Contextualizing Mathematics with Physics Demonstrations

Transcription

1 Contextualizing Mathematics with Physics Demonstrations Bill Spinella San Diego State University Larry Perez Saddleback College

2 Coin Lab Measuring Distances using Similar Triangles Documentation by Patrick Quigley Saddleback College

3 Measuring Distance Using Similar Triangles Introduction In the third century B.C.E. Aristarchus was able to calculate the distance to the Moon by using similar triangles. This activity utilizes similar triangles to calculate a much smaller distance on Earth. Theory If a person holds a coin so that it just barely blocks their view of another object, they can use their knowledge of the the size of the coin, the size of the target object, and the distance from their eye to the coin to calculate the distance between the target object and themselves. D A Observer B C Coin E Target Object The triangle ABC is similar to the triangle ADE and so their corresponding sides are proportional. This means that the proportion AE AC = DE is true. Multiplying both sides of the equation by AC produces BC AE = DE AC BC 2DE AC, and therefore AE = 2BC is also true. Since 2BC is the width of the coin, 2DE is the width of the target object, AC is the distance between the observer and the coin, and AE is the distance between the observer and the target object we get the following formula. (the distance to the target object) = (the width of the target object) (the distance to the coin) (the width of the coin) Apparatus one quarter one meter stick or measuring tape one target object determined by the instructor Experimental Procedure The instructor has selected a target object and has measured its width. The instructor has also marked a viewing position. The observer stands at the viewing position and holds the coin so that it blocks the view of the target object. The observer then moves the coin toward and away from their eye until the coin just barely blocks the view of the target object. If the observer can not hold the coin far enough away from their eye then another participant will hold the coin and adjust its position following instructions from the observer.

4 Page 2 Once the coin is properly positioned, a participant who is not the observer and is not holding the coin will measure the distance from the observer to the coin while being very careful not to injure the observer. Record this distance. Repeat this process two more times and record the results. Lastly the width of the coin must also be measured and recorded. Results and Discussion List the experimental measurements below and calculate the average observer-coin distance from the three trials. Width of the Coin: Width of the Target Object: Distance to the Coin (Trial 1): Distance to the Coin (Trial 2): Distance to the Coin (Trial 3): Average Distance to the Coin: Calculate the distance between the Observer and the Target Object (the distance to the target object) = (the width of the target object) (the distance to the coin) (the width of the coin) = ( ) ( ) ( ) = Conclusions and Recommendations Discuss the likely sources of error for this experiment, and suggest any improvements that could be made to this experiment.

5 Babinet s Principle

6 Babinet s Principle Measuring the width of a hair

7 Babinet s Principle Measuring the width of a hair The Physics According to Babinet s principle, the diffraction pattern of light due to obstruction by an opaque object will be the same as that of a double slit separated by the width of the opaque object.

8 Babinet s Principle Measuring the width of a hair The Physics According to Babinet s principle, the diffraction pattern of light due to obstruction by an opaque object will be the same as that of a double slit separated by the width of the opaque object. Thus, the width of a hair (our opaque object) can be determined using a laser pointer and some beginning algebra

9 Babinet s Principle Measuring the width of a hair The Physics According to Babinet s principle, the diffraction pattern of light due to obstruction by an opaque object will be the same as that of a double slit separated by the width of the opaque object. Thus, the width of a hair (our opaque object) can be determined using a laser pointer and some beginning algebra along with the result of Young s Double Slit experiment.

10 Babinet s Principle Measuring the width of a hair Materials - laser pointer - hair - tape - meter sticks or tape measure - protective eyewear Caution: Laser pointers should never be directed toward a person s head. The safest option is to use red laser pointers in the 3-5 mw range and provide protective eyewear for students.

11 Babinet s Principle Measuring the width of a hair The Physics hair laser pointer wall interference pattern

12 Babinet s Principle Measuring the width of a hair The Physics hair monochromatic light from laser pointer diffracted light interference pattern

