Real time physics lab 2 homework answers

Lab Manual Irina Golub

July 30, 2017

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PART ONE: Photographic Analysis of a Falling Object

INTRODUCTION

With the great advances that have been made in digital imaging and analysis, experimental data is often in the form of photographic images. In this experiment, you will make displacement measurements of a tennis ball dropped from a height using photographic data and your computer’s mouse positioning system. From two displacement measurements and the time between these measurements and one of the five kinematic equations that describe one- dimensional motion, the acceleration due to gravity on earth, “g”, will be estimated. Since you know what the correct answer should be (9.8 m/s2) you will be able to calculate the percent error of your estimate.

Neglecting air resistance, a falling object increases its speed 9.8 meters per second every second that it falls on earth. This is “g”, the acceleration due to gravity. Below you will see snapshots taken of a falling tennis ball at equal intervals of time (0.1 second between frames). You can see that the displacement of the tennis ball increases during each successive time interval. This is due to the tennis ball speeding up in each time interval, i.e., the tennis ball is accelerating. Read University Physics Volume 1 Chapter # 3: MOTION ALONG A STRAIGHT LINE

EQUIPMENT

A PC running MS Internet Explorer web browser. (Other web browsers may not work for this experiment.)

OVERVIEW

The photographic data file shows one composite photo made by splicing six separate images of a tennis ball dropped straight down. Each of the six separate images was taken 0.1 second apart.

Just below the photo in the photographic data file you will see boxes labelled X and Y with numbers that change when you move the mouse over the photo. These numbers are mouse coordinates in what we will call “mouse units.”

You will record the Y-position of the ball in the first and last image (i.e., ball image #1 and ball image #6) (the X-direction is not needed as the object was falling straight down). Since the numbers you record will be in mouse units and not meters, only the difference between these two measurements will be important. You will be able to convert this difference from mouse units into meters because there are two meter sticks joined together vertically in each photo as a reference.

[If you are unable to see the mouse coordinates in your browser and are un- able to get to a BCC lab computer you can measure ruler coordinates instead of mouse coordinates using a plastic ruler held near (but not touching!) the computer display. Substitute the phrase ”ruler coordinates” for ”mouse coordinates” in the procedure and questions. Take your measurements in millimeters. This will not be as accurate as using mouse coordinates which have a higher resolution.]

NOTE: Since mouse units are relative to the top left side of the display, it is important to center the entire photo in your display and not scroll the photo when you take measurements. If you cannot see the entire photo on your monitor without scrolling it, change your display to 800 x 600 by clicking the “Start” button, then “Control Pane’, then “Display”, then “Settings”, and adjusting the Screen Resolution to 800 x 600.

PROCEDURE

1. Determine the conversion factor between mouse units and meters as follows. In each

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photo, two meter sticks are joined together vertically. Read the mouse’s y-coordinate at the top of the upper meter stick by clicking on that point. Read the mouse’s y- coordinate at the bottom end of the lower meter stick by clicking on that point. The difference between these measurements represents 2 meters.

For example, if the top of the top meter stick is at y=50 and the bottom of the bottom meter stick is at y=650, then 650-50 = 600 mouse units = 2 meters. Therefore, one (1) mouse unit would equal 2 meters/600 mouse units = 0.0033 meters [This calculation was just an example of how to perform the calculation. Do not use 0.0033 meters per mouse unit as the conversion between mouse units and millimeters. Measure and perform your own calculation]

2. Open the photographic data file. Measure the y-position (i.e., y-coordinate) of the ball (in mouse units) for the first image and the last (i.e., 6th) image five times. Measure from the same place on the ball in each photograph (e.g., the top of the ball)., and record these values on your data table.

3. Calculate the difference in the y-coordinates (mouse units) of the ball from the first image to the last image. For example, if the y-coordinate of the ball in image #1 was 1369 mouse units and 1748 for the ball in image #2, then the difference would be 1748- 1369 or 379 mouse units.

4. Multiply the difference you found in step #3 by the meters/mouse unit conversion you found in step #2. For example, if you found a difference of 379 mouse units between image #1 and image #6 and a meters/mouse unit conversion of 0.0033 in step #2, the result would be 379 mouse units x 0.0033 meters/mouse unit = 1.25 meters.

5. Using y=0.5 gt2, solve for g where y is the result you found in step #4 (in meters) and t is the time difference between the 1st and 6th photos (0.5 seconds). For example, using 1.25 meters for y and 0.5 sec for t, g is calculated to be 10 m/s2.

6. Compute the percent error between the value for “g” you calculated com- pared to the actual value of “g”.

PART TWO: ESTIMATING REACTION TIME

After you see the rear brake lights on the car in front of you, how much time does it take for you to step on your brake pedal? This time interval is one measure of your reaction time. Reaction time complicates many time measurements because it introduces a time lag between an event and the manual recording of the event (e.g., with a stop watch). Automated measurements employ sensors that have very short response times. Read the Free Falling Object with Air Resistance. 1. Neglecting air resistance, an object in free fall moves through a distance, in meters, given by 0.5gt2.

2. Count the air resistance, an object in free fall moves through a distance, in meters, given by

𝑇𝑇 = �2 𝑦𝑦 𝑔𝑔

(1 + 𝑘𝑘𝑦𝑦 6𝑚𝑚

)

Where k is the constant of proportionality; the numerical value of k depends upon the shape of the object being dropped and the density of the atmosphere. Use the approximal value of the 𝑘𝑘 = 3.46 ∗ 10−3 𝑘𝑘𝑔𝑔

𝑠𝑠

By measuring the distance an object falls between two events, the time interval between the events can be calculated. In this experiment, you will estimate the reaction time for your hand using one-dimensional kinematics and free fall.

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EQUIPMENT Yardstick with metric rulings (i.e., a meter stick) The assistance of another person PROCEDURE 1. Your assistant should hold a meter stick vertically. The top of your outstretched hand should line up with the bottom of the meter stick. Your fingers and thumb should form a “V” around the bottom of the meter stick. Your hand should neither be too open or too closed for best results. 2. At a time chosen by your assistant, your assistant will drop the meter stick through your outstretched hand and you will safely grab the meter stick. 3. After grabbing the meter stick, record the position of the top of your hand on the meter stick in centimeters. 4. Perform five trials of the experiment and record your results into the data table.

5. Determine the weight of the meter stick. 6. Convert the displacements from centimeters to meters. CALCULATIONS 1. Calculate the average displacement for the five trials. Calculate the relative error of your measurement of the displacement. 2. Calculate your average reaction time using the average displacement. If neglecting air resistance using the formula:

y=0.5gt2

3. Calculate your average reaction time using the average displacement count the air resistance using the formula:

𝑇𝑇 = �2 𝑦𝑦 𝑔𝑔

(1 + 𝑘𝑘𝑦𝑦 6𝑚𝑚

)

4. Compere your results.

4. Find in the internet the average reaction time of a human.

5. Calculate the percent error between your results and the theoretical data from the internet.

3. Would you expect the reaction time to move a leg (e.g., pressing the brake pedal in response to a visual stimulus) is greater, the same, or less, than the hand reaction time measured in this experiment? Why?

INTRODUCTION
EQUIPMENT
OVERVIEW
PROCEDURE
2048L/Lab 2/free fall theory lab 2c.pdf
Introduction

If you will drop the object from a fixed height and measure the time t taken to fall between two light beams which are placed a distance y apart.

a

r

b

1st position

2nd position

Object T = 0 T = T0

T = T1

y(t)

t = 0 Y = 0 Y = a

Y = a + r Y = a + r + b

• Figure 1. Note the two types of variables used, the lower case variables, y, r and t are measured values. For example: t, corresponds to the time interval, the y measured in the meter, for the object to travel between the first and second positions. The upper case quantities, Y and T, are variables to be determined in the data analysis. For example: the object has zero velocity at T = 0.

FreeFallingObjectWithoutAirResistance

The equation of motion for a freely falling object is:

m =mgd 2y (1.)

dt2 Integrationyields:

v(t)= v0 + gt (2) where vo is the velocity when the object passes the first position. Integrating this equationyields:

2 0 2

1y(t) = v t + gt (3.)

Chaer FallingObject

These equations account for the presence of a gravitational force only. In practice, air provides a drag force which can have a significant effect on a falling object.

While y(t) represents the actual distance that the object has dropped in a time interval t, (see figure 1), in the actual experiment r is the distance . The distances a and b, (see figure 1), which are on the order of a few millimeters, can not be measured directly because they depend, among other factors, on the exact position. A rough approximation for both values should be found during the the experiment and, as you will see later, very accurate determinations of both a and b will be made.

Keeping in mindthat

y(t)= r(t)+ b (4.) then g can be determined by linearizing equation 3 and applying a least square fit to a straight line. How do you linearize equation 3? Divide it by t; fitting that equation vs. t results in a linear equation with slope g/2 and intercept v0.

What is the effect of b, which at this point is not known with great precision? When we linearize equation 2 weget:

t t 20 +1gty (t )= r (t )+ b =v (5.)

If we introduce a small error, δ, in b and then assume that b' = b + δ we arriveat:

r (t )+ b′ = r (t )+ b + δ= v t t 2 t

+ 1 gt + δ0 (6.)

Figure 2.2 shows a plot of equation 6 vs. t for different values of δ. The asymptotic effect that the δ/t term produces at t = 0 can clearly be seen.

-2

-4

-6

-8

10

8

6

4

2

0 .6

y/t vs. t for different values of delta

0 0.1 0.2 0.3 0.4 0.5 0

0.1 0.05 0 -0.05 -0.1

• Figure 2. y/t vs. t for different values of δ.

Hence, the "best" value for b can be determined by selecting a few well chosen values of b' and calculating the resulting χ2: Only when b' = b, i.e. for δ = 0, will the data approach a straight line at t=0. For all other values of b', the data will deviate from a straight line and the fit will produce a large value of χ2. Therefore, a minimum χ2corresponds to the best value of b. The least square fit automatically also calculates the best values for g/2 and v0.

FreeFallingObjectwithAirResistance

For an object falling in air, the equation of motion, 1 becomes:

m𝑑𝑑 2𝑦𝑦 𝑑𝑑𝑑𝑑2

= 𝑚𝑚𝑚𝑚 − 𝑘𝑘(𝑑𝑑𝑦𝑦 𝑑𝑑𝑑𝑑

)𝑛𝑛 (7.)

where the second term on the right is the drag force and k specifies the strength of the retarding force. The value of n depends on the shape of the object and its velocity. For large velocities, n approaches unity which corresponds to Stoke’s law of resistance. At small velocities, like the ones acquired by the object in this experiment, the resisting force is proportional to the square of the velocity, in which case it obeys Newton’s law ofresistance:

(8.)

This equation can be integrated to yield the solution:

t

g y(t )= V

2

(9.)

where Vt is the terminal velocity which is related to k:

mg ktV =

2 (10.)

If we have a system with Y(T=0) and v0=0 , then equation 9. can be solved for T:

t −1

g V

T = cosh (11.)

Since the effect of the drag force is not very transparent in this equation, the following expansion in Y may be moreilluminating:

(12.)

The correction due to drag appears in second and higher terms. Setting Vt=∞ , which is the case when no retarding media is present, reduces it to the familiar equation 3. (with v0=0 ). Vt is at least on the order of 1000 cm/sec or larger, we only need to consider the first two terms in equation 12. Rewriting this in terms of k we find that T is a function of Y and m, i.e. the mass of the object:

(13.)

This equation shows that when the mass of the object approaches infinity, the second term, which corresponds to the drag force on the object, goes to zero; in that case, a value for g can be determined which no longer is affected by air resistance.

m𝑑𝑑 2𝑦𝑦 𝑑𝑑𝑑𝑑2

= 𝑚𝑚𝑚𝑚 − 𝑘𝑘(𝑑𝑑𝑦𝑦 𝑑𝑑𝑑𝑑

)2

ln(cosh𝑑𝑑𝑑𝑑 𝑉𝑉𝑡𝑡

+ 𝑉𝑉𝑎𝑎 𝑉𝑉𝑡𝑡 𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑𝑑𝑑

𝑉𝑉𝑡𝑡 )

[𝑒𝑒 𝑑𝑑𝑑𝑑 𝑑𝑑𝑡𝑡 2 ]

