Setup Of The Quantitative Description Of Your Rube Goldberg Device Step

The final project for this course is the creation of an analysis report. For Milestone Three, you will submit Setup of the Quantitative Description of Your Rube Goldberg Device Step.

This milestone is due in Module Five. It will provide an additional step towards the completion of the final project. This step should be fully analyzed in the final submission. Your submission will demonstrate the knowledge of how to calculate the values that give a quantitative description of what is going on during the selected step and at the transitions to/from the neighboring steps, using the quantitative description as a starting point.

Specifically, the following critical elements must be addressed:

I. Step Selection: Select a step or stage in the Rube Goldberg device. Provide a concise description of the step.

II. Previous Step A. Description: Analyze the behavior of the object in the interaction between the previous step and the selected step, qualitatively describing the transfer of energy that occurs. Which principles of conservation of energy and momentum can you apply to this behavior? B. Equations: Provide the equations that can be used to describe the transfer of energy and the momentum of the object from the previous step to the selected step. What is the connection between the basic physics concepts in the equations and the interaction of the object and force(s) from step to step? C. Calculations: Using the applicable equations you identified, calculate the transfer of energy and the momentum from the previous step to the selected step. How do these calculations help you predict the object’s location and velocity from the previous step to the step you selected?

III. Selected Step B. Equations: If applicable, provide the equations that can be used to describe the change in type and amount of energy across the selected step. C. Energy Calculation: Calculate the amount of energy that is converted from one form to another form using the changes in mass and height. For example, if appropriate for your selected step, you could calculate the transformation of potential energy to kinetic energy.

IV. Subsequent Step A. Description: Analyze the behavior of the object in the interaction between the selected step and the subsequent step, qualitatively describing the transfer of energy that occurs. Which principles of conservation of energy and momentum can you apply to this behavior? B. Equations: Provide the equations that can be used to describe the transfer of energy and the momentum of the object. What is the connection between the basic physics concepts in the equations and the interaction of the object and force(s) from step to step? C. Calculations: Using the applicable equations you identified, calculate the transfer of energy and the momentum from your selected step to the subsequent step. How do these calculations help you predict the object’s location and velocity from the step you selected to the subsequent step?

Guidelines for Submission:  Submit assignment as a Word document with double spacing, 12-point Times New Roman font, and one-inch margins. Your paper should be 2- to 3-pages.

 

 
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Physics: Projectile Motion Experiment

Projectile Motion Experiment

 

By Monday, August 19, perform the following experiments online at the Projectile Motion – Galileo website. (http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html)

Fill in the tables, and post your answers to the questions.

Question 1

Procedure: Keep the initial velocity fixed at 50 m/s. Perform the experiment for each of the following angles:

  • 15 degrees
  • 30 degrees
  • 45 degrees
  • 60 degrees
  • 75 degrees

Record in table 1:

  • The maximum horizontal distance (Range) traveled by the projectile at various angles
  • The total time of flight at various angles
  • The maximum height attained

Table 1: Experiment 1 – Range & Time of flight

S. No. Initial Velocity Angle of Projection Range Time of Flight  Maximum Height Attained
1.          
2.          
3.          
4.          
5.          

 

Observe and Analyze:

  • What angle produces the maximum range? Why?
  • What angle produces the maximum height? Why?
  • Are there angles which produce the same range? If so, how would you explain this?

Question 2 

Procedure: Set the angle to 45 degrees. Perform the experiment with the following initial velocities.

  • 30 m/s
  • 40 m/s
  • 50 m/s
  • 60 m/s

Record in Table 2: The horizontal and vertical components of the velocity and the ranges for each of the velocities.

Table 2: Experiment 2 – Range at 45°

 

S. No. Initial Velocity Horizontal Velocity Vertical Velocity Range at 45°
1.        
2.        
3.        
4.        

 

Observe and Analyze:

  • As the velocity is increased what happens to the Range?
  • Using the initial horizontal and vertical velocities from your table, verify mathematically that the range is correct for the initial speed of 60 m/s.

Attached are assignment related online lectures and textbook chapters

 
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Newton Second & Third Law

Name

Tutorials in Introductory Physics ©Pearson Custom Publishing McDermott, Shaffer, & P.E.G., U. Wash. Updated Preliminary Second Edition, 2011

Mech HW–39

1. A block initially at rest is given a quick push by a hand. The block slides across the floor, gradually slows down, and comes to rest.

a. In the spaces provided, draw and label separate free-body diagrams for the block at each of the three instants shown.

A quick push by a hand…

1. (Initially at rest)

the sliding block slows…

2.

v

and is finally at rest.

3.

b. Rank the magnitudes of all the horizontal forces in the diagram for instant 1. Explain.

c. Are any of the forces that you drew for instant 1 missing from your diagram for instant 2?

If so, for each force that is missing, explain how you knew to include the force on the first diagram but not on the second.

d. Are any of the forces that you drew for instant 1 missing from your diagram for instant 3? If so, for each force that is missing, explain how you knew to include the force on the first diagram but not on the third.

NEWTON’S SECOND AND THIRD LAWS

Newton’s second and third laws

Tutorials in Introductory Physics ©Pearson Custom Publishing McDermott, Shaffer, & P.E.G., U. Wash. Updated Preliminary Second Edition, 2011

Mech HW–40

2. Two crates, A and B, are in an elevator as shown. The mass of crate A is greater than the mass of crate B.

a. The elevator moves downward at constant speed.

i. How does the acceleration of crate A compare to that of crate B? Explain.

ii. In the spaces provided below, draw and label separate free-body diagrams for the crates.

Free-body diagram for crate A

Free-body diagram for crate B

iii. Rank the forces on the crates according to magnitude, from largest to smallest. Explain your reasoning, including how you used Newton’s second and third laws.

iv. In the spaces provided at right, draw arrows to indicate the direction of the net force on each crate. If the net force on either crate is zero, state so explicitly. Explain.