13 Babinet s Principle Measuring the width of a hair The Math r 1 r 2 y 1 d L

14 Babinet s Principle Measuring the width of a hair The Math Assumptions r 1 r 2 y 1 d L

15 Babinet s Principle Measuring the width of a hair The Math Assumptions L d r 1 r 2 y 1 d L

16 Babinet s Principle Measuring the width of a hair The Math Assumptions L d r1 k r2 r 1 r 2 y 1 d L

17 Babinet s Principle Measuring the width of a hair The Math Assumptions L d r1 k r2 r 1 r 2 y 1 L y 1 d L

18 Babinet s Principle Measuring the width of a hair The Math Assumptions L d r1 k r2 r 1 r 2 y 1 L y 1 d tan() sin() L

19 Babinet s Principle Measuring the width of a hair The Math r 1 r 2 y 1 d L

20 Babinet s Principle Measuring the width of a hair The Math d r 1 r 2 y 1 r 2 r 1 d L

21 Babinet s Principle Measuring the width of a hair The Math d r 1 r 2 y 1 r 2 r 1 d sin() =r 2 r 1 d L

22 Babinet s Principle Measuring the width of a hair The Math d r 1 r 2 y 1 r 2 r 1 d sin() =r 2 r 1 d = n L

23 Babinet s Principle Measuring the width of a hair Putting it all together The Math

24 Babinet s Principle Measuring the width of a hair Putting it all together d sin() =n The Math

25 Babinet s Principle Measuring the width of a hair Putting it all together The Math d sin() =n tan() sin()

26 Babinet s Principle Measuring the width of a hair Putting it all together The Math d sin() =n tan() sin() d tan() =n

27 Babinet s Principle Measuring the width of a hair The Math Putting it all together d sin() =n d tan() =n tan() sin() tan() = y n L

28 Babinet s Principle Measuring the width of a hair The Math Putting it all together d sin() =n d tan() =n dy n L = n tan() sin() tan() = y n L

29 Babinet s Principle Measuring the width of a hair The Math Putting it all together d sin() =n d tan() =n dy n L = n tan() sin() tan() = y n L d = n y n L

30 Babinet s Principle Measuring the width of a hair The Math Putting it all together d sin() =n d tan() =n dy n L = n d = n L y n tan() sin() tan() = y n }We end up with a simple algebraic equation in terms of measured values. L

31 Babinet s Principle Measuring the width of a hair The Experiment Now let s perform the experiment! Typical human hair width between nm Check for on the laser pointers ( 650 nm) Work in groups of 2 or 3 d = n y n L d L n y n - width of hair - wavelength of laser light - distance from laser pointer to wall - number of bright fringe - distance from center of diffraction pattern to center of the nth bright fringe

32 Rotating Tank Find the height as a function of the radius of the parabolic surface formed by water in a thin rotating tank.

33 Rotating Tank The Problem A thin tank of radius R is initially at rest and filled with water to a height h. R h

34 Rotating Tank The Problem A thin tank of radius R is initially at rest and filled with water to a height h. R h The tank is then rotated at a constant angular velocity!.

35 Rotating Tank The Problem A thin tank of radius R is initially at rest and filled with water to a height h. R h The tank is then rotated at a constant angular velocity!. Determine the height of the water as a function of radius, y(r), given h, R,!, and g, the acceleration due to gravity.

36 Rotating Tank The Physics We can examine the forces on a fluid element at the surface.

37 Rotating Tank The Physics We can examine the forces on a fluid element at the surface. At constant angular velocity the fluid is in uniform circular motion.

38 Rotating Tank The Physics We can examine the forces on a fluid element at the surface. At constant angular velocity the fluid is in uniform circular motion.

39 Rotating Tank The Physics We can examine the forces on a fluid element at the surface. At constant angular velocity the fluid is in uniform circular motion. N

40 Rotating Tank The Physics We can examine the forces on a fluid element at the surface. At constant angular velocity the fluid is in uniform circular motion. N

41 Rotating Tank The Physics We can examine the forces on a fluid element at the surface. At constant angular velocity the fluid is in uniform circular motion. N Free Body Diagram N

42 Rotating Tank The Physics We can examine the forces on a fluid element at the surface. At constant angular velocity the fluid is in uniform circular motion. N Free Body Diagram X Fx = N

43 Rotating Tank The Physics We can examine the forces on a fluid element at the surface. At constant angular velocity the fluid is in uniform circular motion. N Free Body Diagram X Fx = Nsin() N

44 Rotating Tank The Physics We can examine the forces on a fluid element at the surface. At constant angular velocity the fluid is in uniform circular motion. N Free Body Diagram X Fx = Nsin() = ma c N

45 Rotating Tank The Physics We can examine the forces on a fluid element at the surface. At constant angular velocity the fluid is in uniform circular motion. N Free Body Diagram N X Fx = Nsin() = ma c = mv2 r

46 Rotating Tank The Physics We can examine the forces on a fluid element at the surface. At constant angular velocity the fluid is in uniform circular motion. N Free Body Diagram N X Fx = Nsin() = ma c = mv2 r = m! 2 r

47 Rotating Tank The Physics We can examine the forces on a fluid element at the surface. At constant angular velocity the fluid is in uniform circular motion. N Free Body Diagram N X Fx = Nsin() = ma c = mv2 r X Fy = = m! 2 r

48 Rotating Tank The Physics We can examine the forces on a fluid element at the surface. At constant angular velocity the fluid is in uniform circular motion. N Free Body Diagram N X X mv 2 2 = m! r Fx = N sin() = mac = r Fy = N cos()

49 Rotating Tank The Physics We can examine the forces on a fluid element at the surface. At constant angular velocity the fluid is in uniform circular motion. N Free Body Diagram N X X mv 2 2 = m! r Fx = N sin() = mac = r Fy = N cos() = 0

50 Rotating Tank The Physics We can examine the forces on a fluid element at the surface. At constant angular velocity the fluid is in uniform circular motion. N Free Body Diagram N X X mv 2 2 = m! r Fx = N sin() = mac = r Fy = N cos() N= cos() = 0

51 Rotating Tank The Physics We can examine the forces on a fluid element at the surface. At constant angular velocity the fluid is in uniform circular motion. N Free Body Diagram N X X mv 2 2 = m! r Fx = N sin() = mac = r Fy = N cos() N= cos() = 0 tan() = m! 2 r

52 Rotating Tank The Math Step 1: Formulate a separable differential equation using: tan() =m! 2 r R h Step 2: Solve the differential equation in terms of an integration constant (requires 2nd semester calculus). Step 3: Determine y(r) fully by solving for the integration constant (requires 2nd semester calculus). Given are h, R,!, and g. N

53 The Solution

54 Rotating Tank The Solution Step 1: Formulate the differential equation starting from: g tan() =! 2 r N

55 Rotating Tank The Solution Step 1: Formulate the differential equation starting from: g tan() =! 2 r N

56 Rotating Tank The Solution Step 1: Formulate the differential equation starting from: g tan() =! 2 r N dy dr Here tan() is actually equal to. N g

57 Rotating Tank The Solution Step 1: Formulate the differential equation starting from: g tan() =! 2 r N dy dr Here tan() is actually equal to. g N dr dy

58 Rotating Tank The Solution Step 1: Formulate the differential equation starting from: g tan() =! 2 r N Here tan() is actually equal to. Thus we have a separable differential equation: dy dr g N dr dy

59 Rotating Tank The Solution Step 1: Formulate the differential equation starting from: g tan() =! 2 r N Here tan() is actually equal to. Thus we have a separable differential equation: g dy dr =!2 r dy dr dy =!2 r g dr g N dr dy

60 Rotating Tank The Solution Step 2: Solve the separable differential equation to get in terms of h, R,!, and g. y(r)

61 Rotating Tank The Solution Step 2: Solve the separable differential equation to get in terms of h, R,!, and g. y(r) dy =!2 r g dr

62 Rotating Tank The Solution Step 2: Solve the separable differential equation to get in terms of h, R,!, and g. Z Z dy =!2 r! 2 g dr r dy = g dr y(r)

63 Rotating Tank The Solution Step 2: Solve the separable differential equation to get y(r) in terms of h, R,!, and g. Z Z dy =!2 r! 2 g dr r dy = g dr y(r) r 2 =!2 2g + C

64 Rotating Tank The Solution Step 2: Solve the separable differential equation to get y(r) in terms of h, R,!, and g. Z Z dy =!2 r! 2 g dr r dy = g dr y(r) r 2 =!2 2g + C Step 3: Determine the integration constant initial conditions. C using the

65 Rotating Tank The Solution Step 2: Solve the separable differential equation to get y(r) in terms of h, R,!, and g. Z Z dy =!2 r! 2 g dr r dy = g dr y(r) r 2 =!2 2g + C Step 3: Determine the integration constant C using the initial conditions. We can use the fact that the area of the water as viewed from the side will not change: R h

66 Rotating Tank The Solution Step 2: Solve the separable differential equation to get y(r) in terms of h, R,!, and g. Z Z dy =!2 r! 2 g dr r dy = g dr y(r) r 2 =!2 2g + C Step 3: Determine the integration constant C using the initial conditions. We can use the fact that the area of the water as viewed from the side will not change: R h Z R y(r) dr = hr 0

67 Rotating Tank The Solution Z R 0 y(r) dr = hr

68 Rotating Tank Z R 0 y(r) dr = hr The Solution Z R 0! 2 r 2 2g + C dr = hr

69 Rotating Tank Z R 0 y(r) dr = hr The Solution Z R 0! 2 r 2 2g + C dr = hr! 2 R 3 6g + CR = hr

70 Rotating Tank Z R 0 y(r) dr = hr The Solution Z R 0! 2 r 2 2g + C dr = hr! 2 R 3 6g + CR = hr C = h! 2 R 2 6g

71 Rotating Tank The Solution Z R y(r) dr = hr 0 2 Z R 0! r +C 2g C=h 2! y(r) = h + 2g h dr = hr 2 3! R + CR = hr 6g R 2 2 r 2 2 R 3! R 6g 2