International Baccalaureate: Extended Essay Investigation on the amount of time required to burn the calories contained in a Mars Bars whilst riding a standard bicycle on a horizontal road Jakub Benicek Candidate Number: 001837-003 May 2010 Physics Word Count: 3864 1
Abstract: My research question is how much time must a person peddle a bicycle on a flat, horizontal road in order to burn the equivalent calories of a Mars Bar. The scope of the practical work essentially involves three experiments: one to determine the total resistive forces acting against the motion of the cyclist. The second one is to find out if the mass of the cyclist makes a difference. This then makes it possible to find an approximate value for the power output of a cyclist and thus the time needed to expend sufficient calories for a Mars Bar. The third experiment is carried out with the aim to calculate the power output of a cyclist necessary to overcome the resistive forces involved when the wheels spin. Thus we can calculate the amount of the power output which is due to the friction in the wheel bearings. In experiment 1 we have found that resistance to motion of the cyclist is constant at low velocities. Experiment 2 demonstrates that air resistance on the bike and cyclist is the dominant opposite force and that wheel resistance is insignificant. Experiment 3 shows that at a constant velocity, the resistive force on the cyclist is constant at any mass. A formula has been derived which allows us to calculate the time required to burn a Mars Bar for any person cycling within a range from 0-8ms -1 (0-29kmh -1 ). For me, a 19 year old male, thus equates to 33.9minutes at 20kmh -1. Word Count: 247 2
Table of Contents: Contents 1. Introduction:... 4 2. The energy of a Mars Bar... 5 3. Breakdown of energy use during cycling... 5 4. Summarizing the experimental strategy:... 7 5. Experiment 1.... 9 6. Experiment 2... 15 7. Experiment 3... 20 8. Conclusion:... 26 9. Evaluation:... 30 10. Appendix... 33 11. Bibliography:... 33 3
1. Introduction: The world s population is getting more obese and people ask why that is. Nowadays people take a piece of chocolate, eat junk food and think that if they walk home from their job, they will remain fit. Therefore the purpose of the investigation is to inform the public, how much time they have to spend in order to get rid of the calories they gain. Cycling is a form of activity through which you can accomplish this. Everyone likes to eat, especially sweets, thus I have chosen a Mars Bar as an example to use. To show the humanity what it takes to do this. I choose cycling because it is a relatively easy form of exercise and is a good example to illustrate the energy distribution over time for an average person. It is friction and air resistance which require a cyclist to change his chemical energy (Mars Bar) to mechanical energy if he wants to keep going at a constant velocity. An average man needs 2200 Calories per day and woman 2000 Calories. This depends on age, height and mass. When you eat a burger i.e. at Burger King it has around 600 Calories that is almost 30% of your daily calories. Therefore now you can imagine how easy it is to exceed this recommended amount of calories. The problem which arises is that people do not realize this and they underestimate the impact of eating such food. Furthermore with lack of exercise instead of burning the calories, they get stored which creates obesity. Cycling is quite healthy way of burning these calories and almost everyone can do it, thus what interests me is the time it takes to burn the amount of calories contained in one Mars Bar. The aim of this investigation is to allocate the distribution of the rate of energy used during cycling processes in order to overcome resistive forces such as wind resistance or friction between the tyres and the road in order to keep your speed constant. Therefore I have decided to investigate the total resistive force which is applied on the cyclist during cycling on a flat road, horizontal at a constant velocity. The purpose of this is to find the amount of energy which is needed to overcome these resistive forces. Hence we can figure out how much energy in total a cyclist has to expend in order to burn the same amount of calories of a Mars Bar and the rate at which a cyclist burns calories. Hence we can find the time needed to burn these calories at a given speed. 4
