25-March-2015 Centripetal Acceleration vs. Angular Frequency

Purpose: To find a relationship between centripetal acceleration and angular speed.

Theory: That a relationship can be found be seen on a graph of acceleration vs. angular speed, to confirm Acceleration_centripetal=radius*angular speed^2 (a_c=r*w^2) .

Procedure: Using a heavy rotating disk, a timer, and logger pro; we measure the centripetal acceleration and time it takes to complete ten rotations, plugging in this data in a acceleration vs. angular spped^2 graph should give a relation that confirms our equation in the theory.


The Apparatus

*At 4.8 volts, the apparatus spun with an a_c of 1.908 m/s^2.

*The time for the first rotation was 0.7405 seconds, and 10 rotations had been completed in 17.234 seconds since the timer began.

*So subtracting the time for the first rotation from the end time is 16.4935 seconds to complete 10 rotations.

Results for all

	4.8 volts	6.2 volts	7.8 volts	8.8 volts	10.9 volts	12.6 volts
before rotation (s)	0.7405	0.2161	0.78729	0.12808	0.44378	0.07418
10th rotation (s)	17.234	12.6919	9.02378	7.13992	6.06639	4.64158
time for 10 rotations (s)	16.4935	12.4758	8.23649	7.01184	5.62261	4.5674
acceleration m/s^2	1.908	3.573	7.891	11.13	17.21	25.89

*angular speed (w) is calculated as 2Pi*number of rotations/time of rotations, so all we have to do to calculate w is multiply 2Pi by ten and divide by the total time for each ten rotations.

Graph

*Setting up an acceleration vs angular speed graph, we see a linear fit line works quite well.

*The slope is 0.1382, given that our model is a_c=r*w^2, 0.1382 must be the radius.

*The actual radius is 13.8 cm, which converts to 0.138, a near perfect prediction for our model.

Conclusion

*Our model was nearly perfect, accurately predicting the radius with the given data.

*I don't see any error with the concept of the experiment, or the equipment.

*The only uncertainty, perhaps, is with the actual measuring of the radius; as it the professor used
a meter stick an those can never be more accurate than +/-0.05.

23-March-2015 Trajectories

Purpose: To use projectile motion equations to predict the impact point of a ball on an inclined board.

Theory: That kinematic equations are adequate for predicting the landing point of a ball, on an inclined board, and general motion of the ball through the air.

Procedure: Set up two V-tracks on top of our desk, so that a steel ball rolls consistently at the same velocity in the horizontal position, off of it; also measure the height of the desk (plus uncertainty). Using carbon paper to track where the ball lands on the ground, five times, measure how far from the desk it landed generally (plus uncertainty). Calculate the horizontal velocity, place a board up against the V-track so that the steel ball will strike the board as it falls, and calculate a prediction of where on the board the ball will strike (also do this five times). Measure the angle (plus uncertainty) of the inclined board, calculate the propagated errors and compare with your prediction.

*The Track was set up.

*The height of the track was measured to be y=94.2 +/- 0.1 cm

*After five trials the general distance of the ball's landing was measured to be x=66.3 +/- 0.2 cm.

*Since the original velocity in the vertical position is always zero in our model, we can calculate the time it takes for the ball to reach the ground using (y=0.5*g*t^2).

*Time is calculate to be 0.438 seconds.

*Since movement in the horizontal direction is independent from the vertical, we can generalize the velocity of the horizontal position using the time it took for the ball to fall and the distance it traveled in the horizontal direction in that time (V=x/t).

*V=1.51 m/s

*We placed a board to lean against the table, so that the ball would strike it before the floor, and measured the angle it created with the floor.

*Angle was 49 +/- 0.5 degrees.

*The distance(d) traveled on the board can be calculated in the x direction using x=d cos (angle), x also equals V*t, so set equal to each other, (t=(d cos angle)/V).

*Similarly d traveled on the board in the y direction can be expressed (y= d sin (angle)), y also equals 0.5*g*t^2, so set equal to each other and replacing t with the equation calculated in the previous bullet point. (d=(sin(angle)*V^2)/(0.5*g*cos(angle)^2)).

*We can now calculate d, its d=0.818 m.

*We performer the experiment five time to find the actual distance of d, its generally d=0.791 m.

*The propagated error can be calculated as (dd=sec(angle)*(uncertainty of x)+csc(angle)*(uncertainty of y)+ ((2V^2)/g)*(sec(angle)^2*sec(angle)+tan(angle)*sec(angle)*tan(angle))*(uncertainty of the angle)

*Its calculated to be dd=+/-0.026 m, which is within my predicted 0.818 m

Conclusion

*The actual was within my prediction and its propagated error, though barely, so I consider it a good model.