Is the magnitude of the net force on crate A greater than, less than, or equal to that on crate B? Explain.

Elevator (moving down

at constant speed)

A

B

Cable

Crate A Crate B

Direction of net force

Newton’s second and third laws Name

Tutorials in Introductory Physics ©Pearson Custom Publishing McDermott, Shaffer, & P.E.G., U. Wash. Updated Preliminary Second Edition, 2011

Mech HW–41

b. As the elevator approaches its destination, its speed decreases. (It continues to move downward.)

i. How does the acceleration of crate A compare to that of crate B? Explain.

ii. In the spaces provided below, draw and label separate free-body diagrams for the crates in this case.

Free-body diagram for crate A

Free-body diagram for crate B

iii. Rank the forces on the crates according to magnitude, from largest to smallest. Explain your reasoning, including how you used Newton’s second and third laws.

iv. In the spaces provided at right, draw arrows to indicate the direction of the net force on each crate. If the net force on either crate is zero, state so explicitly. Explain.

Is the magnitude of the net force on crate A greater than, less than, or equal to that on crate B? Explain.

Crate A Crate B

Direction of net force

Newton’s second and third laws

Tutorials in Introductory Physics ©Pearson Custom Publishing McDermott, Shaffer, & P.E.G., U. Wash. Updated Preliminary Second Edition, 2011

Mech HW–42

3. A hand pushes three identical bricks as shown. The bricks are moving to the left and speeding up. System A consists of two bricks stacked together. System B consists of a single brick. System C consists of all three bricks. There is friction between the bricks and the table. a. In the spaces

provided at right, draw and label separate free-body diagrams for systems A and B.

b. The vector representing the acceleration of system A is shown at right. Draw the acceleration vectors for systems B and C using the same scale. Explain.

c. The vector representing the net force on system A is shown at right. Draw the net force vectors for systems B and C using the same scale. Explain.

d. The vector representing the frictional force on system A is shown below. Draw the remaining force vectors using the same scale.

N BH

N AB

N BA

f AT

f BT

Explain how you knew to draw the force vectors as you did.

A

B

Free-body diagram for system A

Free-body diagram for system B

Acceleration of A

Acceleration of B

Acceleration of C

Net force on A

Net force on B

Net force on C

 
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A Barge Floating In Fresh Water

A barge floating in fresh water (A? = 1000 kg/m3) is shaped like a hollow rectangular prism with base area A =650 m2 and height H= 2.0 m. When empty the bottom of the barge is located H0 =0.4 m below the surface of the water. When fully loaded with coal the bottom of the barge is located H1 = 1 m below the surface. Randomized Variables A = 650 m2 H0 = 0.4 m H1 = 1 m kqxmevrn.phr.png

 

No Attempt No Attempt 14% Part (b) Write an equation for the buoyant force on the empty barge in terms of the known data. No Attempt No Attempt 14% Part (c) Determine the mass of the barge in kilograms. No Attempt No Attempt 14% Part (e) Find the mass of the coal in terms of the given data. No Attempt No Attempt 14% Part (f) Find the mass of the coal in kilograms.
 
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Physics Of Cell HW1

11

2.3 A feeling for the numbers: microbes as the unseen majority

(a) Use Figure 2.1 to justify the assumption that a typical bacterial cell (that is, E. coli) has a surface area of 6µm2 and a volume of 1µm3. Also, express this volume in femtoliters. Make a corresponding estimate of the mass of such a bacterium. (b) Roughly 2–3 kg of bacteria are harbored in your large intestine. Make an estimate of the total number of bacteria inhabiting your intestine. Estimate the total number of human cells in your body and compare the two figures. (c) The claim is made (see Whitman et al., 1998) that in the top 200m of the world’s oceans, there are roughly 1028 prokaryotes. Work out the total volume taken up by these cells in m3 and km3. Compute their mean spacing. How many such cells are there per milliliter of ocean water?

(a) E. coli has (roughly) the shape of a cylinder that is 2 µm in length and 0.5 µm in radius. For those that are so inclined, the bacterium can alternatively be treated as a spherocylinder, though the results will not change in any interesting way. Using these numbers we calculate the area of an E. coli to be:

Acell = ⇡ ⇥ 1 µm⇥ 2 µm ⇡ 6 µm2. (2.28)

Its volume is:

Vcell = ⇡ ⇥ ✓ 1

2 µm

◆2 ⇥ (2µm) ⇡ 1µm3 (2.29)

= 1 fL. (2.30)

If we assume that the density of a bacterium is the same as that of water, the mass of one bacterium is 103kg/m3 ⇥ 10�18 m3 ⇡ 10�15 kg = 1 pg.

(b) The fact that each bacterium has a mass of 1 pg implies that 2�3 kg worth of bacteria in the intestines of one person amounts to 2 ⇠ 3⇥ 1015 bacteria.

Assume that the size of a typical human cell is roughly 10 µm in diameter and has a spherical shape with the same density as that of water. Let’s assume that the mass of a “typical” human body is roughly 80 kg. Further, let’s assume that thirty percent of the human mass corresponds to cells. On the basis of these assumptions, we find that the number of the cells in a human body is approximately

Mhuman Vcell⇢H2O

= 1 3 ⇥ 80 kg

4/3⇡ ⇥ (5⇥ 10�6 m)3 ⇥ 1000 kg/m3 ‘ 5⇥ 1013. (2.31)

By this estimate, the number of bacterial cells outnumbers the number of human cells by more than a factor of ten.

(c) Using (a) we can estimate the volume of 1028 prokaryotes to be about 10�18 m3 ⇥ 1028 = 1010 m3, which is equal to 10 km3.