2. The energy of a Mars Bar Calorie is a pre-si metric unit of energy during which heat is given off by a body. One kcal is 4.184 kj [1]. Calorie 4.184KJ calorie 4.184J kcal 4.184KJ 100g of a Mars Bar contain 449.0 [2] kcal. Therefore for 62.5g of a Mars Bar would equate to 280.6 kcal. Hence 280.6 4.184kJ=1174.0304kJ of energy Therefore to burn this i.e. in hour you would have to do this at rate of Power= 3. Breakdown of energy use during cycling 3.1 Basal metabolic rate (BMR): In order to calculate the Caloric requirements of cycling it is necessary to calculate the total amount of calories needed to maintain person s basic life processes. [7] It is the amount of energy expended per day, while at rest in a neutrally temperate environment in order to enable sufficient functioning of the vital organs, such as the heart, lungs, brain and the rest of the nervous system. NOTE: These figures are kcal per day, hence we can convert them to energy per second where [4] Formula-BMR for Women: BMR = 655 + ( 9.6 x weight in kilos ) + ( 1.8 x height in cm ) - ( 4.7 x age in years ) [4 Formula- ] BMR for Men: BMR = 66 + ( 13.7 x weight in kilos ) + ( 5 x height in cm ) - ( 6.8 x age in years ) i.e. for me this is BMR=66+ (13.7 x 81.0) + ( 5 x 185 ) - ( 6.8 x 19 )=1971.5 kcal per day which is an equivalent of 8248.8kJ per day 5
Overall we can say that since 1day =86400 seconds and BMR is 8248756J, then the power, hence the rate at which a person s body does work in order to maintain basic life processes over 1 second: Power BMR W (Js -1 ). 3.2 Physical work: Secondly the amount of calories required for a particular task in terms of the physical activity. This is the force acting against the bike during cycling on a flat horizontal road. The magnitude of these forces will be explored in experiments 1-3. In order to maintain a steady velocity (v), the rate of energy conversion to maintain this velocity is Power output during cycling=force velocity v= In order for a person to understand the concept of energetic during cycling you have to imagine a world with no resistance. Thus according to the Newton s first law of motion which states: [3] In the absence of force, a body either is at rest or moves in a straight line with constant velocity. To explain this consider the fact that in reality, resistance acts on the body while in a motion, we can apply this to find the energy needed to keep going at constant velocity given that the acceleration(deceleration) of the body due to. Therefore the cyclist needs to constantly apply force in the forward direction to maintain the velocity. Hence, at constant v. However human body is estimated to be 25% efficient at best in converting Calories (notice the capital letter) eaten into Calories delivered as power output. Therefore the Power output chemical necessary to keep the body at a constant velocity is: Power output chemical = experimental power physical 4. 3.3 Kinetic Energy: Once you provide a cyclist with a force to bring it to a constant velocity, it will keep moving unless a force is exerted on the body. On a flat road we can therefore say that Ek =0.5mv 2 which would however mean that at 5ms -1 (18kmh -1 ) with a mass of 81kg+12.8kg of the bike, the biker will only require to use 1173J. That is 0.1% of the total energy contained in 62.5g of a Mars Bar. Therefore we will ignore this figure. 6
3.4 Thermic energy: Thermic effect of food this is the energy required for digestion, absorption and transportation of the Mars Bar energy to the cells around your body. To calculate the rate at which the energy is transported we can say Power Thermic energy = (Power BMR + Power Physical work ) 0.1=energy needs for digestion and absorption in J per second. Thus in terms of power this equates (95.5W+Power Physical work ) 0.1. 3.5 Summary: Calories needed for one day = = Power BMR + Rate of physical work (Power cycling 4) + Power thermic energy 4. Summarizing the experimental strategy: Apparatus Calibration of the speedometer Mass of the bicycle Experiment 1: Determination of resistive forces on bicycle, when in motion at different velocities Experiment 2: Determination of resistive forces on wheel, when in motion at different velocities Experiment 3: Determination of resistive forces on bicycle, when in motion at constant velocity and different mass 7
4.1 Apparatus: mountain bike- Cannondale Scalpel stop watch speed meter BC 1606L DTS spring balance scale iron weights (30kg in total) holder backpack 4.2 Calibration of the speedometer: Measure the length of the tire by rolling the bike over a known distance- Figure 1 Instantaneous speed= = = Figure 1 4.3 Mass of the bike: Attach spring balance to the centre of mass of the bike. This is necessary to do since we need to know the total mass of the body which is the bike and the cyclist. Figure 2 8