*Problems with the experiment itself, that it doesn't account for air resistance slowing the ball.

*Our process probably could have been better, say by averaging the five trials distances in the x direction, and on the board, instead of "generalizing" or approximating their position.

11-March-2015 Modeling Friction Forces

Purpose: By doing five different experiments we will determine the coefficient of static and kinetic friction between a block and our table, as well as the coefficient of static and kinetic friction between a block and a track.

Theory: Static friction (F_s) and kinetic friction (F_k) are equal to their coefficients times the normal force, all forces acting on an object are equal and opposite, the sum of forces acting on an object (the block) is equal to the mass of the object times its acceleration. In experiments:

1) This means the tension from cup suspended with a (assumed) massless pulley will create a tension equal the normal force times the coefficient of static friction.

2) A constant pull force will equal the coefficient of kinetic friction.

3) The coefficient of static fiction times the normal force can be determined from a motionless block on a sloped surface track.

4) The coefficient of kinetic fiction times the normal force can be determined from a sliding block on a sloped surface track.

5) The kinetic friction can be confirmed by a (assumed) massless pulley creating a tension that pulls the block, and a motion sensor records actual acceleration.

Procedure: In experiment

1)

*Get four blocks, string, a pulley, a Styrofoam cup, and some water

*Weigh each block to determine mass

*Tie string to one block, hang the other side of the string off from the pulley and attach the Styrofoam cup to the other side of the string.

*Add drops of water to the cup, until the block begins to slide

*Record mass of cup, place another block on top of the original, and repeat the process until you have used all four blocks

2)

*Get four blocks, string, force gage and logger pull.

*Weigh each block, and zero the force gage.

*Tie one end of the string to a block, and the other to the force, and at a constant speed pull on the force gage to drag the block.

*Add another block on top of the original, repeat, add the third block, repeat, add the fourth block, repeat.

3)

*Get a block and track.

*raise the block up slowly until it begins to slide, and determine the angle of the raised block.

4)

*Get a block, a track, a motion sensor, and logger pro.

*Place the block on an incline that it will slide on down the track, record angle and acceleration of block.

5)

*Get a block, a pulley, string, a hanging mass, a track, and a motion sensor.

*Attach the string to the block and hanging mass, hang mass over pulley, place block on track, and record acceleration with motion sensor.

Experiment 1

*We got the necessary supplies

*Multiplying the mass of the block by acceleration due to gravity (g=9.8 m/s^2), gives us the normal force, and the tension in the string (mass of cup times g) is equal to the friction force.

Results

Mass Blocks (kg)	Mass Cup (kg)	Normal force (N)	Friction force (N)
0.131	0.059	1.284	0.5782
0.252	0.093	2.47	0.9114
0.359	0.121	3.518	1.186
0.485	0.165	4.753	1.617

Graph Friction Force vs Normal

*The slope is the coefficient of static friction, and we get 0.3475 +/- 0.01287

*The reason: F_s*Normal force=Friction Force, solve for F_s

Experiment 2

*We set up equipment

*Zeroed the pull gage, pulled, recorded the run, and repeated four times with increasing mass

Results

Mass Blocks (kg)	Normal force (N)	Friction force (N)
0.121	1.186	0.3909
0.252	2.47	0.7374
0.378	3.704	1.01
0.485	4.753	1.232

Graph Friction Force vs Normal

*Same reasoning in determining F_s, the slope is F_k and that is 0.2712 +/- 0.0098

Experiment 3

*Set equipment up

*The track was raised to 20 degrees, before it started to slip, and the block(m) was 0.105kg

*Equation of net forces in the sloped direction (F_s*N=m*g*cos 20)

*Solving for F_s gives us 0.3640

Experiment 4

*The track was raised to 26 degrees, and set up the motion detector to record acceleration.

Graph

*We average the acceleration, and get 1.789 m/s^2

*Equation for acceleration (F_k*g*cos26 - g*sin26 = a)

*Solving for F_k we get 0.2846

Experiment 5

*We attached a .05kg mass to a string, hang it over a pulley, and the other end of the string to the block.

*Using the equation for acceleration, we determine the block should accelerate 1.877 m/s^2

*Actual was 2.613 m/s^2,  28% off from calculated.

Conclusion

*From my calculations the coefficient of static friction was larger than the kinetic friction, which was expected.

*As for the large error for acceleration in experiment in 5 and 4, I think the acceleration was effected by the track's ramp. As the block slid, it would turn askew and grind against the rallying, instead of being pulled straight.

*I'm satisfied with my models for calculating the coefficient of friction.