HW 1

27

section of this cylinder with the same cross-sectional area. The schematic of the mature virion in fig. 2.31(C) shows that the CA proteins come together to form the capsid with inward pointing “spokes,” in the same way that the GAG polyproteins form the initial outer shell of the virion. This means that CA and GAG have the same surface areas per protein. Because the surface area of the capsid is less than that of the virion and because each CA protein is cleaved o↵ a GAG polyprotein, this fact immediately implies that not all the CA proteins can be used up to form the capsid.

From the micrographs, the capsid can be approximated as a cone with base radius r = 25 nm and side length s = 100 nm. Its surface area is then ⇡rs+⇡r2 ⇡ 1·104 nm2, and the number of CA proteins making up the capsid is then roughly 104

4⇡ ⇡ 800 CA proteins. This result can also be obtained by multiplying the ready-made estimate of 3500 total GAG proteins by the ratio of the surface areas

GAG proteins total· surface area capsid surface area virion

= 3500· 10 4 nm2

4⇡ · 60 nm2 ⇡ 800 CA proteins in capsid, (2.49)

where the virion radius used is 60 nm instead of 65 nm because the outer 5 nm of the virion shell are taken up by a lipid bilayer.

2.9 Areas and volumes of organelles

(a) Calculate the average volume and surface area of mitochondria in yeast based on the confocal microscopy image of Figure 2.18(C). (b) Estimate the area of the endoplasmic reticulum when it is in reticular form using a model for its structure of interpenetrating cylinders of diameter d ⇡ 10 nm separated by a distance a ⇡ 60 nm, as shown in Figure 2.25.

(a) The mitochondria in this yeast are shaped like a cylinder with a diameter of 400 nm approximately (which could be the resolution limit of the microscope). The total extension of this cylinder is about 20 µm. This results in a total mitochondrial volume of ⇡(0.2 µm)2 ⇥ 20 µm ⇡ 2.5 µm3. The total area is 2⇡ ⇥ 0.2 µm⇥ 20 µm ⇡ 25 µm2.

Mitochondria are thus just a small fraction of the total yeast volume, which is around 500 µm3. (b) Each “cross”, the unit that gets repeated in this structure, can be approxi- mated by two cylinders of length a and diameter d. Therefore, its surface area is 2⇥ ⇡ ⇥ 10 nm ⇥ 60 nm ⇡ 4000 nm2. Now, each one of this units occupies a volume a3 ⇡ 0.22⇥ 10�3 µm3.

We assume that a fibroblast has a height of approximately 1 µm and, based on figure 2.15, we approximate the area of the fibroblast were the ER is present to be 25 µm2, around one fourth of the total area of the field of view. Therefore, in this volume of 25 µm3 we can fit 25 µm3/(0.22⇥10�3 µm3) ⇡ 105 such units. This in turn corresponds to a total surface area of 105 ⇥ 4000 nm2 = 400 µm2.

 

 

Chapter 3

When: Stopwatches at Many Scales

3.1 Growth and the logistic equation

In the chapter, we described the logistic equation as a simple toy model for constrained growth of populations. In this problem, the goal is to work out the dynamics in more detail. (a) Rewrite the equation in dimensionless form and explain what units this means time is measured in. (b) Find the value of N at which the growth rate is maximized. (c) Find the maximum growth rate. (d) Use these results to make a one-dimensional phase portrait like that shown in Figure 3.10.

(a) Recall that the logistic equation can be written as

dN

dt = rN(1� N

K ). (3.16)

If we now define ⌧ = rt, this implies that

d

dt =

d

d⌧

d⌧

dt (3.17)

Now we can rewrite the original logistic equation as

dN

d⌧ = N(1� N

K ), (3.18)

with time measured in units of 1/r. (b) By di↵erentiating the right side with respect to N , we have

d

dN [N(1� N

K )] = (1� N

K )� N

K = 1� 2N

K = 0 (3.19)

31

45

proteins by 50% (we assume ideal situation). However, in the case of degrada- tion, the amount of removed protein will depend on its concentration and it can vary from protein to protein.

3.7 The sugar budget in minimal medium

In rapidly dividing bacteria, the cell can divide in times as short as 1200 s. Make a careful estimate of the number of sugars (glucose) needed to provide the carbon for constructing the macromolecules of the cell during one cell cycle of a bacterium. Use this result to work out the number of carbon atoms that need to be taken into the cell each second to sustain this growth rate.

The number of sugars used was already worked out in problem 2.5. In this solution, we consider the implications for fast growing cells. How does all of this sugar get into the bacterium. If we consider fast growth, each cell cycle is approximately 20 minutes (or 1200 seconds), so the intake is:

109 sugars

1200 seconds = 8⇥ 105 sugar molecules/sec

There are approximately 1000 transmembrane proteins that transport sugar in the membrane of an E. coli, so the rate at which each protein must work at is:

8⇥ 105 sugars sec · 1000 proteins ⇡ 800

sugar molecules

transmembrane protein · sec

3.8 Metabolic rates

Assume that 1 kg of bacteria burn oxygen at a rate of 0.006mol/s. This oxygen enters the bacterium by di↵usion through its surface at a rate given by � = 4⇡DRc0, where D = 2µm2/ms is the di↵usion constant for oxygen in water, c0 = 0.2mol/m

3 is the oxygen concentration, and R is the radius of the typical bacterium, which we assume to be spherical. (a) Show that the amount of oxygen that di↵uses into the bacterium is greater than the amount used by the bacterium in metabolism. For simplicity, assume that the bacterium is a sphere. (b) What conditions does (a) impose on the radius R for the bacterial cell? Compare it with the size of E. coli.

