• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19

Analyzing Uniform Circular Motion

Extracts from this document...

Introduction

Lab Report: Analysing Uniform Circular Motion Name: Adam Purpose In places where the velocity of an object is difficult to find, it is reliable to attain the frequency of an object. This is the most-favourable method used when dealing with circular motion. Frequency, as we know, is indirectly proportional to the period of a motion, or more precisely. We can determine the relationship between a spinning objects mass, the radius of its circular motion (if attached to a thread) and the hanging mass in a circular horizontal motion by applying a formula we already know. The following steps will describe the derivation of a formula which will be beneficial for this experiment; Newton's Second Law states: Fnet = msa However; acceleration of a circular motion is given by; But Thus; Substituting back to Newton's second law, Fnet= ms ( But since there is a hanging mass which is hanging down from the thread; the tension in the thread is equal to the weight of the hanging mass. And the only force applied during the motion is by the tension in the thread, thus T = Fnet and T = mhg; therefore Fnet = mhg. Putting it back in the original formula we get; mhg = ms ( where; mh = the mass of the hanging object ms = the mass of the spinning object r = the length of the radius of the horizontal motion. To measure the relationship between frequency and Isolate for the three independent variables; we must isolate for each variable. Isolating for the three manipulating variables in the y = mx + b form we get: ; Where ; Where ; Where These are the manipulated equations which will be utilised throughout the lab report. Materials and Procedure - Refer back to the Lab Sheet provided by the instructor. Variables Although un-noticeable, there are three different experiments being held in this one lab. ...read more.

Middle

2.08 0.006 This equation was derived from the previous equation provided, and helps us determine the relationship between Ms and frequency in a linear manner. A further graphical explanation will be provided in the Graphical Analysis part. Data Table # 8: Frequency for Manipulating Hanging Mass (Mh) Hanging Mass (+/-0.00001kg) Frequency (Hz) Uncertainty (±) 0.013 1.05 0.0011 1.04 0.0011 0.027 1.16 0.0013 1.11 0.0012 0.038 1.27 0.0016 1.31 0.0017 0.051 1.58 0.0025 1.61 0.0026 0.065 1.69 0.0029 1.64 0.0027 Data Table # 9: Frequency2 for Manipulating Hanging Mass (Mh) This equation has also been derived from the original equation provided. It provides a linear relationship between the hanging mass and f^2 which will be further discussed in the graphical analysis section. Hanging Mass (+/-0.00001kg) Frequency ^2 (Hz^2) Uncertainty (±) 0.013 1.11 0.0023 1.08 0.0022 0.027 1.34 0.0031 1.24 0.0028 0.038 1.61 0.0041 1.73 0.0045 0.051 2.51 0.0080 2.58 0.0083 0.065 2.86 0.0097 2.68 0.0088 Graphical Analysis (Notice that due to the precision of the data and its closeness, an average of the trials was taken by the graphing program. By visualizing the data, we see that it is very precise, and its accuracy will be measured later on in this section of graphical analysis. The scatter plot is an average of the two trials which leads to more precise and accurate answers, and eliminates any random errors.) 1. Manipulating Radius This section describes the basic relation between Frequency and radius, as well as the manipulated relation between Frequency2 and 1/r, which is attained from the formula provided. As visible from the graph of Frequency vs. Radius, an indirectly proportional relationship is visible and thus the line of best fit provided for that graph is a rational function. As radius increases, frequency decreases. This is due to the fact that when the radius of a circle motion increases, if the mass of an object and the hanging mass are kept constant, the object must travel a longer distance around the circle, since increasing the radius means increasing the circumference of the circular path taken. ...read more.

Conclusion

This is because a change in reaction time may occur. Reaction time is neglected throughout this experiment, and assumed to be perfect. This is why only the manufacturer's uncertainty was taken for the stopwatch. Since it is a digital device, the smallest unit is the uncertainty. However, reaction time was a major factor and the idea of a bad reaction time could not be eliminated by taking more trials. Although, an attempt to keep it unimportant was made by having the same person and their frame of reference decide, when to start and when to stop. Some non-major sources of error could be those such as air resistance. Although, not much air was there, but the air molecules do interfere with the radius going down at an angle, and more human force being required for the system compared to the force originally caused by the Hanging mass. These errors lead to a systematic error, and the y-intercepts in each manipulated equation reflects that. The experiment could be made better in a sense that, more trials are taken for each different sub-experiment. The person revolving can also make sure that they put enough force behind it to overcome the force of resistance. Also a constant speed would be required for more accurate results. The manipulated graphs could have also had a line of best fit, which touched 0 in order to get a more accurate slope, since there were no y intercepts in the original equation. But if the data was accurate, then the line would have automatically touched 0. All in all, the percent error can be high due to the fact that we deal with quite small numbers. This experiment does consist of some systematic errors which can be improved on. Instruction # 4 Finding the mass of a rubber stopper Radius = 0.53 (+/- 0.0005m) Frequency = 1.22 (+/- 0.001 Hz) Hanging mass = 0.03237±0.00001 kg The original spinning mass of the stopper is 0.013 kg. This means that the error is really low, and that this value is valid. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our International Baccalaureate Physics section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Physics essays

  1. Forces Lab. I decide to investigate the relationship between the propelling force exerted on ...

    Gradient 1, therefore the value of is 3.0667m/kg. The uncertainty in gradient 1 can be found from the steepest and least steep lines in graph 1: Maximum value Minimum value Uncertainty Therefore Graph 2: The gradient of the line in graph 2 gives me the value of where h represents

  2. Aim: ...

    I believe that the reason for this is due to the relationship between force and acceleration. If the force is larger, then the acceleration is also larger. The forces that were most important in this experiment were the forces of gravity and friction.

  1. Circular motion lab

    first length L1 = 49.00 ± 0.05 cm g = = = > Time average = Sum total of average time / Number of trials = 138.58/ 10 = 13.858 = 13.86 seconds > Time range = (Time maximum - Time minimum )

  2. Investigate the Size of Craters in Sand Due to Dropped Object.

    * Mean of the Reading (in 1 decimal place) Height, cm Uncertainties: ± 0.05 Depth, cm (d) Uncertainties: ± 0.05 2.00 0.28 4.00 0.43 6.00 0.58 8.00 0.73 10.00 0.88 12.00 1.03 Centroid: Height (x - axis) = 2.00 + 4.00 + 6.00 + 8.00 + 10.00 + 12.00 6 = 7.00 ± 0.05 Depth (y - axis)

  1. In this extended essay, I will be investigating projectile motion via studying the movement ...

    The metal ball is then released and moves along the cylinder until it leaves it and undergoes projectile motion afterwards. The projection range can be obtained by measuring the distance between the point directly under the muzzle of the cylindrical container and the center of crater formed when the metal ball hits the sand.

  2. Simple Harmonic Motion

    V(x02 - x2)]2 * Ek = 0.5m?2(x02 - x2) * Et = Ep + Ek * Et = 0.5m?2x02 * If you graph Ep and Ek: * NOTE: - Ep: Maximum at x0 (furthest away) Minimum at 0 (equilibrium position)

  1. Centripetal Force

    with mass 10 g: Table 4.2 Data of Centripetal Force for mass 20 g Time for 10 revolutions (s) Period (T/s) f = 1/T (Hz) f2 Fc (N) 6.47 0.647 1.545595 2.388864 0.528535 0.101778 0.0103587 7.47 0.747 1.338688 1.792086 0.396498 -0.030259 0.0009156 6.91 0.691 1.447178 2.094324 0.463368 0.036611 0.0013404 7.44

  2. Physics Spinning Stopper Lab

    of mark by the marker was used on the string to indicate where the length of the spring was supposed to be kept at when the stopper was spinning. * (Average Time uncertainty) = ((Max Time)-(Min Time))/2 = (3.2-2.8)/2 = 0.2s.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work