Mass of the bike: 12.8±0.05kg Mass of the cyclist: 81.1kg±0.05kg % uncertainty= then =0.4% Figure 2 5. Experiment 1. 5.1 Comparing the stopping time of the cyclist at different initial velocities in order to find acceleration (deceleration) and hence the size of the resistive force: The cyclist will start at an initial velocity and continue moving in a direction until the cyclist stops with no further intervention from the cyclist. The dependent variable is the acceleration of a cyclist in terms of measuring the time it takes for the bike to stop at different initial velocities. Note that the acceleration is negative in all of the experiments. Therefore if a cyclist goes at a certain speed, according to the Newton s first law the body will kept moving or stationary unless a force acts on it. We can say that this deceleration period is due to the resistive forces. 5.2 Procedure-Experiment 1: 1. vary the initial velocity while riding a bike 2. start from range between 2 and 10ms -1 9
3. choose a flat, horizontal road 4. reach your initial velocity and keep pedaling at least for 3 seconds maintaining velocity 5. stop pedaling and start the stop watch, remember u 6. continue in straight line until the bike comes to halt then stop the stop watch 7. record the time and u 8. repeat for different values of u Figure 3 The independent variable is the change in the initial velocity at which the bike starts to decelerate. Therefore by varying the initial velocity with a constant mass, t will change. There is large number of variables that need to be taken into a consideration for this experiment (starting with the major once): 5.3 Other variables: air resistance not possible to control in this experiment, wind is a major variable Mass of the body-mass of the bike & mass of the cyclist will not be altered Forces on the bike: -Rolling Resistance: resistance caused due to the deformation of two objects sliding over each other, therefore in order to keep this constant, the tires used in the experiment will be the same throughout. Carry out the experiment on the same road, however the road will vary due to changes in weather conditions. Rolling resistance on a dry surface is smaller than on a wet road. 10
-position of the centre of mass: sitting at the same position every time to assure that the pressure on each tyre is going to be constant as well as the shape of the cyclist- same clothes (thus the frontal area will be the same) -slope of the road: will be effectively 0 0 (horizontal) gravitational potential energy-since the road is flat then effectively the h=0, thus E p =mgh=0j Velocity ±1.0(km/ Time (s) Velocity ±1.0(km/ Time (s) 10.0 9.8 20.0 28.9 10.0 14.3 20.0 30.7 10.0 14.2 20.0 33.9 10.0 14.8 21.0 36.9 10.0 16.3 22.5 41.6 10.0 17.4 22.0 36.7 11.5 16.9 22.5 29.8 11.5 17.5 23.0 36.2 14.0 25.1 25.0 38.8 14.0 25.5 25.0 28.4 14.5 26.5 25.0 31 15.0 19.7 25.0 33.7 15.0 25.6 25.0 31.4 15.0 25.3 25.0 38.2 15.0 20.6 28.0 35.8 15.0 28.8 28.0 39.0 17.5 25.7 28.5 42.8 19.0 31.6 30.0 38.5 20.0 25.8 34.5 46.5 20.0 28 40.0 53.5 7.4 Raw Data: The underlined data represent the times with abnormal headwind and the data in bolt are used in the average time. NOTE: applies as well for 7.5 5.5 Processed Data-used in the conclusion as well: 11
Graph 1-the time taken for a bike to come to halt starting at an initial velocity u 12
Time (s) to come to a halt International Baccalaureate-Extended Essay: 60 55 50 How initial veloicty effects the stopping time of a cyclist Small wind resistance Line of best fit 45 40 35 y = 5.407x + 0.4652 R² = 0.9604 High wind resistance 30 25 20 15 10 See below for the explanation of dark red points 5 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 Initial velocity (ms -1 ) of the cyclist Graph 1 shows the graph of time (s) versus velocity ms -1 for a constant mass of 93.9kg. We can conclude from the data that the initial velocity is directly proportional to the time taken for the cyclist to stop. The dark red points represent the times corresponding to the initial velocities which were repeated more than once on the same day in order to estimate for the uncertainties in time. The rest of the points are randomly chosen velocities with corresponding times at different days. From the line of best fit you can see that it goes through all the dark red points. 5.6 Mathematical analysis of graph 1: 13
