Rotational Kinematics and Point Mass Analysis

Introduction

Today's lecture focuses on using relationships between rotational kinematic variables in graphical and mathematical forms to analyze the motion of point masses. We will explore several examples to illustrate the conversion between linear and rotational variables.

Example 1: Centripetal Acceleration

Problem: A wheel with a radius of 1 meter spins at a constant angular velocity of 2 radians per second. What is the centripetal acceleration of a point on the wheel's rim?
Solution:
- Draw a diagram of the wheel with a point on the rim.
- The point is 1 meter from the center and moves at an angular speed of 2 rad/s.
- Centripetal acceleration ac is given by: ac = \frac{v^2}{r}, where v is the linear speed and r is the radius.
- The relationship between linear speed and angular speed is: v = \omega r, where \omega is the angular velocity.
- Calculate the linear speed: v = (2 \text{ rad/s}) (1 \text{ m}) = 2 \text{ m/s}.
- Calculate the centripetal acceleration: a_c = \frac{(2 \text{ m/s})^2}{1 \text{ m}} = 4 \text{ m/s}^2.
- The centripetal acceleration is 4 m/s² towards the center of rotation.

Example 2: Angular Velocity Vector Direction

Problem: A bicycle travels north. What is the direction of the angular velocity vector of the front wheel?
Solution:
- Visualize the wheel rotating clockwise from the observer's point of view.
- Apply the right-hand rule: curl your fingers in the direction of rotation (clockwise), and your thumb points in the direction of the angular velocity vector.
- The angular velocity vector points into the screen.
- Using cardinal directions, if the bicycle travels north, the angular velocity vector points west.

Example 3: Angular Displacement

Problem: A spot of paint on a tire moves in a circular path of radius 0.42 meters. Through what angle must the tire rotate for the spot to travel 1.48 meters?
Solution:
- Use the relationship between linear and angular positions: \Delta s = r \Delta \theta, where \Delta s is the arc length, r is the radius, and \Delta \theta is the angle in radians.
- Rearrange to solve for \Delta \theta: \Delta \theta = \frac{\Delta s}{r}.
- Substitute the given values: \Delta \theta = \frac{1.48 \text{ m}}{0.42 \text{ m}} = 3.52 \text{ radians}.
- Convert radians to degrees: 3.5 \text{ radians} \times \frac{180 \text{ degrees}}{\pi \text{ radians}} \approx 200 \text{ degrees}.
Note: When converting from arc length to an angle, the result is in radians.

Example 4: Linear Velocity and Angular Velocity

Problem: A merry-go-round has a child at 4 meters from the center with a linear speed of 2 m/s. What is the linear velocity of a child at 6 meters from the center?
Solution:
- The merry-go-round rotates at a constant angular velocity, \omega.
- First, find the angular velocity using the child at 4 meters: v = \omega r \implies \omega = \frac{v}{r}.
- \omega = \frac{2 \text{ m/s}}{4 \text{ m}} = 0.5 \text{ rad/s}.
- Now, find the linear speed of the child at 6 meters using the same angular velocity: v = \omega r.
- v = (0.5 \text{ rad/s}) (6 \text{ m}) = 3 \text{ m/s}.

Example 5: Angular Acceleration and Graphical Representation

Problem: A ladybug starts with zero angular velocity and has an angular acceleration of -8 rad/s². Sketch the angular velocity vs. time and angular position vs. time graphs.
Solution:
- Angular Velocity vs. Time:
  - Since the angular acceleration is constant, the angular velocity vs. time graph will be a straight line with a negative slope equal to -8 rad/s².
  - The line starts at zero and decreases linearly.
- Angular Position vs. Time:
  - Assuming the ladybug starts at an angular position of zero, the angular position vs. time graph will be a parabola that is concave down.
  - The angular position changes more quickly as time goes on because the angular velocity becomes increasingly negative.

Example 6: Area Under Angular Velocity vs. Time Curve

Problem: A ladybug's angular velocity starts at 10 rad/s and decreases to 5 rad/s over 2 seconds, then remains constant at 5 rad/s for another 2 seconds. How many radians has the ladybug traveled in 4 seconds?
Solution:
- The area under the angular velocity vs. time curve represents the angular displacement.
- First 2 seconds:
  - The area can be split into a triangle and a rectangle.
  - Triangle area: A_\text{triangle} = \frac{1}{2} bh = \frac{1}{2} (10 - 5 \text{ rad/s}) (2 \text{ s}) = 5 \text{ radians}.
  - Rectangle area: A_\text{rectangle} = bh = (5 \text{ rad/s}) (2 \text{ s}) = 10 \text{ radians}.
  - Total area for the first 2 seconds: 5 + 10 = 15 \text{ radians}.
- Next 2 seconds:
  - The area is a rectangle: A = bh = (5 \text{ rad/s}) (2 \text{ s}) = 10 \text{ radians}.
- Total angular displacement:
  - 15 \text{ radians} + 10 \text{ radians} = 25 \text{ radians}.

Conclusion

Today, we used relationships between rotational kinematic variables in graphical and mathematical form to analyze the motion of point masses. Several examples performed illustrated the conversion of linear and rotational variables. We also used basic concepts of graphical models for linear variables to solve angular variable problems. Next lecture, we will discuss how rotational kinematic equations apply to a point mass over time.