Centripetal Force Lab Activity

Centripetal Force Lab Activity Analysis: 1. A) Average Percent Difference: 50g: (values expressed in newtons) Step 1: Calculate the average value of the two variables Average Value= Value 1+ Value 2 /2 = 0. 49+ 0. 61/2 = 1. 1/2 = 0. 55 Step 2: Calculate the difference between the two variables Difference= Value 2- Value 1 = Fc- Fg = 0. 61- 0. 49 = 0. 12 Step 3: Calculate % difference % difference= difference of the variables / average of the variables x 100 = 0. 12/ 0. 55 x 100 = 21. 81% 100g: (values expressed in newtons)Step 1: Calculate the average value of the two variables Average Value= Value 1+ Value 2 /2 = 0. 98+ 1. 84/2 = 2. 82/2 = 1. 41 Step 2: Calculate the difference between the two variables Difference= Value 2- Value 1 = Fc- Fg = 1. 84- 0. 98 = 0. 86 Step 3: Calculate % difference % difference= difference of the variables / average of the variables x 100 = 0. 86/ 1. 41 x 100 = 60. 99% 150g: (values expressed in newtons) Step 1: Calculate the average value of the two var iables Average Value= Value 1+ Value 2 /2 = 1. 47+ 2. 19/2 = 3. 66/2 = 1. 83Step 2: Calculate the difference between the two variables Difference= Value 2- Value 1 = Fc- Fg = 2. 19- 1. 47 = 0. 72 Step 3: Calculate % difference % difference= difference of the variables / average of the variables x 100 = 0. 72/ 1. 83 x 100 = 39. 34% 200g: (values expressed in newtons) Step 1: Calculate the average value of the two variables Average Value= Value 1+ Value 2 /2 = 1. 96+ 2. 66/2 = 4. 62/2 = 2. 31 Step 2: Calculate the difference between the two variables Difference= Value 2- Value 1 = Fc- Fg = 2. 66- 1. 96 = 0. 70 Step 3: Calculate % difference difference= difference of the variables / average of the variables x 100 = 0. 70/2. 31 x 100 = 30. 30% 250g: (values expressed in newtons) Step 1: Calculate the average value of the two variables Average Value= Value 1+ Value 2 /2 = 2. 45+ 3. 57/2 = 6. 02/2 = 3. 01 Step 2: Calculate the difference between the two variables Difference= Value 2- Valu e 1 = Fc- Fg = 3. 57- 2. 45 = 1. 12 Step 3: Calculate % difference % difference= difference of the variables / average of the variables x 100 = 1. 12/ 3. 01 x 100 = 37. 20% Average % difference: = Sum of all 5 averages/5 21. 81+ 60. 99+ 39. 34+ 30. 30+ 37. 20/ 5 = 189. 64/ 5 = 37. 92% B) Slope Calculations (Graph is displayed on a separate sheet) 50g: Slope= Rise/Run = 0. 61/0. 49 = 1. 25 100g: Slope= Rise/Run = 1. 84/0. 98 = 1. 877 150g: Slope= Rise/Run = 2. 19/1. 47 = 1. 489 200g: Slope= Rise/Run = 2. 66/1. 96 = 1. 357 250g: Slope= Rise/Run = 3. 57/2. 45 = 1. 457 After calculating the slope of each section of the graph (each section corresponds to a certain mass used in the lab activity) it is evident that it varies from it’s expected value by a great amount.The expected value of the slope was 1 as the rise and the run were supposed to be equal. However in our case the rise and the run varied greatly and therefore because they were different numbers the slope did not turn o ut to be 1 (the only way to get a slope of 1 is if both the numerator and denominator are equal, as a number divided by itself is always 1 and a number divided by a different number can never equal 1). 2. Yes the data collected did verify the equation Fc=42Rmf2. This is because the only varying value in this case â€Å"f†, had a direct relationship with the value of Fc.The only other values that had to be determined in this lab was the radius and the mass of the rubber stopper but they were constant variables (constant at 0. 87m and 12. 4g respectively) meaning that they had no varying effect on the value of Fc. For there to be a relationship between Fc and 42Rmf2 when the value of any of the variables changes the value of Fc has to change as well Because the value of â€Å"f† had a direct relationship with the value of Fc, when the value of â€Å"f† changed the value of Fc changed as well. In this particular case when the value of â€Å"f† grew so did th e value of Fc.For example, during the 50g test the frequency was 1. 2Hz and the Fc was 0. 61N, and during the 100g test the frequency was 2. 08Hz and the Fc was 1. 84N. This shows that as the frequency increases so does the Fc acting on the system. This therefore shows the relationship between Fc and 42Rmf2. 