(a) To estimate the amount of oxygen di↵using into a bacterium we assume that all the oxygen arriving at the cell membrane is absorbed. Furthermore, assuming that the bacterium is a sphere of radius R = 1µm (the typical size of an E.coli cell), the rate of oxygen entering the cell is

� = 4⇡DRc0 = 3⇥ 109 1/s . (3.68)

Given that bacteria burn oxygen at a rate of r = 0.2mole/kg s and that the

 
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Physics Momentum

CONSERVATION OF MOMENTUM Name INTWO DIMENSIONS

1. Two objects are arranged on a level, frictionless table as shown. Two experiments are conducted in which object A is launched toward the stationary block B. The initial speed of object A is the same in both experiments; the direction is not. The initid and final velocities of object A in each experiment are shown.

The mass of block B is four times that of object A (m” = 4m^).

Top views Velocity Yectors (drawn to scale)

Mech HW-63

{ do,

Experiment 1 – before collision

ld”,l= o ilur: ?

– after collision

Tn, . uei

Direction of Lfio

Experiment2 – before collision 2 – after collision

a. In the space provided, draw separate arrows represeiting the direction of the change in momentum vector of object A in the two experiments.

Is the magnitude of the change in m.omcntum of object A in experiment I greater than, less than, or equal to that in experiment 2? Explain.

Experiment I Experiment 2

b. In the space provided, draw separate arrows representing the direction of the change in momentum vector of block B in the two experiments.

Afier the collisions, is the magnitude of the momentum of block B in experiment I greater than, less than, or equal to that in experiment 2? If the momentum of block B is zero in either case, state that explicitly. Explain.

Direction of Al^

Experiment I Experiment 2

Tutorials in Introductory Physics McDermott, Shaffer, and the P,E.G., U. Wash.

I arson Custom zod EA.,for U,CO, Boulder

 

 

Mech HW-64

Conseroation of momentum in two dimensions

Two objects collide on a level, frictionless table. The mass of object A is 5.0 kg; the mass of object n is f .O tg. The objects stick together after the collision. The initial velocity of object A and the final velocity of both objects are shown.

2.

Before collision dltAi

After collision 1do, = d”r)

(One side of a square represents 0.1 m/s)

In the space provided, draw separate arrows for object A and for object B representing the direction of the change in momentum vector of the object.

Is the magnitude of the change in m,omentum of object A greater than, less than, or equal to that of object B? Explain your reasoning.

Direction of Al Object A Object B

b. System C is the system of both objects A and B combined. How does the momentum of system C before the collision compare to the momentum of system C after the collision? Discuss both magnitude and direction.

Construct and label a vector showing the momentum 6f system C at an instant before the collision. Show your work clearly.

c, Construct and label a vector showing the initial velocity of object B. Show your work clearly.

(Each side of a square represents 0.4 kg’m/s)

(Each side of a square represents 0.1 m/s)

Tutorials in Introductory Physics McDermott, Shaffer, and the P.E.G., U, Wash.

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Consentation of momentum in two dimensions Name

3. Object A collides on a horizontal frictionless surface with an Frictionless horizontal surface

Mech HW-65

initially stationary target, object X. The initial and final velocities of object A are shown. The final velocity of object X is not given.

a, At an instant during the collision, is the net force on object A zero ot nbn-zero?

b. During the collision, is the momentum of object A conserved? Explain.

Is the momentum of the system consisting of objects A and X conserved? Explain.

c. On the same horizontal surface, object C collides with an initially stationary target, objectZ, The initial speeds of objects C and A are the same, ild trtx= trlz) tne.= tltc, After the collisions, object C moves in the direction shown and has the same final speed as object A.

i. In the space below, copy the vectors d6; and d6l with their tails togbther. Use these vectors to draw the change in velocity vector for glider C, AAc.

Top view

Before collision

A X oH

ax at rest

Aftercollision

AX€o 6o,

Arter collision

ii. Is the magnitude of the change in velocity vector of object A greater than, less than, or equal to the magnitude of the change in velocity vector of object C? Explain.

iii. Is the magnitude of the change in momentum vector of object A greater than, less than, or equal to the magnitude of the change in momentum vector of object C? Explain.

iv. Is the final speed of object X greater than, less than, or equal to the final speed of objectZ? Explain.

d. Consider the following incorrect statement: ‘6liders A ond C have the some chonge in momentum. They hove the some moss, ond because they have the some initiql speed ond same f inol speed, Av is the some for eoch of ,them.”

Discuss the error(s) in the reasoning.

Tutorials in Introductory Physics McDermott, Shaffer, and the P.E.G., U. Wash.

earson

Custom 2nd Fd., for U.CO, Boulder

Before collision

Z o

At rest

 

 

DYNAMICS OF RIGID BODIES Name Mech HW-69

1. Energy analysis of the block-and-spool problem

A block and a spool are each pulled across a level, frictionless surface by a string, as illustrated at right.

The string pulling the block is tied to a small hook at the center of the front face of the block (not shown). The string pulling the spool is wrapped many times around the spool and may unwind as it is pulled.

The block and the spool have the same mass. The strings are pulled with the szlme constant tension and start pulling at the same instant.

Make the approximation that the strings and the hook are massless.

a. Does the spool cross the finish line before, afier, or at the sante instant as the block? Explain.

Tbp view

Start Finish k– -_-_- d — *–4

Bloc

Spooll

u Same

b. Consider the following dialogue between two students:

Student l: “f think thot there’s the some omount of work done on block ond spool os they ore pulled from the stort to the f inish since they both move the some distonce.”

Student 2: ‘f disogree. f think thot the hond pullirg the spool does more work thon the hond pulling the block since the string unwinds qs the spool is pulled.”

With which student, if either, do you agree? Explin.

When each crosses the finish line, is the total kinetic energy of the spool greater than, Iess than, or equal to that of the block? Explain. (Hint: Use the work-energy theorem.)

d. When each crosses the finish line, is the translational kinetic energy of the spool greater than,less than, or equal to that of the block? Explain.

