Lab 5 Centripetal Acceleration (Report on Exercise 1)

docx

School

University of Southern Mississippi *

*We aren’t endorsed by this school

Course

111L

Subject

Physics

Date

Dec 6, 2023

Type

docx

Pages

Uploaded by DeaconPanther2799

Lab 05: “Centripetal Acceleration”; Exercise 1 Date Performed: Oct 9, 2023 Date Submitted: Oct 12, 2023 Student: Kyra Pell (Ann) USM ID: w10132479 TA: Shwe Oo

Purpose Uniform circular motion occurs when objects travel in a circular path at a constant speed, while their direction of velocity changes. This change in velocity indicates acceleration, the acceleration found in this motion is called centripetal acceleration. The purpose of this experiment was to use a Centripetal Force Apparatus kit to determine how altering the length of the radius of a rotating mass will affect the revolution period (T). The measured physical quantities of this experiment are the total number of washers in the kit, the mass of all the washers, the average mass of a washer, the rotating mass, the hanging mass, the radius of the uniform circular motion (r), and finally the time needed to travel 10 revolutions. The calculated physical quantities are the time of one revolution (T), The circumference (C), the velocity (v), the centripetal force (Fc), the gravitational force of the hanging mass (Fg), and finally the percent error (% Error). The physical principles used to calculate these values are the uniform circular motion and the centripetal acceleration. Introduction As previously stated, I used a Centripetal Force Apparatus kit to determine how altering the length of the radius of a rotating mass will affect the revolution period. To do this I first Counted the number of washers in my kit, weighed their total mass and then calculated the avg mass of a single washer in kilograms (see Equation 1). I then measured my hanging and rotating masses also in kilograms. I then proceeded to set up my apparatus with a hanging weight of approximately 30 (g). I then used a stopwatch to time 10 revolutions of my apparatus at 4 different measured radiuses, with said radiuses shortening by 10 cm after every 10 revolutions. These radiuses were measured with a tape measure. Using these values, I then calculated the times of 1 revolution (see Equation 2), the circumferences (see Equation 3), the velocities (see Equation 4), centripetal forces (see Equation 5), the gravitational force of the hanging mass (see Equation 6), and finally the percent error (see Equation 7), in that order. The physical assumptions of this experiment are that I made accurate measurements with my scale, tape measure, and stopwatch. Also, that I used my apparatus properly. Equation 1 Equation 5 Equation 3 Equation 2 Equation 6 Equation 4 Equation 7

Occurs when objects travel in a circular path at a constant speed 1 Uniform circular motion Change in velocity when an object travels in a circular path at a constant speed 2 Centripetal acceleration Points toward the rotational axis of a rotating object, and causes the velocity to change directions without changing the magnitude 3 Centripetal force The distance traced out in a circular path 4 Arc length Force transmitted through a string or similar material that allows a force to act over a distance 5 Tension Data, Calculations, and Questions Physics 111L Centripetal Acceleration Final Report Student Name Kyra Pell Student ID 291291 Lesson Centripetal Acceleration Institution University of Southern Mississippi Session Fall 2023 Course Physics 111L Instructor Sidney Gautrau Test Your Knowledge Match each term with the best description.

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

Identify each statement as true or false. True False 1 2 The tension of a swinging string can The magnitude and direction of the provide the centripetal force needed to tangential velocity of an object in keep an object on its circular path. uniform circular motion change over time. The speed of an object undergoing uniform circular motion can be The arc length is equal to the subtended determined using the circumference and angle divided by the radius. the period of the motion. Exploration During uniform circular motion, the magnitude and direction of the velocity change. True False The term centripetal means . circular center-seeking

towards the margin None of the above

velocity is the velocity of an object moving in a circular path in uniform circular motion. Tangential Centripetal Circular None of the above The centripetal force always points in the same direction as the centripetal acceleration. True False For circular motion calculations, the angle is measured in units of . degrees radians revolutions None of the above The of an object undergoing uniform circular motion can be determined using the circumference and the period of the motion. acceleration

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

radius speed None of the above

velocity is equal to two times pi times the radius, (v = 2πr), v and r are directly proportional. Meaning an increased r will lead to an increased v, as well as a decrease in r will lead to a decreas of the velocity, as shown in the equation Fc=m moving v2/r, implying that a doubling of speed will require four times the centripetal force to keep the motion in a circle. This means that the cent provides the centripetal force needed to keep an object swinging on a string on its circular path. Gravity Tension Velocity None of the above Exercise 1 What effect does radius have on the velocity of a rotating object? What effect does the velocity of a rotating object have on the centripetal force exerted on it? The Moon orbits Earth at an average distance of 3.84 x 10 8 m in a path that takes 27.3 days to complete. What is the centripetal acceleration of the Moon? (Remember, you must convert time into seconds.) The centripetal acceleration can be found with the equation ac=v 2 /r. We have to radius 3.84 x 10 8 (m) but we need to find the velocity. To do so we need to use the equation (2πr)/T, where T is the time, it takes to travel the distance. After first converting the time to seconds, I then received an approximant value of 1025 (m/s) for velocity. We can now plug in the velocity and radius in the centripetal acceleration equation. This gives us an approximant value of 2.74 x 10 − 3 (m/ s 2 ).

