Lab Report 3

docx

School

Northeastern University *

*We aren’t endorsed by this school

Course

1141

Subject

Mechanical Engineering

Date

Apr 3, 2024

Type

docx

Pages

Uploaded by GeneralFlyMaster440

Report for Experiment #5 Uniform Circular Motion Abstract In this experiment, we used it the equipment to find the centripetal acceleration of an object. In this experiment, we found the acceleration, velocity and period by means of timing and calculations. These data were obtained to find their role in the centripetal force. In the experiment, by using a rubber band and a spring to generate centripetal force, we investigated how different tensions affect centripetal acceleration and the force of rotation around the apparatus. Introduction The main heading of this experiment is rotational motion. The most important thing in this is the centripetal force. Centripetal force is the force that causes an object moving in a circle to change its direction constantly but to move steadily around the circle as a whole. However, the constant change of direction means that the object moving in a circle always receives a force towards the center. According to Newton's second law, a centripetal force is needed and always a centripetal acceleration is needed. The magnitude of this centripetal force can be calculated using the following equation F = m v 2 r The main objects of this were 1. to understand the velocity of an object in uniform circular motion, and acceleration in uniform circular motion 2. to measure the force on an object in uniform circular motion 3. to apply Newton’s second law to

an object in uniform circular motion. In investigation1, we measured the rotational velocity and centripetal force of the iron block by using the equipment. For this experiment, the centripetal force was tension, since it was perpendicular to the velocity. The following equation can be used to express the centripetal force: F ( t )= mg Through several experiments, we can calculate the period based on the data obtained and the following formula. T = 2 π ω in this equation, ω is the angular speed which could be calculated by the times of rotation over how long did these rotation take. Once the period is calculated, we can calculate the speed by using the following equation: V = 2 π r T After completing all the tests, we inserted the compiled data into an Excel sheet and plotted scatter plots of force vs. velocity and force vs. v². Investigation 1 In investigation 1, the equipment that we used were Rubber bands, Assorted springs, Stopwatch or timer program, Plastic pail, washers, String, paper clips, Ruler, and Digital scale. The way how they were assembled is showed in the graph on the right. By unscrewing the screws on the top of the column, we could adjust the front and rear position of the horizontal arm of the instrument until the bob is directly above the pointer. String was used to provide different values of centripetal force to the bob. The amount of centripetal force was determined by the mass of the bucket and washer. The bucket was attached to the other end of the rocker by a string across the pulley, then the washer was added to the bucket until the rocker is realigned with the pointer, and the weight of the washer and bucket are recorded in the data sheet. However, after the bucket was removed, the speed and period were calculated by artificially providing the bob with a rotational force and then using the phone to record the time and number of revolutions. The following two tables were the data that we collected. Since the 17 column table was too big to implement in one table, I divided them into two pieces.

Table : Time Trials, Mass, Force, revolutions, period, velocity, and their errors Trials M (kg) Error M (kg) Centripet al Force (N) Error (N) Revolution s t1 t2 t3 Averag e Time (s) 1 0.13 0.009 1.274 0.088 2 2 2.79 2.97 2.91 2.89 2 0.133 0.018 1.3034 0.176 4 2 2.72 2.77 2.72 2.75 3 0.568 0.032 5.5664 0.313 6 5 4.08 3.59 3.86 3.84 4 0.482 0.034 4.7236 0.333 2 5 3.43 4.08 3.61 3.707 5 0.214 0.014 2.0972 0.137 2 5 5.9 5.95 6.15 6 6 0.947 0.039 9.2806 0.382 2 5 3.17 2.84 2.91 2.973 7 0.637 0.039 6.2426 0.382 2 5 4.18 3.91 3.68 3.923 Error t (s) Period T Error T (s) v (cm/s) Error v v^2 (cm/s) Error v^2 0.09165 1.445 0.0458 80.877 2.565 6541.09 103.73 0.02887 1.375 0.0144 84.994 0.893 7223.98 37.97 0.2454 0.769 0.0491 151.973 9.701 23095.79 737.12 0.3356 0.741 0.0671 157.716 14.287 24874.33 1126.61 0.1323 1.2 0.0265 97.389 2.148 9484.62 104.59 0.1739 0.595 0.0348 196.526 11.487 38622.47 1128.72 0.2503 0.785 0.0501 148.939 9.497 22182.83 707.24 m of bob (kg) 0.434kg ± 0.00005 kg radius (cm) 18.6cm ± 0.01cm

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Before doing the experiment, we first let the rope supporting the rocker remain vertical. In this case, the bob will not receive a horizontal force. Therefore, we can get the tension of the rope equal to the weight of the sphere mg. And when the experiment starts, we will give a force to this support rocker so that it starts to rotate. In this case, the rocker also receives a centripetal force needed to maintain the circular motion of the rocker. We keep the rope vertical by using a spring connected between the post and the rocker, which means that the horizontal net force is 0. The centripetal force is equal to the tension of the spring. By looking at the table we recorded, we can see that the larger the mass of the object, the more tension is needed to align the rocker. The following two graphs show the relationship between force and velocity and the relationship between force and velocity squared. In Figure 1, we added the error part and made it to be a curve graph. In Figure 2, we added the tenderline to the error because of the equation

F = m v 2 r and we also called out the y-intercept to make the data more accurate. Since the bob did not have any speed at the beginning, the initial centripetal force was zero. Therefore, the y-intercept can be adjusted to make the image more accurate. The slope equals to F/ v 2 =m/r. According to appendix A, the error in slope is δ slope = √ ( δ v 2 v 2 ) 2 +( δ F F ) 2 =1.6575e^-5N/ cm 2 /s the average of (m/r) is 0.0002389 which is a little bit larger than the slope in the graph. I agree with the experimental error because they are the same if choose four valid digits. Conclusion In this experiment, we tested the weight of the string with different strengths and recorded the time required to complete two revolutions and calculated the period. The data obtained was then entered into Excel to obtain a table and two images were created. The two graphs represent the relationship between centripetal force and velocity, and the square of centripetal force and velocity, respectively. The relationship between centripetal force and velocity is curvilinear, while the relationship between centripetal force and velocity squared is linear. By analyzing our data, we can conclude that as the mass of an object increases, its force and velocity also increase. At the beginning, some of the data we calculated are very unrealistic. For example, the slope in our image is 0.00002 and the data we calculated is 100 times larger than the image, and the error of the square of the velocity is much larger than the expected value. Later we found out that it was a problem with our choice of units. In the next experiment, we need to improve our unit selection. Adjust the units of multiple data to the most common combination in SI units. Questions 1. Slope = F / v² =(kg*m/s²) / (m² s²) = kg/m 2. The lines pass through the origin because the object initially starts at rest, so there is no force or velocity. 3. Path B. Since the rate of marble is always tangential to the circle when it travels in the circle. 4. v²/r is the answer of centripital force divided by mass. Since the precision of v = 1%, the precision of v² is 2x1 = 2%. The precision of r = 1%. Thus, precision of (v²/r) = 2% +1% = 3% 5. In circular motion, a=v^2/r. If the v is doubled, r has to multiply 4 in order to keep the centripetal force zero.