Kinematics Unit. Measurement

Transcription

1 Kinematics Unit Measurement

2 The Nature of Science Observation: important first step toward scientific theory; requires imagination to tell what is important. Theories: created to explain observations; will make predictions. Observations will tell if the prediction is accurate, and the cycle goes on. Physics- The study of the natural world

3 The four fundamental forces Electromagnetic Gravity Nuclear Strong Nuclear Weak Any force in nature must be one of these four fundamental forces. There have been many attempts to unify these forces (Grand unified theory). The work is ongoing.

4 How does a new theory get accepted? Predictions agree better with data Explains a greater range of phenomena

5 Physics is needed in both architecture and engineering. Other fields that use physics, and make contributions to it: physiology, zoology, life sciences,

6 Communication between architects and engineers is essential if disaster is to be avoided.

7 Models, Theories, and Laws Models are very useful during the process of understanding phenomena. A model creates mental pictures; care must be taken to understand the limits of the model and not take it too seriously. A theory is detailed and can give testable predictions. A law is a brief description of how nature behaves in a broad set of circumstances. A principle is similar to a law, but applies to a narrower range of phenomena.

8 Measurement and Uncertainty; Significant Figures No measurement is exact; there is always some uncertainty due to limited instrument accuracy and difficulty reading results. The photograph to the left illustrates this it would be difficult to measure the width of this 2x4 to better than a millimeter.

9 Estimated uncertainty is written with a ± sign; for example: Percent uncertainty is the ratio of the uncertainty to the measured value, multiplied by 100:

10 Significant Figures The number of significant figures is the number of reliably known digits in a number. It is usually possible to tell the number of significant figures by the way the number is written: cm has 4 significant figures cm has 2 significant figures (the initial zeroes don t count) 80 km is ambiguous it could have 1 or 2 significant figures. If it has 3, it should be written 80.0 km. Using scientific notation for a number avoids any ambiguity. (e.g. 8.0x10 1 km has 2 sig figs)

11 When multiplying or dividing numbers, the result has as many significant figures as the number used in the calculation with the fewest significant figures. Example: 11.3 cm x 6.8 cm = 77 cm In this example 11.3 has 3 sig figs, and 6.8 has 2 sig figs. The final result must have 2 sig figs according to our rule. When adding or subtracting, the answer is no more accurate (number of decimal places) than the least accurate number used. See the example on the next page.

12 Example: Add 5.67 J, 1.1 J, and J 5.67 J (two decimal places) 1.1 J (one decimal place) J (four decimal place) 7.7 J (one decimal place)

13 Example with multiple operations: (25.23 m+ 0.5 m m)/ (1.5 s)=? Solution: Do the addition m/ 1.5 s Carry an extra digit in the number. This number is actually only significant to 28.4 ( 3 SF), according to the addition/subtraction rule. Now do the division. According to the multiplication/ division rule the final result must have 2 SF (significant figures). Final answer = 19 m/s *One generally rounds up if the adjacent number is >= 5

14 Keep One Extra Digit in Intermediate Answers When doing multi-step calculations, keep at least one more significant digit in intermediate results than needed in your final answer. For instance, if a final answer requires two significant digits, then carry at least three significant digits in calculations. If you round-off all your intermediate answers to only two digits, you are discarding the information contained in the third digit, and as a result the second digit in your final answer might be incorrect. (This phenomenon is known as "round-off error.") The Two Greatest Sins Regarding Significant Digits Writing more digits in a final answer than justified by the number of digits in the data. Rounding-off, say, to two digits in an intermediate answer, and then writing three digits in the final answer don t round mid-calculation

15 Vector and Scalar Quantities Measurable physical quantities can be either a scalar or a vector quantity. Scalars have no spatial direction associated with them while vectors do. Examples of scalars: mass, temperature, speed, time Examples of vectors: velocity, momentum, force.

16 Units, Standards, and the SI (MKS) System Quantity Unit Standard Length Meter Length of the path traveled by light in 1/299,792,458 second. Time Second Time required for 9,192,631,770 periods of radiation emitted by cesium atoms Mass Kilogram Platinum cylinder in International Bureau of Weights and Measures, Paris

17 These are the standard SI prefixes for indicating powers of 10. Many are familiar; Y, Z, E, h, da, a, z, and y are rarely used.

18 We will be working in the SI system, where the basic units are kilograms, meters, and seconds. Other systems: cgs; units are grams, centimeters, and seconds. British engineering system has force instead of mass as one of its basic quantities, which are feet, pounds, and seconds.

19 Converting Units Converting between metric units, for example from kg to g, is easy, as all it involves is powers of 10. Converting to and from British units is considerably more work. For example, given that 1 m = ft, this 8611-m mountain is feet high.

20 Conversions Example: What is the equivalent of 500 mm in metres? Example: 500 mm X 1 m = m 1000 mm Convert 120 km/h to m/s Solution: 120 km h 1000 m 1 km 1 h 60 minutes Converson factors: 1 km= 1000 m 1 h = 60 minutes 1 minute = 60 s 1 minute 60 s = 33.3 m/s Notice that the units cancel to give the correct units. Each ratio has a value of 1.