Tutorials in Introductory Physics McDermott, Shaffer, and the P.E.G., U. Wash.

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Mech Dynamics of rigid.bodies HW-70

2. Three identical rectangular blocks are at rest on a level, frictionless surface. Forces of equal magnitude that act in the same direction are exerted on each of the three blocks. Each force is exerted at a different point on the block (indicated by the symbol “Xo’), as shown in the top-view diagram below. The location of each block’s center of mass is indicated by a small circle.

For each of the blocks, draw an affow on the diagram above to indicate the direction of the acceleration of the block’s center of mass at the instant shown. If the magnitude of the acceleration of the center of mass of any block is zero, state that explicitly. Explain.

b. Rank the blocts according to magnitude of center-of-mass acceleration, fromlargest to smallest. If any two blocks have the same magnitude center-of-mass acceleration, state so explicitly. Support your ranking by drawinga point free-body diagram for each block.

3, A uniform rigid rod rests on a level, frictionless surface. The diagram below indicates four different combinations of (1) net force on the rod and (2) net torque on the rod about its center of mass. In each box, draw vectors that represent one or two forces that achieve the given combination of net force and net torque. If any combination is not possible, state so explicitly.

For example: In the second case, indicate one or two forces that could be exerted on the rod so that at the instant shown the net force on it is zero, but the net torque on it is not zero.

Tbp view

li,,l = o, l(-,1 = o l{,,*l = o, l(,,1 * o l{,,,1 + o, li,,l = o lF”.,l * o, l?”.,1 * o

.,,

Block I

Tbp view

Block 2 Block 3

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@Pearson

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Dynamics of rigid bodies Name Mech HW-71

4. Three objects of equal mass, A, B, and C, are released from rest at the same instant from the same height on identical ramps. Objects A and B are both blocks, and they slide down their respective ramps without rotating. Object C rolls down the ramp without slipping. Its moment of inertia is unknown.

Objects A, B, and C are made of different materials, thus the coefficients of friction between the objects and their coffesponding ramps are not necessarily the same.

Object A reaches the bottom of its ramp first, followed by objects B and C, which reach the bottom at the same instant.

a. Rank the objects according to magnitude of center-of-mass acceleration,from largest to smallest. If any objects have the same magnitude center-of-mass acceleration, state so explicitly. Explain.

b. Rank the net forces exerted on the three objects according to magnitude, from largest to smallest. If the net force on arly two objects is the same, state so explicitly. Explain.

c. In the spaces provided, draw and label a (point) free- body diagram for each object.

Free-body diagraur for object i\

F-ree-bocly dia*uram

for object B Free-body diagram

for object C

d. Rank the frictional forces exerted on the three objects according to magnitude, from largest to smallest. If the magnitude of the frictional force is the same on any two objects, state so explicitly. Explain your reasoning.

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All three objects are relea.sed

 

 

Mech Dynamics of rigiilbodies HW:72

5. Energy analysis of falling-spools experiment

The modified Atwood’s machine shown at right consists of two identical spools connected by a massless, ‘ ” inextensible thread that runs over an ideal pulley. The thread is wrapped around spool A many times, but it is attached to a fixed point on spool B, so that spool B will not rotate.

The spools are released from rest from the same height at the same instant.

a. In tutorial, you observed the motion of the spools after they were released. Ignoring small dffirences in their nntions:

. In which direction did each spool move?

. Did spool A hit the ground before, after, or at the same instant as spool B?

Is the mngnitude of the center-of-nutss acceleration of spool A (while it is fallin g) greater than,less than, or equal to that of spool B? Explain.

Is the translational kinetic energy of spool A just before it hits the ground greater thsn, less than, or equal to that of spool B? Explain.

d. Is the total kinetic energy of spool A just before it hitsothe ground greater than, less than, or equal to that of spool B? Explain.

Consider the system consisting of all of these objects: spool A, spool B, the thread, the pulley, and the Eafth.

i. Explain how you can tell that the total energy of this system(i.e., Ugnv,o+ Ug^”,n * Kton,,e * K,*nr. n * Kro’ a, * K,o’ s) is constant as spools A and B fall.

b.

c.

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Dynamics of rtgtd bodies Narne Mech HW-73

Suppose that this system starts with Ugo”, o = f/srr, B = 9 J. Just before the spools hit the ground, which is where the zero for gravitational potential energy is chosen, spool A has translational kinetic energy K,on,”a = 4 J. Determine the value of the rotational kinetic energy of spool A at this instant. Show your work.

6. A third identical spool, spool C, is added to the falling-spools experiment described in the preceding problem.

As above, all spools are released from rest from the same height at the same instant. Spool C is not in contact with any other objects as it falls.

a, Rank the spools according to magnitude of center -of-mas s acc eleration (while falling), from largest to smallest. If any spools have the same center-of-mass acceleration, state so explicitly. Explain.

b. As in the preceding problem, suppose that Ugrrr. n = Ugr,, s = 9 J before the spools are released. Just before the spools hit the ground, which is where the zero for gravitational potential energy is chosen, spool A has trarrslational kinetic energy K,*,, e,= 4J.

i. Rank the spools according to maximum tanslatianal kinetic energy, from largest to smallest. If any spools have the same maximrrtn translational kinetic energy, state so explicitly. Explain. (Use the definition K*, = lma”^’.)

ii. Rank the spools according to maximum total kinetic energy, from largest to smallest. If any spools have the same maximum total kinetic energy, state so explicitly. Explain.

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CONSERVATION OF ANGULARMOMENTUM

Name Mech HW-77

1. In tutorial, you observed the following three experiments involving a student sitting at rest on a stool, holding a spinning bicycle wheel as shown at right:

Experiment 1: The student places his arm against the side of the wheel, slowing it to half its initial angular speed.

Experiment 2: The student places his arm against the side of the wheel, bringing it to a stop.

Experiment 3: The student quickly flips the wheel over (so that it is spinning clockwise when viewed from above, with the same angular speed it had initially).

Student initially at rest

Initial sense of wheel’s

rotation

a. You observed that the final angular speed of the student in experiment 3 is greater than that in experiment2, Account for this result using the ideas developed in the tutorial.

b. Rank the experiments according to final kinetic energy of the wheel, from largest to smallest. If the final kinetic energy of the wheel is the same in any two experiments, state so explicitly. (Hint: Can kinetic energy ever be negative?) Explain.

Rank the experiments according to final kinetic energy of the student, from largest to smallest. If the final kinetic energy of the student is the same in any two experiments, state soexplicitly. Explain.

d. Rank the followingfour quantities from largest to smallest: the initial kinetic energy of the wheel (K*i) and the final kinetic energy of the student-wheel system in experiments 1 ,2, and 3 (Kstr, etc.). Explain. (Hint: It may be helpful to think about changes in energy other than mechanical energy.)

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HW-78 Conseruation of angular momentum

Z. The diagram below illustrates four hypothetical collisions that take placg on a level, frictionless surface.- The collisions are shown from a top-view perspective. All pucks are identical. If a linear or angular velocity is not specified, it is zero. If distances appear to be equal, assume that they are.

For each hypothetical collision:

a. Specify the direction of the angular momentum of the rod-puck(s) system with respect to the center of the rod both before and after the collision. If necessary, use the convention that a vector into the page is represented by the symbol I and a vector out of the page is represented by the symbol O.

b, Specify the direction of the linear momentum of the rod-puck(s) system, both before and after the collision.

c. On the basis of your answers above, state whether each hypothetical collision could ot could not occrfi. If a particular hypothetical collision could not occur, state whether it violates (l) the principle of conservation of linear momentum, (2) the principle of conservation of angular momentum, or (3) both.

Before collision

Case I

Case 2

Case 3

Case 4

Ball sticks to

rod

Sense of rotation

After collision

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Medical ICD-PCS Coding

M132 Module 02 Coding Assignment

Find the correct code and explain your rationale for each case study below.

1. Case Study:

PREOPERATIVE DIAGNOSIS:

1. Gangrene right foot.

POSTOPERATIVE DIAGNOSIS:

1. Gangrene right foot.

OPERATION:

1. Right below the knee amputation.

ANESTHESIA: General LMA.

PROCEDURE: The patient was brought to the operative suite where a general LMA anesthesia was induced.

A Foley catheter was inserted. The right foot was s secluded in an isolation bag and the right lower extremity circumferentially prepped and draped in its entirety. Beginning on the right side the skin was marked with a marking pen 4 fingerbreadths below the tibial tuberosity anteriorly with a long posterior flap. The skin was incised circumferentially and the anterior musculature sharply divided exposing the tibia The tibia was cleaned with a periosteal elevator and then transected with the Stryker saw. The fibula was exposed and transected with the bone cutter and the amputation completed by sharply incising the posterior musculature. Bleeding vessels were ligated with 2-0 silk Ligature. There appeared to be adequate bleeding at this level for primary healing. The tibia was then cleaned with a bone rasp and the fibula with a rongeur. The wound was irrigated and ultimately closed without significant tension utilizing interrupted 2-0 vicryl sutures for reapproximation of the fascia and skin staples for reapproximation of the skin.

The right side was dressed with sterile gauze fluff dressings and a Kerlix roll. Estimated blood loss throughout the procedure was approximately 150 mL. The patient received one unit intraoperatively of packed cells because of preoperative anemia. She was transported in stable condition to the recovery room.

Code: Click here to enter text.

2. Case Study:

PROCEDURE: Open reduction and internal fixation of bilateral tibial plateau fractures.

INDICATIONS: This 23-year old was involved in a serious accident and sustained bilateral tibial plateau fractures

DESCRIPTION OF OPERATION:  The patient was brought to the operating room and placed on the operating room table in the supine position. General anesthesia was induced, and after this both lower extremities were prepped and draped in the usual sterile fashion. Attention was first directed towards the left tibial plateau. A standard lateral procedure to reduce the lateral tibial plateau fracture was performed. After a submeniscal arthrotomy was performed, the joint was visualized via the lateral approach. The posterolateral fragments were reduced and the lateral tibial plateau was elevated, restoring the articular surface. K-wires were placed to provisionally hold this reduction. C-arm fluoroscopy was used to confirm good reduction of the joint surface. Next, a 6-hole lateral plateau locking plate from the Stryker sets was selected. This locking plate was advanced down the tibial shaft. Screws were placed to secre the plate to the bone. Four screws were placed in the distal shaft fragments and 4 locking screws in the proximal fragment. A kickstand screw was also placed in the locking mode. After all screws were placed, x-rays exhibited good reduction of the fracture, as well as good placement of all hardware. Next, the wound was thoroughly irrigated with normal saline. The meniscal arthrotomy was closed with the 0 PDS suture, including the capsule. Next, the IT band was closed with 0 Vicryl suture, followed by 2-0 Vicryl sutures for the skin and staples. Attention was then directed toward the right tibial plateau. A similar procedure was performed on the right side. Then, the lateral approach to the lateral tibial plateau was performed, exposing the fracture. The incision was approximately 4 cm on the right side. A 6-hole LISS plate was advanced down the tibial shaft. Four screws were placed in the distal fragments followed by four screws in the locking mode and proximal metaphyseal fragment. Excellent fixation was obtained. The C-arm fluoroscopy was used to confirm excellent reduction of the fracture on both the AP and lateral fluoroscopic images. Next, the wound was thoroughly irrigated and closed in layers. Sterile dressings were applied All wounds were dressed with sterile dressing and the patient was placed into knee immobilizers. The patient was then awakened from anesthesia, and transferred to recovery. The patient will be nonweightbearing for approximately three months on bilateral lower extremities. The patient will receive DVT prophylaxis during this time.