Graph shows t α u where t-stopping time & u-initial velocity If we assume that acceleration is constant, then: v=u+at but v=0ms -1, since we end up stationary 0=u+at thus t= So if a is constant, t α u The graph is consistent with this, so we can deduce that the gradient of the graph is Graphical analysis (using Excel) showed that the equation of the line is: T=5.407v + 0.465 Hence 5.407 = thus a= NOTE: We can assume that the 0.465 is a small consistent error. We can now use this to calculate the resistive force that must be on the cyclist during motion: F=ma F=93.9kg = -17.4N Finally, we can compute the power output in overcoming this force at a given velocity. Power=Fv Power=17.4N u i.e. at 4.5ms -1 P=17.4N 4.5=78.3W Since human body is 25% efficient at best in converting Calories eaten into Calories delivered as power output as I have mentioned before, the rate of converting the energy from chemical to mechanical is 313.2W (Js -1 ) inside the body. Thus P chemical = 4 P mechanical 5.7 Comparison to an external model: Assumptions for the [6] Power output during motion: a level road no head wind constant speed i.e. no acceleration or deceleration ideal road or mtn. bike and rider External data 5 mph - 37 Cal/hr 10 mph - 133 Cal/hr 1 mile per hour = 0.45 ms -1 5 mph= 2.25ms -1 10 mph=4.5ms -1 1 Cal/hr= 37 Cal/hr=43.0W 133 Cal/hr=154.6W According to an external source [6], the value for power output at 4.5ms -1 is estimated to be 154.6W 14
It is only 49.5% the value which we have obtained, thus I can conclude that my equation does work to a certain extent, since their model does not take into an account the headwind resistance. Experimental: at 2.25ms -1 P chemicall =4 17.4N 2.25=157.5W External model P=43.0W This is only 27.3% similar. It is important to consider their assumptions (no head wind, different mass of the cyclist), thus we can argue that my value is closer to the actual P chemical expended by the cyclist due to the presence of headwind. 6. Experiment 2 6.1 Finding the stopping time of the front wheel at different initial velocities in order to find acceleration (deceleration) and hence the size of the resistive force acting on the wheels: For experiment number 2 we will use the same concept as in experiment 1. That is that the front wheel will be spun in order to reach an initial velocity u i.e. 5ms -1 and let come to halt in time t with an acceleration of a. We will obtain a value for power which will indicate how much energy per second (P mechanical ) has to be used to overcome the friction in the wheel bearings in terms of the total P mechanical determined in experiment 1. Thus total P mechanical from experiment 1 = P mechanical due to the friction in wheel bearings + P mechanical due to the air resistance Thus we can make a judgment, whether or not friction in the wheels has a significant impact on the deceleration. 6.2 Procedure-Experiment 2: 1. place the bike on the holder as shown in Figure 4 2. spin the wheel to reach an initial velocity (u ms -1 ) and wait at least for 3 seconds for the speedometer to synchronize 3. record initial velocity and start the stopwatch 4. once the wheel comes to halt record the time 5. repeat for different values of u ( u in range between 2 and 10ms -1 ) Figure 4 15
The independent variable in experiment 2 is the different initial velocity. Therefore by varying the initial velocity, we can calculate the deceleration of the wheel as in experiment 1 where deceleration=, hence a is constant.we assume that the rear wheel has the same friction in terms of the bearings. The dependent variable is the time taken for the wheel to come to halt starting at u. 6.3 Other variables: air resistance air resistance is present, however the headwind is eliminated Mass of the body-mass of the wheel is constant Forces on the wheel-friction due to the ball bearings we can ignore gravitation due to torque, since the wheel has a momentum and therefore the force on the wheel tends to rotate the object around its axis 6.4 Raw Data: 16
Velocity ±0.5 (km/h) Trial 1 Time (s) Trial 2 Trial 3 20.0 194.3 192.5 194.1 18.5 189.5 187.3 192.1 18.0 185.2 192.7 186.9 16.5 179.6 180.6 180.7 14.5 172.8 170.6 171.4 11.0 164.3 152.6 155.2 10.0 143.0 143.4 143.5 9.0 132.7 135.5 139.8 8.0 131.9 133.3 124.8 7.0 120.7 120.3 121.7 6.0 147.2 118.9 106.3 4.5 145.9 86.3 87.2 6.5 Processed Data: Clearly the two data points 147.2 & 145.9 are outliers. Thus they will be ignored in the calculation for the average stopping time of the wheel Velocity % uncertainty Velocity Square Root Uncertainty Uncertainty of Time (s) Average Uncerntainty ±0.5 (km/h) of 'v' (m/s) of v of 'v' (m/s) 'square root of v' Trial 1 Trial 2 Trial 3 Time(s) for time (s) 20.0 2.5 5.6 2.4 0.14 0.06 194.3 192.5 194.1 193.6 1.8 18.5 2.7 5.1 2.3 0.14 0.06 189.5 187.3 192.1 189.6 4.8 18.0 2.8 5.0 2.2 0.14 0.06 185.2 192.7 186.9 188.3 7.5 16.5 3.0 4.6 2.1 0.14 0.06 179.6 180.6 180.7 180.3 1.1 14.5 3.4 4.0 2.0 0.14 0.07 172.8 170.6 171.4 171.6 2.2 11.0 4.5 3.1 1.7 0.14 0.08 164.3 152.6 155.2 157.4 11.7 10.0 5.0 2.8 1.7 0.14 0.08 143.0 143.4 143.5 143.3 0.5 9.0 5.6 2.5 1.6 0.14 0.09 132.7 135.5 139.8 136.0 7.1 8.0 6.3 2.2 1.5 0.14 0.09 131.9 133.3 124.8 130.0 8.5 7.0 7.1 1.9 1.4 0.14 0.10 120.7 120.3 121.7 120.9 1.4 6.0 8.3 1.7 1.3 0.14 0.11 147.2 118.9 106.3 112.6 12.6 4.5 11.1 1.3 1.1 0.14 0.12 145.9 86.3 87.2 86.8 0.9 Graph 2-stopping time of the wheel at different initial velocities 17