The measured length from my shoulder to the center of a bucket is approximately 82 (m). So, using the equation, velocity =√rg, I find I have an approximant critical velocity of 28.36 (m/s). When a bucket of water is swung in a vertical circle, the water will remain in the bucket if the velocity is high enough. If you let the bucket slow down too much, the water will spill out. The critical velocity is the slowest velocity necessary to keep the water in the bucket. What is the critical velocity if the formula is velocity = , where r is the radius (the radius of your arm swing including the bucket) and g is the acceleration of gravity ( g =9.81 m/s 2 )? Hint: You must measure or estimate the length from your shoulder to the center of the bucket to determine the radius. Data Table 1: Constants for Varying Radius Number of Washers 20 Mass of all Washers (kg) 0.10156 Average Mass of Single Washer (kg) 0.005078 Rotating Mass (kg) 0.01699 Hanging Mass (kg) 0.03009

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

al force required to keep an object in circular motion depends on the rotating object's mass, the radius of the circular path, and the square of the angular velocity. However, it can indirectly aff Yes, increasing the mass of a rotating object does affect its velocity. As Data table 4 shows, as the mass increases, the velocity decreases to keep the momentum constant, and vice versa. Trial Radius (m) Time - 10 rev (s) Time T – 1 rev (s) Circumference (m) Velocity (m/s) F c (N) F g (N) % Error 1 0.70 9.72 0.97 4.40 4.53 0.50 0.29 72.4% 2 0.60 9.18 0.92 3.77 4.10 0.48 0.29 65.5% 3 0.50 8.80 0.88 3.14 3.57 0.43 0.29 48.3% 4 0.40 8.08 0.81 2.51 3.10 0.41 0.29 41.4% Data Table 2: Varying Radius – Measurements and Calculations Exercise 2 What was the effect of increasing the hanging mass on the centripetal force in Part 1 of this exercise? Did increasing the mass of the rotating object affect its velocity in Part 2 of this exercise? Explain your answer by referencing Data Table 4.

otating mass increases so does its velocity, its tangential velocity to be specific. This relationship holds because points farther from the axis of rotation need to cover a larger distance in one ro easured my radiuses, revolutions, and masses incorrectly or not precise enough. In the case of incorrect calculations, I could have made a mistake in my calculations leading to an error, or I wa nd precise enough. Finally, If I had more time, I could have practiced using my apparatuses so that I could get the most accurate measurements. oing uniform circular motion and the radius, r, of the circular path. The simulations in Exercise 2 have contributed to my understanding of how the velocity and the radius of centripetal accele ng a centripetal force apparatus lets you see first-hand the effects of centripetal acceleration. Therefore Exercise 1 was just as effective. What effect did increasing the radius of the rotating mass have on its velocity? Reference Data Table 5 in your answer. How do these results compare to those recorded in Exercise 1? What factors introduced error in both Exercise 1 and Exercise 2? How could the error have been reduced? 5. How did using the simulations in Exercise 2 contribute to your understanding of centripetal acceleration? Did you find them more or less effective than the physical experiment in Exercise 1? Explain your answer.

Data Table 3: Varying Hanging Mass - Measurements and Calculations Rotating Mass (kg) 0.025 Radius (m) 2.00 Circumference (m) 12.6 Trial Hanging Mass (kg) Time - 10 rev (s) Time T - 1 rev (s) Velocity (m/s) F c (N) F g (N) % Error 1 0.10 14.2 1.42 8.87 0.98 0.98 0.00% 2 0.13 12.4 1.24 10.16 1.29 1.27 1.57% 3 0.20 10.0 1.00 12.60 1.98 1.96 1.02% 4 0.26 8.8 0.88 14.32 2.56 2.55 0.39% Data Table 4: Varying Rotating Mass - Measurements and Calculations Hanging Mass (kg) 0.13 Radius (m) 2.00 Circumference (m) 12.6 Trial Rotating Mass (kg) Time - 10 rev (s) Time T - 1 rev (s) Velocity (m/s) F c (N) F g (N) % Error 1 0.025 12.4 1.24 10.16 1.29 1.27 1.57% 2 0.047 17.1 1.71 7.37 1.28 1.27 0.79% 3 0.056 18.6 1.86 6.77 1.28 1.27 0.79% 4 0.065 20.1 2.01 6.27 1.28 1.27 0.79% Data Table 5: Varying Radius - Measurements and Calculations Rotating Mass (kg) 0.025 Hanging Mass (kg) 0.13 Trial Radius (m) Time - 10 rev (s) Time T - 1 rev (s) Circumference (m) Velocity (m/s) F c (N) F g (N) % Error

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

1 0.25 4.4 0.44 0.39 0.89 0.08 1.27 93.7% 2 0.50 6.1 0.61 1.57 2.57 0.33 1.27 74.0% 3 0.75 7.1 0.71 3.53 4.97 0.82 1.27 35.4% 4 1.00 8.7 0.87 6.28 7.22 1.30 1.27 2.4% Competency Review Uniform circular motion occurs when objects travel in a circular path at a constant . velocity direction speed All of the above The centripetal acceleration points towards the center of the circle. True False The centripetal acceleration is always to the tangential velocity.

perpendicular tangential central All of the above The centripetal force is proportional to the . mass velocity squared inverse of the radius All of the above The force provides the tension force in the centripetal force apparatus. spring gravitational incline All of the above The tension force is greater than the centripetal force in the centripetal

force apparatus. True False An object moving in uniform circular motion with a radius of 0.6 m and a period of 0.4 s has a tangential velocity of m/s. 2 4 9 None of the above The centripetal force acting on a 0.30 kg object moving with a tangential velocity of 12 m/s in a 0.80 m radius circle is N. 4.5 54 380 None of the above The centripetal acceleration acting on a 0.30 kg object moving with a tangential velocity of 12 m/s in a 0.80 m radius circle is m/s 2 . 15 180

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

direction of the center of the turn, causing a frictional force that keeps the car moving in a curved path. , If the child themselves are the one holding the pole, then the pulling force of the child arms are what provides the centripetal force. Both the tension force and the pulling force acts toward force between the two. The normal force is the force exerted by a surface to support an object resting on it. While the frictional force occurs due to the movement of the carousel. I 480 None of the above In Exercise 1, the theoretical centripetal force was calculated from the . tension velocity weight None of the above Extension Questions For each of the following scenarios, describe the force providing the centripetal force for the motion: a. a car making a turn b. a child swinging around a pole c. a person sitting on a bench facing the center of a carousel d. a rock swinging on a string e. the Earth orbiting the Sun.

which allows the rock to maintain its circular path as it swings back and forth. wards the center of the circular orbit, keeping the Earth in its path around the Sun. Me and my lab project

My Calculations

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

Results In this experiment I had to count the total number of washers in my kit, weight their total mass, and calculate the avg mass of a single washer. This led me to report the corresponding values of 20, 0.10156 (kg), and 0.005078 (kg). I also measured 4 different radiuses of an object in uniform circular motion, weighted the hanging and rotating masses, and time 10 revolutions of the corresponding radiuses. This led me to report a rotating mass of 0.01699 (kg), a hanging mass 0.03009 (kg), the radiuses of 0.70 (m), 0.60 (m), 0.50 (m), and 0.40 (m), along with their corresponding 10 revolutions times of 9.72 (s), 9.18 (s), 8.80 (s), and 8.08 (s). The time of 10 revolutions values led me to report the correlating values of 0.97 (s), 0.92 (s), 0.88 (s) and 0.81 (s) for 1 revolution. After this, Using the radiuses values, I was able to report the correlating circumferences of 4.40 (m), 3.77 (m), 3.14 (m), and 2.51 (m). Then using the time of 1 revolution values and the circumference values I was able to report their correlating velocities of 4.54 (m/s), 4.10 (m/s), 3.57 (m/s), and 3.10 (m/s). After this, using the rotating (moving) mass, the velocities, and their corresponding radiuses I was able to report the associating centripetal forces of 0.50 (N), 0.48 (N), 0.43 (N), and 0.41 (N). Then using the hanging mass and the acceleration of gravity (g), 9.80 (m/ s 2 ), I was able to report a value of 0.29 (N) for the gravitational force of the hanging mass. Finally using the centripetal forces and the gravitational force I was able to calculate the following precent error values of 72.4%, 65.5%, 48.3%, and 41.4% respectfully. In this experiment, in theory, the centripetal forces should equal the gravitational force of the hanging mass. As shown above, when comparing these two forces, my precent error of all four trials are far above the accepted 5%, thus making this experiment invalid. Conclusion When I compare my centripetal forces with my gravitational force, I find that there are at least three possible sources of error. The most likely being my improper handling of my centripetal force apparatus. I could have very well misused my apparatus leading to an error in the results. There is also the possibility that I incorrectly measured my measured physical quantities, once again, leading to an error in my results. I also could have miscalculated my calculated physical quantities, again, leading to an error in my results.

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

Lab 5 Centripetal Acceleration (Report on Exercise 1)

Related Documents