9-March-2015 Modeling Air Resistance

Purpose: Determining the relationship between air resistance force and speed, and seeing if we can model the tall of an object including air resistance.

Theory: Once an object reaches terminal velocity while falling, the force of friction from the air and the weight of the object are the net force, which is equal to the mass and its acceleration. We can use this relation to create an expression for air resistance with velocity as a term, and can test how accurate our model is by comparing the predicted velocity vs. actual measured velocity. Hopefully with only 10% of error.

Procedure: Take five coffee filters, weigh them to determine mass, drop one from a certain height, record the fall, place another coffee filter atop the previous one, and repeat dropping and recording until you have used all five filters. Load videos into logger pro, in order to determine velocity of each trial run, record velocity, create a velocity vs mass times gravity, use it to create an equation/model for air resistance, test model on excel and compare it to the recorded velocity to determine how accurate it was.

*First we dropped a coffee filter from a balcony.

*Loading the video into logger pro, we were able to determine the velocity by plotting points at each frame of where in relation to the balcony (roughly 1 meter) the filter was while falling.

*We repeated the process, always adding another coffee filter to increase the mass, five times.

*We determined the terminal velocity of each run, recorded the mass of each run, and calculated the force of mass times acceleration due to gravity of each run.

Recordings

V (m/s)	M (kg)	F (N)
0.9578	0.0008947	0.008768
1.323	0.001758	0.01723
1.605	0.002684	0.0263
1.922	0.003579	0.03507
2.253	0.004473	0.04384

*We created a velocity vs force of mass*gravity graph, since the only two forces acting on the falling filter are air resistance (with velocity as a factor in the equation) and the filter mass times acceleration due to gravity, we hope to see some type of relation between the two on a graph.

Graph

*The graph is exponential, and when fitted to give an equation it gives us the force of the mass times gravity is equal to the velocity times a constant taken to a power (mg=Av^B).

*A and B are constants, and represent factors that would effect air resistance, like surface area and shape and friction of the air.

*The value for A given from logger pro is 0.01105 and B is 1.725, and v is the terminal velocity of a falling coffee filter

*We now test our model using excel.

*Our model is the net force(the mass as it accelerates downwards) will equal the force of weight minus the force of air resistance

                                                                  (ma=mg-Av^B)

*In time increments of only .001 seconds, the change in velocity will be the acceleration determined from our model multiplied by the time increment.

*Eventually the changes in velocity will become so minute, the velocity will be virtually unchanged by them, this is our terminal velocity.

*For a single filter of mass 0.0008947kg

Results

t=	0.001
A=	0.01105
B=	1.725
m=	0.0008947
t (s)	a (m/s^2)	∆v (m/s)	v (m/s)	∆x (m)	x (m)
0	9.8	0	0	0	0
0.001	9.79576796	0.0098	0.0098	0.0000098	0.0000098

*Scrolling down

t (s)	a (m/s^2)	∆v (m/s)	v (m/s)	∆x (m)	x (m)
0.485	0.00216469	2.2074E-06	0.8743945	0.00087439	0.36720737
0.486	0.00212285	2.1647E-06	0.87439666	0.0008744	0.36808177
0.487	0.00208182	2.1229E-06	0.87439879	0.0008744	0.36895617
0.488	0.00204158	2.0818E-06	0.87440087	0.0008744	0.36983057
0.489	0.00200212	2.0416E-06	0.87440291	0.0008744	0.37070497
0.49	0.00196342	2.0021E-06	0.87440491	0.0008744	0.37157938

*At close to 0.486 of second the velocity to four significant figures "settles" at 0.8744 m/s, all further changes to velocity will be so miniscule, it can be argued that the filter is now at terminal velocity.

*For the first run the "predicted" velocity was 0.8744 m/s, actual 0.9578, error 8.7%

*For the first run the "predicted" velocity was 1.304 m/s, actual 1.323, error 1.4%

*For the first run the "predicted" velocity was 1.653 m/s, actual 1.605, error 2.9%

*For the first run the "predicted" velocity was 1.953 m/s, actual 1.922, error 1.6%

*For the first run the "predicted" velocity was 2.222 m/s, actual 2.253, error 1.4%

Conclusion

*My model worked very well, all "predicted" terminal velocities were only off by less than 10%, so I am satisfied with the values of A and B, and the formula Av^B as a model for the force of air resistance.

*Perhaps the greater error for the first run was because of how more "malleable" a single coffee filter is, its surface area and shape could change easier while it fell vs. multiple coffee filters holding each others shape in place.

*The whole process might be time consuming, as the constants A and B change depending on the shape of the object, its surface area, and air temperature effecting air density. All contribute to air resistance, so the entire experiment would have to be done again for another object of differing shape and multiple outside temperature.