𝑇𝑇 = 2 𝑌𝑌 𝑔𝑔

(1+ 𝑔𝑔𝑌𝑌 6𝑉𝑉𝑡𝑡2

+ 𝑔𝑔 2𝑌𝑌2

120𝑉𝑉𝑡𝑡2 +…)

𝑇𝑇 = 2 𝑌𝑌 𝑔𝑔

(1+𝑘𝑘𝑌𝑌 6𝑚𝑚

)

t g

g − g ′ V ≅ v (15.)

for an object dropped from a maximum height ymax is:and observing that the average velocity v

2 g ′ymaxv ≅ (16.)

Additional Information

g = 980.616 − 2.5928Cos(2ϕ)+ 0.0069Cos2(2ϕ)− 3.086×10−6H where ϕ is the latitude and H the elevation above sea level in centimeters. (Latitude for Minneapolis is approximately 47 degrees and its elevation is about 700 feet above sea level.)

A recent measurement by the geophysics department determined a value of g here at the University as g = 980.58cm/sec2.

TerminalVelocity

Equation 8. can be also be written as:

where g’ represents the acceleration of an object due to gravity in a resistive medium and Vt is the terminal velocity for the particular object. An average value of g’, i.e. g ′ The corresponding values of Vt can then be found by solving equation 14 :

𝑚𝑚′=g(1- 𝑉𝑉 2

𝑉𝑉𝑡𝑡2 )

(14.)

ü Dropping things from the Empire State Building (for real)

What about throwing things from tall structures in air?

We set up Newton’s Second Law equations for this system as a differential equation. The key difference is that we are moving through a fluid that has low density and low viscosity (think of this as a measure of the syrupy-ness of the fluid) we need to modify the drag expression:

FG - FD =m g - a v2 =m „

„t v

or

„

„t v = g - a

m v2

Remember, at t=0 the velocity is 0. So the slope of the velocity is, again, simply g. This looks familiar. As time goes on the velocity increases so that the slope decreases. The difference is that the slope changes more rapidly as time goes on. As the speed increases, the slope decreases more swiftly. Eventually, at some place close to (D), the slope reaches 0 and no longer changes significantly.

The slope of the velocity reaches 0 when we reach “terminal velocity.” For our system described above

this point is vterm = m

a g .

We can see that the curve is going to look like the previous expression for our aqueous exploration but with a steeper rise sharper turn around:

„v = Ig - a m v2M „ t

For each tick of the clock the change in the velocity is getting smaller rapidly as it incrementally adds to the starting velocity (0) and increments toward the terminal velocity. Integrating this expression is not as simple as the previous efforts only because the result doesn’t look like anything we will find on the reference sheet. As before we will factor out a êm leaving vterm behind.

„vIvterm2- v2M = am „ t This time we can’t simply transform the variables. Try if you let u = vterm2 - v2 you get „u = - 2 v „v. This doesn’t make the integral look like anything we recognize. We have to go to a book on integral tables. After scouring the tables we find the following indefinite integral:Ÿ 1

k2-x2 „ x =

tanh-1J x k N

k + C

We can adapt this fairly quickly to our expression and get the following result…‡ vHtL0 „vIvterm2- v2M = ‡ t0 am „ tï tanh-1J vHtLvterm Nvterm = am t

We can look at the dimensions of a m vtermand it is easily to confirm algebraically that this combination of constants has

units of 1@tD In other words: TanhATanh-1@xDE = x So we get for the velocity…

vHtL = vterm tanhIvterm am tM

This is as far we will take this. You can see that the Tanh function has all the right characteristics as the previous function only it rises more steeply and bends sharply to asymptotically arrive at the terminal velocity a bit sooner. The graph above is scaled so that the two different cases focus on the different shapes.

free fall theory for lab 2
Falling Object
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free fall theory
Labs/free fall/freefall.html
PHOTOGRAPHIC DATA
Y X
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Table of Contents.html

PHY2048L PH CALC I LAB ONLINE 622068 - Unit 2 : Lab - Free Fall and Reaction Time
1. Lab 2 - Free Fall and Reaction Time

2. Free Fall Theory

3. Estimating g from Photograpic Analysis of Freefall