21 Precision and Accuracy Precision: The degree of exactness to which the measurement of a quantity can be reproduced. For instance if the value of acceleration due to gravity is measured to be 8.83 ms -2 and 8.79 ms -2 in a second trial, what is the precision? The average value is 8.81 ms -2, thus the acceleration due to gravity can be stated as ( ) ms -2. The precision is 0.02 ms -2 - a precise measurement, but inaccurate (see below). Accuracy: The extent to which a measured value agrees with a standard value. In the experiment to measure the acceleration due to gravity, the accuracy is the difference between the student s measurement and the defined value (the defined value is 9.81). Let s assume that the measured value was 9.83 ms -2. Then the accuracy would be = 0.02 ms

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23 Errors-Systematic and Random Random Error: These are random errors inherent in measurements. The effect would be on the spread (precision) of the data. The more tightly grouped the data the smaller the random error Systematic Error: These kind of errors skew the data (either high or low). A systematic error will shift the data away from the true value. It has and effect on the accuracy of the data.

24 Order of Magnitude: Rapid Estimating A quick way to estimate a calculated quantity is to round off all numbers to one significant figure and then calculate. Your result should at least be the right order of magnitude; this can be expressed by rounding it off to the nearest power of 10. Diagrams are also very useful in making estimations.

25 Graphical Analysis Intro A relationship between two variables can be shown by a graph. Often a graph is a convenient and powerful way to represent the data. The shape of the graph can give insight into the nature of the relationship. The general convention is to have the independent variable on the horizontal (x) axis, and the dependent variable on the vertical (y) axis. It should noted that this convention is often ignored. The independent variable is usually the variable that is initially changed and the dependent variable responds.

26 These are the steps to make a graph: Collect data. Measure quantities and place in a table. Plot data on a graph (scatter plot) after picking appropriate scales. One picks a scale (units per division) that spreads the data out to make good use of the page. Often the line of best fit is drawn through the data points. The line of best fit is the best straight line through the points. The slope of the line of best fit is often useful to determine. When graphing you should be aware of these points: Make sure your graph, and axes are properly labeled. Make sure you include units on axes and on data tables Give your graph a title even if it is Y versus X.

27 An example of a relationship is how the weight (W) added to a vertical spring is related to the stretch (x)of the spring. The amount of stretch (measured from its original unstretched length) turns out to be directly proportional to the weight. The relationship is given by W= kx, where k is some constant. This is a linear relationship (y = mx +b) with a y intercept of zero. On would expect the data to be linear if one plotted W versus x. The degree of random and systematic error can be observed by examining the data. If the points are more spread out this indicates more random error. If the data is shifted (y intercept 0) it indicates a systematic error. If for some reason the data appears curved perhaps the assumptions of the relationship being linear is not correct.

28 Kinematics in One Dimension

29 Reference Frames and Displacement Any measurement of position, distance, or speed must be made with respect to a reference frame. For example, if you are sitting on a train and someone walks down the aisle, their speed with respect to the train is a few miles per hour, at most. Their speed with respect to the ground is much higher.

30 Distance and Displacement We make a distinction between distance and displacement. Displacement or position (blue line) is how far the object is from a given reference point, regardless of how it got there. It is a vector. Distance traveled (dashed line) is measured along the actual path. It is a scalar.

31 The change in displacement is written: Left: Change in Displacement is positive. Right: Change in Displacement is negative.

32 Average Velocity and Speed Speed: how far an object travels in a given time interval Velocity includes directional information: average velocity = change in displacement change in time = Δx Δt

33 Example: Bob walks to the right a distance of 8.00 m, then left a distance of 6.00m in a total time of 4.00 s. What was his average velocity and average speed over this time? Solution: The average speed is total distance divided by time: s = = 3.50 m/s The average velocity is change is position divided by time: υ = = m/s to the right

34 Instantaneous Velocity The instantaneous velocity is the average velocity, in the limit as the time interval becomes infinitesimally short. These graphs show (a) constant velocity and (b) varying velocity.

35 Acceleration Acceleration is the rate of change of velocity.

36 Acceleration is a vector, although in onedimensional motion we only need the sign. The previous image shows positive acceleration; here is negative acceleration:

37 There is a difference between negative acceleration and deceleration: Negative acceleration is acceleration in the negative direction as defined by the coordinate system. Deceleration occurs when the acceleration is opposite in direction to the velocity.

38 The instantaneous acceleration is the average acceleration, in the limit as the time interval becomes infinitesimally short.

39 Graphical Analysis of Linear Motion (1-D) The following graphs represent the motion of an object in one dimension (a line) This is a graph of x vs. t for an object moving with constant velocity. The velocity is the slope of the x-t curve.

40 On the left we have a graph of velocity vs. time for an object with varying velocity; on the right we have the resulting x vs. t curve. The instantaneous velocity (on the position time graph) is slope of the tangent line to the curve at any given point. The slope of the line on a velocity- time graph is the acceleration.