Time (s) for the wheel to come to halt Time (s) for the wheel to come to halt International Baccalaureate-Extended Essay: 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0 Stopping time of the wheel versus the wheel's initial velocity 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 Initial velocity (ms -1 ) Graph 3-average stopping time of the wheel at different initial velocities 200.0 190.0 180.0 170.0 160.0 150.0 140.0 130.0 120.0 110.0 100.0 90.0 80.0 70.0 60.0 50.0 40.0 30.0 20.0 10.0 0.0 Time against velocity - front wheel average 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 Initial Velocity (ms -1 ) Graph 4- stopping time of the wheel at different 18
Time (s) for the wheel to come to halt from an initial velocity International Baccalaureate-Extended Essay: 200.0 190.0 180.0 170.0 160.0 150.0 140.0 130.0 120.0 110.0 100.0 90.0 80.0 70.0 60.0 50.0 40.0 30.0 20.0 10.0 0.0 6.6 Mathematical analysis of graph 4 Graph 4 shows t α where t-stopping time & - Using the assumption from experiment 1, average deceleration= Graphical analysis (using Excel) showed that the equation of the line is: t= thus when we substitute it into the equation: average deceleration= Stopping time of the wheel versus the wheel's initial velocity 1/2 y = 82.334x + 4.7351 R² = 0.9831 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 Inintial Velocity 1/2 NOTE: We can assume that the 4.7 is a small consistent error and ignore it. We can now use this to calculate the resistive force that must be on two wheels during motion: =2ma 19
=2 2.0kg i.e. at 5.56ms -1 =2 2.0kg =0.11N P= v = 0.11N 5.56ms -1 =0.61W Thus P chemical = 4 P mechanical Thus P chemical = 4 0.61W = 2.44W NOTE: 5.56ms -1 is 20kmh -1 7. Experiment 3 7.1 How does varying mass affect the total resistive force: The experiment is a simulation to investigate the time taken for a cyclist of different masses to come to halt at 5.56ms -1. This should show whether or not there is a significant difference between a lean cyclist and an obese cyclist in terms of the energy distribution during cycling. 7.2 Procedure-Experiment 3: 1. vary the extra mass of the cyclist between 0kg and 30.0kg in range of 5.0kg 2. choose a flat, horizontal road 3. reach 20ms -1 and keep pedaling at least for 3 seconds to maintain this velocity 4. stop pedaling and start the stop watch 5. continue in straight line until the bike comes to halt then stop the stop watch 6. record the time and the extra mass Figure 5 20
The independent variable is the extra mass due to the weights in the backpack and the dependent variable is the time taken for the body to come to halt at a deceleration a. 7.3 Other variables: air resistance not possible to control in this experiment, wind is a major variable Forces on the bike -Rolling Resistance: see Experiment 1 for rolling resistance -position of the centre of mass: sitting at the same position every time to assure that the pressure on each tire is going to be constant. However, with the addition of the backpack including the mass will alter the centre of mass and thus change the pressure exerted on each tyre. gravitational potential energy-see Experiment 1 7.4 Raw Data: The underlined values represent anomalies. They were ignored in the calculations. 21
Mass Time (s) (kg) Mass (kg) Time (s) 124.1 27.0 109.1 22.7 124.1 28.0 109.1 23.6 124.1 27.5 104.1 22.8 124.1 30.0 104.1 24.2 124.1 37.8 104.1 22.3 124.1 43.2 99.1 25.2 124.1 15.4 99.1 26.4 119.1 30.9 99.1 23.6 119.1 24.9 93.9 20.7 119.1 26.6 93.9 22.4 114.1 24.4 93.9 21.5 114.1 21.1 93.9 33.9 114.1 25.2 Data from experiment 1 114.1 24.5 93.9 28 109.1 22.1 93.9 28.9 109.1 22.7 93.9 30.7 7.5 Processed Data 1: The underlined values represent anomalies. They were ignored in the calculations. 7.5 Processed Data 2: 22