3. A) When the string was pulled down and the stopper was still spinning, the stopper started spinning at a faster rate (took less time to complete 1 cycle around the trip) B) This happens simply because the radius is being shortened.Because the stopper on the end of the string is moving around the horizontal circle at a constant speed it is therefore being acted upon by a constant net-force. In this case the net-force acting upon it (the stopper) is Fc, therefore because it is Fc acting upon it, the force can be calculated by the formula 42Rmf2 as that is equal to Fc. In this case because the string with the stopper on the end was being pulled down this means that the radius of the entire circle was decreasing (less string= smaller distance= smaller radius).In that formula if the radius is smaller that means that the centripetal force will be larger. In this case that larger the centripetal force acting on the rubber stopper, the faster the rubber stopper rotates around the horizontal circle. C) The laws of conservation of energy state that the total energy in the system stays the same but simply takes on different forms (kinetic and potential being examples). Therefore this case is not contrary to the laws of conservation of energy simply because when the radius is decreasing the rubber stopper speeds up.In the laws of conservation of energy when an object is speeding up the object is gaining kinetic energy. However in this case while the stopper is speeding up the hanging mass (along with some of the string) is falling to the ground. From a conservation of energy perspective when an object loses height it loses potential energy. Therefore in this case t he object at the top gains kinetic energy while the mass loses potential energy. Because of this energy transfer no energy is lost in the system as hen the object is losing potential energy the other object in the same system is gaining kinetic energy, therefore the energy stays the same. D) In figure skating the skaters do the exact same thing as what was done in this lab experiment. In order to spin faster they bend low (get low to the ground) and tuck their arms and legs in. This causes them to spin much faster than they were originally spinning and follows the same principles that the rubber stopper experiment followed. When they get low they lose potential energy but getting low causes them to tuck in (tuck in their legs and arms) and ultimately have a smaller radius.This smaller radius causes them to have a much greater centripetal force and ultimately causes them to spin faster and causes them to gain kinetic energy. This follows the laws of conservation of energy as when the y lose potential energy they gain kinetic energy (theoretically no energy lost- only transferred) Sources of Error: In this particular lab activity there were not very many potential sources of error simply because it was not as complicated an activity as many others. Therefore all errors that were made were simply human measurement errors.The main source of error in this lab activity was measuring the period/frequency. This was a challenge simply because the person measuring had to do many different things in a very small amount of time. That one person was responsible for firstly choosing a spot along the path of the horizontal circle to begin the measurement from, then that same person had to start the watch during the very small time frame in which the rubber stopper passed by that specific point on the circle. From there the person had to count the stopper pass by 5 times and stop the watch when it passed by the 5th time.This made it very difficult to get a completely accurate measurement for the period and the frequency, as it was very difficult to get an exact measurement of that time period. These slight miscalculations of the frequency caused the calculation of the centripetal force to be slightly wrong as well because the calculation of centripetal force depended on the frequency. This is evident because our â€Å"Fg† and â€Å"Fc† calculations are way off, as they were supposed to be nearly the same number as Fg= Fc. – X-axis= Fc – Y-axis= Fg – point 1= 50g – point 2= 100g – point 3= 150g – point 4= 200g – point 5= 250g Data: Mass of stopper: 12. 4g Radius of Rotation: 87cm Mass of suspended masses| Time for 5 cycles| Period (T)| Frequency (f)| FgFg=mhg| FcFc=42Rmf2| 50g| 4. 2s| 0. 84| 1. 2Hz| 0. 49N| 0. 61N| 100g| 2. 44s| 0. 48| 2. 08Hz| 0. 98N| 1. 84N| 150g| 2. 23s| 0. 44| 2. 27Hz| 1. 47N| 2. 19N| 200g| 1. 99s| 0. 4| 2. 5Hz| 1. 96N| 2. 66N| 250g| 1. 65s| 0. 34| 2. 9Hz| 2. 45N| 3. 57N|