41 Example: d(m) Describe the object s motion over the 55 s t(s) 0-10s: Constant velocity of 6 m/s, or constant speed of 6 m/s in positive direction 10-15s : Stationary 15-40s: Constant velocity of -4 m/s, or constant speed of 4 m/s in negative direction. Also it comes back to its starting point at 40s s: Constant velocity of 2.67 m/s or constant speed of 2.67 m/s in positive direction.

42 Example: d (3) (0,0) (1) (2) (4) t How would you describe the motion of an object given by the position time graph above? Use the tangent line to the curve.

43 Solution: It is initially stationary at a negative position till (1) It starts to speed up in the positive direction crossing the origin and reaching a max velocity near (2) It starts to slow down and has a velocity of zero at (3) and immediately changes direction. It starts to speed up in the negative direction and reaches a constant negative velocity for a short time It starts to slow down before (4) and comes to a stop. It remains stationary at a positive position The above deductions are made by reading directly off the graph and by using the slope of the tangent line (velocity).

44 Example: Using this position-time graph, answer the following questions on the next slide

45 a) What was the average speed from t=0 to t=21 s? b) What was the average velocity over the same time interval? c) At what times was the velocity zero? d) At what times was the object moving in the positive direction? e) At what times was the object moving in the negative direction? f) What was the object's velocity at t=9.0 s? g) What was the object's velocity at t=21 s? h) What was the object's greatest speed? i) Estimate the object's velocity at t=30 s? j) When did the object first pass through d=0 m?

46 a) 1.33 m/s b) m/s c) 0 to 6 s, 11 to 18 s, 27 s, and 34.5s d) s, s e) 6-11 s, s f) -4 m/s g) 2.67 m/s h) 4 m/s i) approx m/s j) t=9 s

47 Acceleration Position time graphs Acceleration occurs when there is a changing velocity. A positive change in velocity indicates a positive acceleration, and a negative change in velocity indicates a negative acceleration. Using the slope of the tangent line on a position-time graph the nature of the acceleration can be deduced. In figure 1, one can see the slope of the tangent line going from -, to flat, to +. This indicates a positive acceleration. One way to remember is that a happy smile is positive J. In figure 2, one can see the slope of the tangent line going from +, to flat, to -. This indicates a negative acceleration. One way to remember is that an unhappy smile is negative L. d d figure 1 figure 2 t t

48 The change in displacement is the area beneath the v vs. t curve. If the line is below the x axis (v<0) then that area would give an negative change in displacement. By calculating areas one can determine both total distance travelled and the net change in displacement.

49 Example: Using this velocity-time graph, answer the following questions on the next slide

50 a)at what times did the object have zero velocity?. b)find the object's change in displacement over the first 15 s. c)what was the average velocity over the first 15 seconds? d) What was the average speed over the first 15 s? e) What was the object's average acceleration from t=20 to t=30s? f) Over what time period(s) was the object's acceleration +? g) Over what time period(s) was the object's acceleration -?

51 a) t= 12.2 s, and t=25 s b) d= = m c) v= 137.5/15 = 9.17 m/s d) s= ( )/15 = 11.0 m/s e) a= (13- (-10))/10 = 2.3 m/s 2 f) t greater than 20 s g) t= 8 to 15 s

52 Motion at Constant Acceleration & the Kinematic Equations The average velocity of an object during a time interval t is, change in position (displacement) divided by change in time. The acceleration, assumed constant, is

53 In addition, as the velocity is increasing at a constant rate, we know that Combining these last three equations, we find: Or, often x-x 0 is synonymous with d for change in displacement d = υ 0 t at 2

54 We can also combine these equations so as to eliminate t: We now have all the equations we need to solve constant-acceleration problems.

55 Summary (Equivalent) Relationships: d = υ 0 t at 2 υ = υ 0 + at υ 2 = υ ad υ = υ 0 +υ 2 d = υt d= change in displacement (m) v 0 = initial velocity(m/s) v= final velocity (m/s) a= acceleration (m/s 2 ) t= change in time (s) υ = average velocity (m/s) Only true for constant a. *These equations can only be used over periods of a constant acceleration

56 Solving Problems 1. Read the whole problem and make sure you understand it. Then read it again. 2. Decide on the objects under study and what the time interval is. 3. Draw a diagram and choose coordinate axes. Define a positive x direction. 4. Write down the known (given) quantities, and then the unknown ones that you need to find. 5. What physics applies here? Plan an approach to a solution.

57 6. Which equations relate the known and unknown quantities? Are they valid in this situation? Solve algebraically for the unknown quantities, and check that your result is sensible (correct dimensions). 7. Calculate the solution and round it to the appropriate number of significant figures. 8. Look at the result is it reasonable? Does it agree with a rough estimate? 9. Check the units again.

58 Falling Objects Near the surface of the Earth, all objects experience approximately the same acceleration due to gravity. This is one of the most common examples of motion with constant acceleration.

59 In the absence of air resistance, all objects fall with the same acceleration, although this may be hard to tell by testing in an environment where there is air resistance.

60 The acceleration due to gravity at the Earth s surface is approximately 9.80 m/s 2.