Lab 5/8: Linear Speed and Acceleration in Circular Motion

Lab 5/8: Linear Speed and Acceleration in Circular Motion

Purpose

The purpose of this assignment was to determine the linear speed and acceleration of a point on a rotating beam as a function of its distance from the axis of rotation.

Procedure

A horizontal beam was spun on a rotational apparatus to simulate satellite motion. Video analysis was used to measure motion. The camera was positioned directly above the center of the beam to minimize distortion. Videos were recorded to capture at least two full revolutions of the beam at various angular speeds. 3cm, 8cm, and 15cm marks along the beam were selected from the center. The perpendicular components of the velocity were measured using the video to calculate the linear speed and corresponding centripetal acceleration.

                                                                            Analysis

1. Use the velocity components to determine the direction of the velocity vector. Is it in the expected direction?

By analyzing the graphs of the x and y positions over time, we see that the velocity vector is always tangential to the circular path. This is confirmed by the sinusoidal shape of the x and y graphs, which are out of phase by 90 degrees—exactly what we expect for circular motion. The direction of the velocity is perpendicular to the radius vector at every point, which is consistent with the theoretical prediction for uniform circular motion. Therefore, yes—the direction is in the expected direction.

2. Analyze enough different points in the same video to make a graph of speed of a point as a function of distance from the axis of rotation. What quantity does the slope of this graph represent?

Radius (m)	Speed (m/s)
0.03	0.0315
0.08	0.110
0.15	0.336

Plotting speed vs. radius gives a straight line. The slope of this line represents the angular speed (ω), based on the relationship $v = \omega r$. Calculating the slope between the largest and smallest data points:

$$\omega \approx \frac{\Delta v}{\Delta r} = \frac{0.336 - 0.0315}{0.15 - 0.03} = \frac{0.3045}{0.12} \approx 2.54 \, \text{rad/s}$$

This confirms that the slope is indeed the angular speed of the rotating system.

3. Calculate the acceleration of each point and graph the acceleration as a function of the distance from the axis of rotation. What quantity does the slope of this graph represent?

Using the formula $a = \frac{v^2}{r}$, the centripetal acceleration for each radius is:

Radius (m)	Speed (m/s)	Acceleration (m/s²)
0.03	0.0315	0.0331
0.08	0.110	0.1513
0.15	0.336	0.752

Plotting acceleration vs. radius results in a nonlinear graph, but if we plot acceleration vs. radius², the data becomes linear, which confirms the relationship$a$$=$${\omega }^{}$$2^{}$$r$a=ω2r. So, the slope of the acceleration vs. radius graph represents${\omega }^{}$$2^{}$ω2, and taking the square root of the slope gives the angular speed (ω).

Approximating the slope between the largest and smallest values:

$$\omega^2 \approx \frac{0.752 - 0.0331}{0.15 - 0.03} = \frac{0.7189}{0.12} \approx 5.99 \quad \Rightarrow \omega \approx \sqrt{5.99} \approx 2.45 \, \text{rad/s}$$

This aligns closely with the value calculated from the speed-radius graph.

The experimental results support the theoretical predictions for circular motion:

• The direction of the velocity is tangential, as expected.

• A graph of speed vs. radius yields a straight line whose slope is the angular speed.

• The centripetal acceleration increases with radius, and a plot of acceleration vs. radius shows a slope related to $\omega^2$.

How do your results compare to your predictions?

Our predictions were that:

• Velocity would always be tangent to the circle.

• Speed would be directly proportional to radius.

• Acceleration would increase with radius due to $a = \omega^2 r$.

3in:

8in:

15in: