lab 10

docx

School

Prince George's Community College, Largo *

*We aren’t endorsed by this school

Course

272

Subject

Physics

Date

Dec 6, 2023

Type

docx

Pages

6

Uploaded by CoachFerretMaster288

Report
Lab #10: Centripetal Force - Lab REPORT
Intro Centripetal Forces implies "center-seeking," gravitational and electrical interactions are given, correspondingly, for both of these situations. The World orbits around the Sun, for instance. The electrons that pass around the nucleus are another example. The centripetal force is also what holds certain objects in space. As it goes along the object has a constant speed, but momentum varies in terms of position when it moves in circular motion, so the direction's velocity is perpendicular to the circular motion. This transition is the product of centripetal acceleration due to centripetal force velocity. Objective In this lab you will swing an object of mass m on the end of a string in a horizontal circular path of radius r . The string tension will provide the centripetal force F that produces the centripetal acceleration and keeps the object revolving in its circular path. The centripetal force will be determined by using a spring scale or force sensor that will measure the tension on the string. By timing the revolutions and calculating the distance traveled by the object, you will determine the velocity v of the object. You will demonstrate the relationships described in between the centripetal force F , the mass m , the radius of revolution r , and the velocity v . F = mv 2 /r v = 2πr/T m/r. Procedures In Part 1, the centripetal force of an object moving in a circular direction at a constant radius of rotation will vary and the velocity of the object will be determined. You will illustrate the dependency of the centripetal force on the velocity square. A string moving through a glass tube mounted at the top of a vertical rod links a spinning target to a force sensor mounted on the rod.
The ball will spin in a circle with a radius of about the distance from the glass tube to the center of the ball mass. For four experiments, complete the following measures using the centripetal force F values of 1.00 N, 2.00 N, 3.00 N and 4.00 N. For each trial, estimate and record the value of the revolution time T. Measure and record the value of v for each trial. For each trial, determine and record the value of v2. Construct a graph of the centripetal force F values vs the velocity square v2 values you entered in the table. Divide the value of m by the value of r to determine the known value of m/r. Calculate and report the percent error of your experimental value (m/r)exp relative to the known value of m/r. Use these procedures to repeat for exceroise 2 and 3. Results Centripetal Force Part 1: Variation of Force Lab Report Name: 12:07 PM, 11/20/2020 m = 0.100 kg r = 0,01 m Trial F (N) n t (s) T (s) v (m/s) v 2 (m 2 /s 2 ) 1 1 15.9 14 20 6.74 7.45 2 2 20 19 26 5.63 8.5 3 3 19 17 24 8.43 10 4 4 12 20 29 9.6 13 Graph Title: Centripetal Force X -Axis Label: force Y -Axis Label: time Expand Report F = 0.5193 kg/m v 2 + -2.56 N Slope = ( m / r ) exp = 2.3 kg/m m / r = 1.32 kg/m % error = 45 % Centripetal Force
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
Go to Part 1 Centripetal Force Part 2: Variation of Radius Lab Report Name: 6:19 PM, 11/20/2020 m = 1000 kg F = 2 N Trial r (m) n t (s) T (s) v (m/s) v 2 (m 2 /s 2 ) 1 15.9 2 4.34 6.42 2.34 1.3 2 19.3 2 4.31 4.23 3.32 1.5 3 28.3 2 4.5 4.32 4.21 2.34 4 37.3 2 7.5 4.65 3.32 2.12 Graph Title: Variation of Radius X -Axis Label: error Y -Axis Label: time Expand Report V 2 = 0.0431 m/s 2 r + 0.73 m 2 /s 2 Slope = ( F / m ) exp = 2.45 m/s 2 F / m = 4.23 m/s 2 % error = 2.34 % Variation of Radius
Go to Part 1 Centripetal Force Part 3: Variation of Mass Lab Report Name: 8:05 PM, 11/20/2020 r = 0.100 m F = 1 N Trial m (kg) 1/ m (1/kg) n t (s) T (s) v (m/s) v 2 (m 2 /s 2 ) 1 15.9 2.78 2.34 8.7 4 4.34 5.4 2 19.3 2.79 3.21 9.8 4 2.33 4.3 3 28.3 2.31 3.26 6.5 5.42 3.23 4..19 4 37.3 2.34 3.2 4.3 2.31 3.56 5.42 Graph Title: Variation of Mass X -Axis Label: mass Y -Axis Label: erroe Expand Report V 2 = 0.3781 Nm 1/ m + 3.81 m 2 /s 2 Slope = ( Fr ) exp = 12.3 Nm Fr = 4.50 Nm % error = 3.23 % Variation of Mass CONCLUSION In this experiment, I studied the centripetal acceleration of an object in a uniform circular motion and checked its relationship between force and as a result, centripetal acceleration. The second section of my error analysis explicitly describes what kind of difference the amount of force applied to a given mass will influence its direction and acceleration. When we do such an experiment, we assume the mass to have the defined radius following the process, and that our measured values and experimental values to all be the
same. This is not the case in this lab where the examination of possible errors will shed light on whether the data deviates from the predicted values.
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