Time (s) for the bike to come to halt International Baccalaureate-Extended Essay: Average Mass (kg) Mass uncertainty Mass % uncertainty Average Time (s) Average Time uncertainty (s) Average % Time Total resistive Force (N) Uncertainty for total resistive (N)Force (5% for 124.1 0.6 0.5 28.1 3 10.7 24.5 4.0 119.1 0.6 0.5 27.5 6 21.8 24.1 6.6 114.1 0.6 0.5 23.6 4.1 17.4 26.9 6.2 109.1 0.6 0.5 22.8 1.5 6.6 26.6 3.2 104.1 0.6 0.6 23.1 1.9 8.2 25.0 3.4 99.1 0.6 0.6 25.1 2.8 11.2 22.0 3.6 93.9 0.1 0.1 21.5 0.8 3.7 24.2 2.0 Data which were ignored for the average 93.9 0.1 0.1 29.2 2.7 9.2 17.9 2.4 124.1 0.6 0.5 37.8 3 7.9 18.2 2.3 Graph 5-Time taken for a cyclist of different mass to come at halt at constant velocity 50.0 48.0 46.0 44.0 42.0 40.0 38.0 36.0 34.0 32.0 30.0 28.0 26.0 24.0 22.0 20.0 18.0 16.0 14.0 12.0 10.0 8.0 6.0 4.0 2.0 0.0 Time versus Mass at a constant velocity at 5.56ms -1 90 92 94 96 98 100 102 104 106 108 110 112 114 116 118 120 122 124 126 128 Total Mass of the cyclist (kg) 23
Time (s) International Baccalaureate-Extended Essay: Graph 6- Average time taken for a cyclist of different mass to come at halt at constant velocity 44.0 42.0 40.0 38.0 36.0 34.0 32.0 30.0 28.0 26.0 24.0 22.0 20.0 18.0 16.0 14.0 12.0 10.0 8.0 6.0 4.0 2.0 0.0 Average time to come to halt versus total mass of the cyclistat a constant velocity at 5.56ms -1 y = 0.1782x + 5.0875 90 92 94 96 98 100 102 104 106 108 110 112 114 116 118 120 122 124 126 128 Mass (kg) 24
Total resistance force (N) International Baccalaureate-Extended Essay: Graph 7- Total resistive force versus total mass of the cyclist 36.0 34.0 32.0 Total resitance force versus total mass Average values of total resistive force from the graph Average- Time versus total mass at a constant velocity of 5.56ms -1 30.0 28.0 26.0 24.0 22.0 20.0 18.0 16.0 14.0 12.0 10.0 8.0 Values obtained from experiment 1: Overall the wind speed was smaller than in experiment 3 6.0 4.0 2.0 0.0 90 92 94 96 98 100 102 104 106 108 110 112 114 116 118 120 122 124 126 128 Total mass (kg) 7.6 Mathematical analysis of graph 7 Graph 7 shows that the total resistive force for an obese person and lean person is independent of the total mass. Since the argument is that the reason why the stopping time for an obese person is greater is due to the fact that it has a greater momentum at the same velocity, since Therefore an increase in ρ causes an increase in t for the stopping time. Hence, mass does not affect the resistive force, since gradient is 0. 25
Total Power necessary to keep going at a constant velocity (W) International Baccalaureate-Extended Essay: 7.7 Comparison to experiment 1 at the same velocity Graph 7 show two lines of best fit. The bottom one represents experiment 1 at generally smaller wind resistance. Thus P mechanical =Fv P mechanical for experiment 3 =25.1 5.56ms -1 =139.6W P mechanical for experiment 1 =17.4 5.56ms -1 =96.5W NOTE: due to an abnormal wind resistance in experiment 3 the value of P mechanical is greater than the value estimated for experiment 1 which is 96.5W. Thus wind resistance has a great impact on the P mechanical. 8. Conclusion: Graph 8-Total P mechanical for a cyclist to maintain a constant velocity 220.0 210.0 200.0 190.0 180.0 170.0 160.0 150.0 140.0 130.0 120.0 110.0 100.0 90.0 80.0 70.0 60.0 50.0 40.0 30.0 20.0 10.0 0.0 Total Power necessary for the cyclist to maintain a constant velocty versus initial velocity Maximum Power Minimum Power y=19.1x Line of best fit y = 17.36x R² = 1 0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00 5.50 6.00 6.50 7.00 7.50 8.00 8.50 9.00 9.50 10.00 10.50 11.00 11.50 12.00 y=15.7x Initial velocity ms -1 of the cyclist From experiment 2 we have found out that 2.44W are required to overcome the total force due to the friction in the wheel bearings and the air resistance when both wheels would be spinning. Thus we can conclude that it is only a fraction of the total P mechanical for a cyclist to maintain a constant velocity, thus air resistance is the major variable. 26
8.1 Processed Data 1: % uncertainty Height ±0.05m 1.85 2.70 Mass ±0.05kg 81 0.06 Processed Data 1: Velocity ±1.0(km/ Velocity uncertainty 1/Velocity 1/Velocity uncertainty Mars Bar time Uncertainty total Power %(W) Mars Bar time uncertainty± (min) 10.0 1 0.10 0.010 64.6 24.7 37.45 10.0 1 0.10 0.010 64.6 24.7 37.45 10.0 1 0.10 0.010 64.6 24.7 37.45 10.0 1 0.10 0.010 64.6 24.7 37.45 10.0 1 0.10 0.010 64.6 24.7 37.45 10.0 1 0.10 0.010 64.6 24.7 37.45 11.5 1 0.09 0.008 58.8 22.1 31.53 11.5 1 0.09 0.008 58.8 22.1 31.53 14.0 1 0.07 0.005 51.2 19.0 24.98 14.0 1 0.07 0.005 51.2 19.0 24.98 14.5 1 0.07 0.005 49.9 18.5 24.00 15.0 1 0.07 0.004 48.7 18.0 23.09 15.0 1 0.07 0.004 48.7 18.0 23.09 15.0 1 0.07 0.004 48.7 18.0 23.09 15.0 1 0.07 0.004 48.7 18.0 23.09 15.0 1 0.07 0.004 48.7 18.0 23.09 17.5 1 0.06 0.003 43.4 16.1 19.52 19.0 1 0.05 0.003 40.7 15.2 17.92 20.0 1 0.05 0.003 39.1 14.7 17.02 20.0 1 0.05 0.003 39.1 14.7 17.02 20.0 1 0.05 0.003 39.1 14.7 17.02 20.0 1 0.05 0.003 39.1 14.7 17.02 20.0 1 0.05 0.003 39.1 14.7 17.02 21.0 1 0.05 0.002 37.6 14.2 16.22 22.5 1 0.04 0.002 35.6 13.6 15.19 22.0 1 0.05 0.002 36.2 13.8 15.52 22.5 1 0.04 0.002 35.6 13.6 15.19 23.0 1 0.04 0.002 34.9 13.4 14.89 25.0 1 0.04 0.002 32.6 12.7 13.81 25.0 1 0.04 0.002 32.6 12.7 13.81 25.0 1 0.04 0.002 32.6 12.7 13.81 25.0 1 0.04 0.002 32.6 12.7 13.81 25.0 1 0.04 0.002 32.6 12.7 13.81 25.0 1 0.04 0.002 32.6 12.7 13.81 28.0 1 0.04 0.001 29.7 11.8 12.56 28.0 1 0.04 0.001 29.7 11.8 12.56 28.5 1 0.04 0.001 29.3 11.7 12.38 30.0 1 0.03 0.001 28.0 11.4 11.89 34.5 1 0.03 0.001 24.9 10.5 10.74 40.0 1 0.03 0.001 21.8 9.7 9.76 27
