefore, every point on the system has:
- The same angular velocity \( \mathrm{\omega} \)
- The same angular acceleration \( \mathrm{\alpha} \)
However, since \( \mathrm{v = r\omega} \) and \( \mathrm{a_t = r\alpha} \), the linear velocity and tangential acceleration increase with distance from the axis of rotation.
Key Idea: Angular quantities are the same for all points in a rigid system; linear quantities depend on the distance from the rotation axis.
Example
A solid disk rotates with an angular velocity of \( \mathrm{8\,rad/s} \). Compare the linear velocities of two points located \( \mathrm{0.05\,m} \) and \( \mathrm{0.10\,m} \) from the center.
▶️ Answer / Explanation
Step 1: Use \( \mathrm{v = r\omega} \) for each point:
For \( \mathrm{r_1 = 0.05\,m} \): \( \mathrm{v_1 = (0.05)(8) = 0.40\,m/s} \)
For \( \mathrm{r_2 = 0.10\,m} \): \( \mathrm{v_2 = (0.10)(8) = 0.80\,m/s} \)
Step 2: Compare results:
The point twice as far from the axis moves twice as fast linearly.
Result: Both points share the same angular velocity (\( \mathrm{8\,rad/s} \)), but the outer point has a greater linear velocity.