Lab 114_ Uniform Circular Motion

pdf

School

New Jersey Institute Of Technology *

*We aren’t endorsed by this school

Course

111A

Subject

Physics

Date

Dec 6, 2023

Type

pdf

Pages

9

Uploaded by MegaRockDinosaur11

Report
Lab 114: Uniform Circular Motion Physics 111A Section 014 Group H Instructor: Date of Experiment: 03/29/23 Date of Submission: 04/05/23
Equipment List: Centripetal force apparatus set Standard weight set 50 g Weight hanger Stopwatch Vernier caliper Ruler Bubble level Digital scale on the lab room counter Introduction: Objectives: In this lab, we looked to study the motion of a body traveling with constant speed in a circular path (uniform circular motion). We also sought to verify the expression for centripetal acceleration and centripetal force. Background: As can be seen in Figure 1, a body with mass (m) is traveling in a circular path of radius r around a point (O).
At any point along the path, the body has a linear velocity tangent to the circle and according to Newton’s First Law of Motion it would retain this linear velocity if not acted upon by external forces. If the body is traveling with a constant speed, it has acceleration normal to the path of motion which changes the direction of the velocity. This acceleration points towards the center and is referred to as centripetal acceleration. As we know, Newton’s Second Law states: (1) ∑ 𝐹 = ?𝑎 This force (F) represents the centripetal force. Referring back to Figure 1, the linear velocity at point A has the same magnitude as point B but in traveling from A to B the direction of motion has changed. The change in velocity can be represented by ∆𝑣 = 𝑎∆? . By similar triangles we can say: (2) ∆𝑣 ∆? = 𝑎∆? 𝑣∆? = 𝑣 ? (3) 𝑎 = 𝑣 2 ? The relationship between linear velocity (v) and angular velocity ( ) can be written as: (4) 𝑣 = ?⍵ = ?(2π?) Once r and n are determined, centripetal force can be calculated as: (5) , where m is the mass of the rotating body 𝐹 ? = 4π 2 ?? 2 ? The external force F necessary to produce a deformation d in an elastic body can be expressed as: (6) 𝐹 = 𝑘?
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
Experimental Procedure: We ensured the apparatus base was level using the bubble level and then continued on with the experiment We began by measuring the mass of the revolving object three times: once with no weight added, second time with an additional 50 g, and third time with 100 g weight added. We then adjusted the position of the radial indicator rod and using a ruler we measured the radius to be 0.135 m. We then adjusted both the position of the crossarmand the counterweight and ensured it was level again with the bubble level. We then proceeded to determine the static force required to displace the mass to the same radial position by passing a string over the pulley. One end of said string was hooked onto the revolving mass and the other end attached to a weight hanger. We measured the weight to be 1 kg. We then began to rotate the system by applying torque with our fingers on the knurled portion of the shaft. We got the rotating mass to keep passing directly over the indicator and began to measure the rate of rotation (n). We measured how long it took for the revolving mass to make 50 complete revolutions while passing over the indicator. We then conducted the trial two more times by keeping the same radius and increasing the mass by 50 g both times.
Results: Trial No. Mass (kg) Radius (m) Time of 50 Revs Avg. Time of 50 Revs RPS* (n) Force Computed Force measured % Diff 1 0.4495 0.135 29.79 29.07 1.72 7.09 9.8 32% 28.74 28.67 2 0.4990 0.135 29.70 32.02 1.56 6.47 9.8 40% 32.04 34.31 3 0.5490 0.135 31.69 31.74 1.58 7.30 9.8 29% 32.02 31.5
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
Discussion: 1. What is causing the centripetal force on the body of mass m when it undergoes the circular motion in this experiment? In this experiment, we observed a mass m rotating thus the force pulling the mass towards the center of the system while in motion was the spring attached to the mass itself. As we spun the mass increasingly faster, the spring stretched more and more thus showing that the force increased in order to try and pull the object back towards the center. 2. Is the expression (eq. 5) for centripetal acceleration verified? Is the expression (eq. 6) for centripetal force verified? Both equations are verified through completion of this experiment. While it can be observed that there were percent differences between the forces measured vs. computed, the percent differences weren’t too drastic. It can be observed that the centripetal acceleration of equation 5 is used in equation 6. 3. Is it possible for an object to have acceleration when the velocity of the object is constant? How about when the speed of the object is constant? No it is not possible as the shear definition of acceleration is rate of change of velocity. Thus, if velocity is not changing, the object is not accelerating. However, when an object has a constant speed it is possible for it to have an acceleration. If said object is moving in a circle, it has acceleration normal to the path of motion.
4. Consider two people, one at the equator, and another at the North Pole. Which has the greatest centripetal acceleration? The person at the equator because that individual has a much larger radius than the person at the North Pole. A greater radius results in a greater tangential velocity and a greater tangential velocity results in a larger centripetal acceleration. 5. Suppose we have two identical balls connected to two identical strings. The arrangement is whirled around in a horizontal circle as shown below. As it whirled faster and faster, which string is likely to break first? As the balls continued to spin, Ball B would break first because as they spun faster and faster, the tension would continue to build and the tension of the string attached to Ball B is twice as much as the other string causing it to break first. 6. The moon constantly accelerates toward the earth, yet it never falls into the earth. How can this happen? This happens because the force of gravity balances out with the centripetal force acting on the moon. The gravitational force pulling on the moon keeps it in this circular path and the force pulling the moon away thus balances out. Conclusion: In conclusion, through this lab we were able to verify the centripetal acceleration expression by comparing the computed value of the centripetal force using equation 6 with the static force required to displace the mass to the same radial position based on equation 7. We
were able to visually see the spinning increase which represented the centripetal acceleration over 3 separate trials. After completing the calculations, I noticed that there was a percent difference in our values but I conquered that this could have been due to incorrect reading of radius or stop watch, not counting exactly 50 rotations, or just a simple algebra error in the 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