Time required to burn the calories from the Mars Bar (minutes) Time required to burn the calories from the Mars Bar (minutes) International Baccalaureate-Extended Essay: Graph 9-Time required to burn the amount of calories contained in a Mars Bar using my BMR data for a constannt velocity 70.0 65.0 60.0 55.0 50.0 45.0 40.0 35.0 30.0 25.0 20.0 15.0 10.0 5.0 0.0 Time required to burn the calories contained in a Mars Bar at a constant velocity 0.0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0 Constannt velocity (km -1 ) Graph 10-Time required to burn the amount of calories contained in a Mars Bar using my BMR data for a constannt velocity 1/2 70.0 65.0 60.0 55.0 50.0 45.0 40.0 35.0 30.0 25.0 20.0 15.0 10.0 5.0 0.0 Time required to burn the calories contained in a Mars Bar at a constant velocity 0.000 0.010 0.020 0.030 0.040 0.050 0.060 0.070 0.080 0.090 0.100 0.110 Constannt velocity 1/2 Graph 10 shows that there is direct inverse relationship between the time (min) required to burn the calories in a Mars Bar to the (km -1 ), thus we can say: 28
Time to burn the calories contained in a Mars Bar α NOTE: The processed data for the time required to burn the calories contained in a Mars Bar at a constant velocity contained in section 8.1 are an estimate of the uncertainty. This is due to the number of calculations which are necessary to compute for the time required to burn the calories contained in a Mars Bar at a constant velocity. From all the experiments carried out I can estimate the uncertainty to be at most 10% of t. Graph 10 shows that there is an consistent error of 6 minutes. This could explain why points (t=58,64) are off by 10%. To draw a conclusion from the collected data we can estimate the time needed to burn the same amount of calories in a Mars Bar to be: Calories needed for one day = =Power BMR + Rate of physical work (Power cycling 4) + Power thermic energy given in part 3.5 where Mars Bar is 1174030.4J and physical work let s say at 5.56ms -1 is 96.5W according to graph 8, hence = + + ( + ) 0.1 = BMR (95.5W) + physical work (96.5W 4) + thermic effect (96.5W + 95.5W) 0.1 =529.65W (Js -1 ) t= =2216.6s=36.9minutes Uncertainty of 10% (estimate)=3.69miutes Thus t=36.9±3.7 minutes In terms of the mass, according to the experiment 3, it is important to realize that the total force exerted on the cyclist is constant throughout the activity. Thus we can conclude that the power is only dependable on the constant velocity and independent on the mass of the cyclist as this value is constant at each velocity. This equation does not take into the consideration the energy required to get to certain velocity and only applies for velocities in range 0-8ms -1 (29kmh -1 ) Power mechanical =Fv Power mechanical =kv Power mechanical α v -graph 8 represents this relationship 29
8.1 Summary: To calculate the time required for any person to burn any amount of calories at any constant velocity between 0-29kmh -1 we can do so by using the derived formula below: = + + ( + ) 0.1 thus for a female: = + + + 0.1 time to burn the energy gained= = (any other energy) + + + 0.1 for a male: time to burn the energy gained= = (any other energy) + + + 0.1 This formula only works at a constant velocity, thus acceleration is 0ms -2 on a flat, horizontal road. 9. Evaluation: External source: At cycling speeds 6.7ms -1, the energy required to overcome air resistance greatly exceeds those of the rolling and mechanical resistance in your bike (see other variable in 7.1 for more information). For example, in going from 3.3ms -1 to 8.8ms -1 : [5] mechanical resistance increases by 225% [5] rolling resistance by 363% 30