ICD-10-PCS Code: Click here to enter text.

3. Case Study:

PREOPERATIVE DIAGNOSES:

1. Pelvic pain.

2. History of previous pelvic surgery and ovarian cyst.

POSTOPERATIVE DIAGNOSES:

1. Pelvic pain.

2. History of previous pelvic surgery and ovarian cyst.

OPERATION PERFORMED:  Laparoscopic adhesiolysis.

SURGEON:  Susan Smith, MD

ANESTHESIA:  General endotracheal.

ESTIMATED BLOOD LOSS:  Less than 10 mL.

URINE OUTPUT:  70 mL.

IV FLUIDS:  750 mL.

DESCRIPTION OF OPERATION:  After informed consent was obtained, the patient was taken to the operating room. She was placed in the dorsal supine position and general anesthesia was induced and prepped and draped in the usual sterile fashion. A Foley catheter was placed to gravity and speculum was placed in the posterior and anterior vagina and the cervix was grasped with a single-toothed tenaculum. A Hulka clamp was then inserted through the cervix into the uterus for uterine manipulations and the tenaculum was removed and attention was then turned to the abdomen.

A supraumbilical incision was made with a scalpel and elevated up with towel clamps. A long Veress needle was then placed and CO2 gas was used to insufflate the abdomen and pelvis. A 10-12 trocar and sleeve were then placed and confirmed via the laparoscope. The dense greater omental adhesions to the anterior abdominal wall were noted immediately. At this time, we were not able to see into the pelvic region. A second 5 mm trocar and sleeve were placed in the left mid quadrant under direct visualization. The ligature device was then placed developing a plane between the omentum and the anterior abdominal wall.

The adhesiolysis took place and it took approximately 25 minutes to release all of the omental adhesions from the anterior abdominal wall. We were then able to visualize the pelvis and a blunt probe was placed through the port. The ovary was visualized and photos were taken with no evidence of any ovarian cyst or ovarian pathology or of pelvic endometriosis. The uterus also appeared normal and the left tube and ovary were surgically absent. The appendix was easily visualized and noted to be noninflamed, normal in appearance, and there were no adhesions in the right lower quadrant. The upper abdominal exam was unremarkable. The procedure was terminated at this time. The ports were removed. CO2 gas was allowed to escape. The incisions were closed with 4-0 Vicryl suture. The Hulka clamp was removed. The vagina was noted to be hemostatic. The patient’s anesthesia was awakened from anesthesia, the Foley catheter was removed, and she was taken in stable condition to the recovery room.

ICD-10-PCS Code:

 
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Phet Light And Sound Wave Simulation Answer Guide

SET 110

Light and Sound Wave Simulation

Name:______________________________________________

Click on the “Wave Interference” simulator and click “Run Now”

A. Sound Wave Simulator

Click on the top tab that says sound.

 

Press the Particle button

1)What happens when you press the particle button?

2)What happens to the particles as you alter the frequency and amplitude?

Click show graph.

3)What happens to the graph as frequency and amplitude are changed?

Click on grayscale.

4) What happens when you rotate the view?

Turn the speaker off.

5)What happens when you turn the speaker off?

Turn the speaker back on after all the particles become still.

Turn on the sound.

6) What happens to the sound as you vary the frequency and amplitude?

Put the sliders for frequency and amplitude in the middle.

Add a barrier

7) What happens if you add a barrier?

8) What if you add two barriers?

Light Wave Simulator

Click the top tab that says “light”

Click “show graph” and “add detector”

Vary the amplitude

  1.  What happens when you vary the amplitude?

   2. What happens if you put the amplitude at 0?

Change the color

3) What happens to the wavelength and frequency as you change the color?

Add a slit.

Move the slit back and forth.

5. What happens as the slit gets closer to the source.

6. What happens as the slit gets further away from the source?

Change the Color

7. What effect dose changing the color have?

Set the slit width at 1050 and the barrier location at 2950. Using the target on the detector compare the amplitudes on each side of the barrier.

8)Describe what happens what happen directly behind the opening,

8. Describe what happens behind the barrier?

Click “Show Screen” and “Intensity Graph.”

Interpret and discuss what you think these represent and are showing?

Conclusions

Using the information you have observed lost your observations and conclusions.

How do you think altering factors like wavelength and frequency affect they way we see and hear?

How does this help us understand how things like lenses and walls alter the way we see certain lights and hear certain sounds?

 
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Physics Conservation Of Momentum

EQNS_E_RV.AUON OF MOMENTUM Name IN ONE DIMENSION MechHW-:

1. Two gliders, A aad B, collide on a level, frictionless track, as shown below,

The mass of glider A is less than the mass of gliderB (i.e., mr< ma). Thefinal speedof glider A is greater than the final speed of glider B (i.e., ztn > as). ”

Boforc collieion Affer collision

Is the magnitude of the final momentum of glider ,\ lFd, greater than, less than, or equal to the magnitude of the final momentum of glider B, lft/? Diaw a monrentum vector diagram to support your answer. (An example of a momentum vector diagram can be found on tha second page of the tutorial.)

a

2. Two gliders, c and D, collide on a level, frictionless track, as shown below. The mas er D (i.e., fitc 1mo). The initial speed of glider C D (i.e,, uci )opi). Arter the collision, Gliders C and D m final speed, ur.

Before collision After collision

Is the magnitude of the initial momentum of glider C_rlF./, greater than, less than, or equal to the magnitude of the initial momentum of glider D, lFD/? Draw a momentum vector diigram to support your answer.

