Video Blog Analysis

  Nikki Montlick
Video Blog Analysis

The video to the right is the video my group used to collect the majority of our data for this lab. The yellow line against the blue wall represents a meter within the context of the video. My group uploaded our video to capstone and identified the points of motion the ball in the air. The proceeding image is the points of motion for the ball along the arc. The balls movement is forward and up.

Next, using capstone, my group created position versus time graphs and velocity versus time graphs and position versus time graphs in both the x and y for the motion of the ball.

3. Analysis:

Fg is the only force in the vertical direction. The greatest possible acceleration is -10m/s^2. When I highlighted my points on my graph my acceleration was within a a tenth of 10m/s^2. The FBD is the same as a. because the same object is being thrown. The Fnet=0 because there is no force in the horizontal. Therefore the acceleration is equal to 0m/s^2. I know this because it is the same as the velocity- since there is no acceleration, the velocity would remain constant. The velocity of the object will remain constant throughout its path because there is no force acting in the horizontal direction. 

c
   c. The initial velocity in the x direction is 0.6m/s. I know because I looked on the vty graph and on the equation of the line, b is always the initial velocity. I can identify the initial velocity in the x and y direction by looking at the equation on the graph. The equation is y=mx+b, and I know that b is equal to the initial velocity, then the initial velocity based off my equation of the graph is 0.6m/s. The initial velocity is .3m/s. I know this because on the velocity versus time graph in the y direction, the b in the equation of the line is .3m/s.The velocity at the top of its x path is .6m/s. 

f.      The velocity at the top of its path in the y direction is 0m/s. I can distinguish this by looking at my v versus t graph in the y direction. By looking to find a straight diagonal on the graph, I can see that the line cuts through the y value of 0. This represents the point where the ball hits the vertex at 0m/s.

  

The final velocity in the x direction is 2.94 m/s. This can be found be looking at the equation on the graph and identifying the number plugged in for b in y=mx+b. The final velocity in the y direction is -2.06   m/s. The ball traveled 1.13m total in the y direction. The ball traveled .71 meters.  How much time did it take for the ball to get to the top of the path? It took the ball 4.04 seconds to get to the top of the path.The ball was in the air for 4.348 seconds.

4. Conclusions

This project introduced a new perspective into thinking about force for our class. I split forces into vertical and horizontal directions. The vertical direction in this project incorporated the CAPM and UFPM model. The vertical direction used the following equations: (Keep in mind that any of these equations could solve for any of the variables used in them. Algebraically you could solve for whatever variable you didn't know by plugging in the known factors.)

Vf=at+Vi                 (equation for final velocity)

x=1/2(a)(t^2)+Vit    (equation for displacement)

Vf^2=Vi^2+2a        (equation for final velocity)

In the horizontal direction, there is no acceleration. Since there is constant velocity and no acceleration the following equation can be easily used. The horizontal direction applied the BFPM and CVPM models.

The vertical forces could be found using the following equations:

V= change in X

     change in t

UFPM: Nikki and Elise

Materials:

-cart

-buggy

-pulley

-Atwood machine

-500 kg weight

First we took the mass in grams of all variables in the experiment, and converted them to Newtons.

In this experiment, the goal was to predict the point that the buggy and the pulley would intersect. To make this prediction, we had to find the following variables:

•   net force

•   velocity of the buggy

mass and net force on the Atwood Machine

In this experiment, we did not take the acceleration of the buggy became the force was balanced since it moved at constant velocity.

Item	Mass (kg)
Buggy	.477
Cart	.9892
Pulley	.0509
Total system	1.04

First, we plugged in Fg, which was -10.04N, and could conclude that the diagram was balanced in the vertical direction, so therefore Fn was also 10.4N.  We knew that it was frictionless along the surface, so we concluded friction was 0N. Next, to solve for Ft, we used the equation: Fnet= -Ff+ Ft (which is equal to ma).

Using the data we found for the mass of the pulley, we found the force of this mass due to gravity.

.0509 kg

Fg2= 0.509N

After I found the Fnet in the horizontal direction, I plugged it and the mass into A=Fnet.

              Mass

Then I used the equation for displacement to solve for time of the hanging mass to drop.

ΔX= 1/2at^2+Vit

.86 = 1/2at^2 + Vit

(2) .86 = .5 t^2

1.8547 = t

Prediction:

1.85s =t

I also timed how long it would go for the buggy to go one meter. Since it is moving at constant speed, acceleration is equal to 0. So the equation becomes:

ΔX= Vit

Vi= ΔX  = 1 m= 0.25 m/s

       T         4

Now that we found the velocity, we can use the time it took for the hanging mass to drop and solve for the displacement of the buggy.

Δx= Vit

(.25)(1.85)

ΔX= .4625m

=46.25cm back from where the pulley will land.

Our pulley landed on the cart, so the experiment had 0% error.