[5] air resistance by 1800%. This contradicts my investigation, since from the collected data I have obtained a relationship where the t is directly proportional to v. Due these circumstances, I suppose that wind resistance is a major issue and thus I suggest doing a further research to determine the relationship between the initial velocity and the time taken for the bike to stop. This could be done in a wind tunnel with a known headwind. To collaborate further on this issue, I suggest to find the distance traveled, hence Power= work done distance I find headwind to be the major issue while I was conducting the experiments. The graph obtained for t versus u shows that due to the fluctuation in the wind velocity, the stopping time was effected. Therefore this limits the time required to burn the calories gained from a Mars Bar, giving the final time to burn the calories in a Mars Bar a smaller value. Second issue was with the speedometer, since when you measure the time taken for the bike to stop then it is very hard to reach the initial velocity and at the same time, start the stop watch. Furthermore, it is impossible to measure the exact stopping time since when you get to 1-1.5ms -1 then it is very difficult to keep balance and thus the last 2-3 seconds is more of an estimate. Thus it would be better to measure time for a distance between points A and B where at point A you would record the initial velocity.i.e. through light gates and at point B final velocity 2ms -1 including time taken for the distance in between A and B. Figure 6: a= gives an average velocity. Thus you can plot a against u. If you know m then P=Fd where Word done=fd, F=ma Figure 6 31
During cycling the temperature of the wheel changes and therefore the longer you cycle, the greater the temperature of the air molecules in the tyre is. Thus we can say that if you cycle for longer period of time, the friction between the tire and the ground decreases since the area of the tire which is in contact with the ground at any point is smaller thus less friction occurs due to the increase in pressure. This means that in the long-run, you would have to cycle longer in order to burn the same amount of energy which a Mars Bar contains, however this is debatable and further research would have to be done. The down side to the experiment 1 & 3 is the fact that I was unable to measure the velocity of the wind. Since this wind velocity was changing significantly and it was obvious when the wind was blowing that the bike was decelerating at a much faster rate. Experiment 3 proves my point since the difference due to the wind fluctuation is 7 seconds in the stopping distance as graph 6 shows. Predominantly this investigation has a great benefit to the society since it gives them an estimate of the time required to cycle in eating a Mars Bar which can be compared to other food sources. Furthermore, research could be conducted to find the net force acting against the cyclist while cycling uphill or in different positions in a peloton (appendix 1) which could be used in sports. 32
10. Appendix Appendix 1 [8] Peloton: pack is the large main group in a road bicycle race. Riders in a group save energy by riding close to each other (particularly behind) other riders. The reduction in drag is dramatic. In the middle of a well-developed group it can be as much as 40%. Thus the power needed to overcome the net resistive force at a certain velocity decreases. 11. Bibliography: [1] http://health.howstuffworks.com/health-illness/wellness/physical-fitness/weightloss/calorie1.htm Author: Julia Layton Title: How Calories Work Data of publication: 2008 Publisher: 1998-2009 HowStuffWorks, Inc. [2] http://www.weightlossresources.co.uk/calories/calorie_counter/chocolate_sweets.htm Author: unknown Title: Calories in Chocolate and Sweets Data of publication: 04.03.2009 Publisher: Weight Loss Resources Ltd [3] http://www.grc.nasa.gov/www/k-12/airplane/newton.html Author: Tom Benson Title: Newton's laws of motion Data of publication: 23.05.2009 Publisher: NASA.gov [4] http://www.hussmanfitness.org/bmrcalc.htm Author: John P. Hussman, Ph.D Title: John s BMR Calculator Data of publication: 2001 33
Publisher: Copyright 2001-2002 John P. Hussman, Ph.D [5] http://www.cptips.com/energy.htm Author: Richard Rafoth MD Title: CYCLING PERFORMANCE TIPS Data of publication: August 1, 2008 Publisher: Unknown [6] http://www.cptips.com/formula.htm Author: Richard Rafoth MD Title: CYCLING PERFORMANCE TIPS Data of publication: August 1, 2008 Publisher: Unknown [7] http://www.kristensguide.com/health/weight_loss/bmr.asp Author: Kristen Brooke Beck Title: Know Your Basal Metabolic Rate Data of publication: 24.11. 2007 Publisher: 2002-2009 Kristen Brooke Beck Company [8] http://www.sheldonbrown.com/gloss_p.html#peloton Author: Sheldon Brown Title: Peloton Data of publication: 04.05.1996 Publisher: Copyright 1996, 2008 Sheldon Brown Programs: -Microsoft Office Word 2007 -Microsoft Office Excel 2007 34