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Mech Conseroation of momentum in one dimension HW{0

3. Two astronauts, A and B, participate in three collision experiments in a weightless, frictionless environment. In each experimenl, astronaut B is initially at rest, and astronaut A has initial momentumi*= 20 kg-m/s to the right. (The velocities of the astronauts are measured with respect to a nearby space station.)

Before After

The astronauts push on each other in different ways so that the outcome of each experiment is different. As shown in the figure at right, astronaut A has a different final momentum in each experiment.

a. Determine the magnitude of the final momentum of astronaut B in each experiment. Explain.

b. Rank the three experiments according to the final kinetic energy of astronaut B, from largest to smallest. Explain.

c. Is the totalkinetic energy after the collision in experirfient2 greater than, less than, or equal to the total kinetiCenergy after the collision in experiment 3? (In this case, total kinetic energy means the sum of the kinetic energies of the two astronauts.) Explain.

d. Consider the following statement: “The momentum of the system is conserved in eoch experiment becouse there is no net f orce on the system. ff momentum is conserved, then kinetic ene?gY must olso be conserved, becouse both momentum ond kinetic energy ore mode up of moss ond velocity.”

One of the sentences above is completely correct. Discuss the error(s) in reasoning in the other sentence.

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Experirnents 1,2, and3

At rest

 

 

Consentation of momentum in one ditnenston Name

e. In the boxes below, draw the initial mompntum, the change in momentum, andthe final momentum for each astronaut in the three experiments. The initial momentum is shown for astronaut A. Draw the other vectors using the same scale.

Mech HW-61

Initial

If the net force on a system of two colliding objecls is zero, how does the change in momentum of one object compare to the change in momentum of the other object:

. in magnitude?

. in direction?

Bxplain how your answers to part f are consistent with Newton’s third law and the impulse- momentum theorem ( 4″t N = [F) for:

. each astronaut considered separately, and

. for the system of both astronauts together.

‘,

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Mech Conseroation of momentum in one dimension HW-62

4. A pyrotechnician releases a 3-kg firecracker from rest. At t = 0.4 s, the firecracker is moving downward with speed 4 mls. At this same instant, the firecracker begins to explode into two pieces, “top” and “bottom,” with masses nt op 1 kg and ffibotto – 2kg. At the end of the explosion (/ = 0.6 s), the top piece is moving upward with speed 6 m/s.

The mass of the explosive substance is negligible in comparison to the mass of the two pieces.

The questions below can serve as a guide to completing the diagram below right:

a. Determine the magnitude of the net force on the firecracker system before the explosion. (Use I = 10 m/s’.) ExPlain.

b. Determine’the magnitude of the net force on the firecracker system at an instant during the explosion. (Hint: Does the net force on a system depend on forces that are internal to that system?)

Determine the magnitude and direction of the net impulse (\”rN) on the firecracker system during the explosion (i.e., over the interval from t =0.4 s until r = 0.6 s). Explain.

Use the impulse-momentum theorem to determine the magnitude and direction of the change in momentum of the firecracker system during the explosion. Enter this vector in the table using the scale set by the initial momentum of the system.

Determine the final momentum of the firecracker system and enter it in the table. (Hint: Is it the same as the initial momentum?)

Complete the vector diagram at right.

Initial l: 0.4 s

Final /: 0.6 s

Change Final /=0.6s

Initial /=0.4s

Prop

Pbuno*

d.

/rr*,”,n

12 kg-m/s

i_l_r._l:

m=3kg

lfrl = 4 mls

,onl= 6 m/s

rurot, = I kg

ffiborton – 2 kg

A -t”bottom – :

I !

i1tirt -Tj -t–j

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Kinematics

Kinematics Carolina Distance Learning

Investigation Manual

2

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Table of Contents

Overview …………………………………………………………………………………………… 3

Objectives …………………………………………………………………………………………. 3

Time Requirements ……………………………………………………………………………. 3

Background ………………………………………………………………………………………. 4

Materials ……………………………………………………………………………………………. 8

Safety ………………………………………………………………………………………………… 9

Alternate Methods for Collecting Data using Digital Devices. ……….. 10

Preparation ……………………………………………………………………………………… 11

Activity 1: Graph and interpret motion data of a moving object ….. 11

Activity 2: Calculate the velocity of a moving object ……………………. 12

Activity 3: Graph the motion of an object traveling under constant

acceleration ……………………………………………………………………………………. 16

Activity 4: Predict the time for a steel sphere to roll down an incline 23

Activity 5: Demonstrate that a sphere rolling down the incline is

moving under constant acceleration …………………………………………….. 26

3

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Overview

Kinematics is the branch of physics that deals with the analysis of the motion of objects

wihout concern for the forces causing the motion. Scientists have developed

equations that describe the movement of objects within certain parameters, such as

objects moving with a constant velocity or a constant acceleration. Using these

equations, the future position and velocity of an object can be predicted. This

investigation will focus on objects moving with a constant velocity or a constant

acceleration. Data will be collected on these objects, and the motion of the objects

will be analyzed through graphing these data.

Objectives

 Explain linear motion for objects traveling with a constant velocity or constant

acceleration

 Utilize vector quantities such as displacement and acceleration, and scalar

quantities such as distance and speed.

 Analyze graphs that depict the motion of objects moving at a constant velocity

or constant acceleration.

 Use equations of motion to analyze and predict the motion of objects moving at

a constant velocity or constant acceleration.

Time Requirements

Preparation …………………………………………………………………………………5 minutes

Activity 1 …………………………………………………………………………………….15 minutes

Activity 2 …………………………………………………………………………………….20 minutes

Activity 3 …………………………………………………………………………………….20 minutes

Activity 4 …………………………………………………………………………………….10 minutes

Activity 5 …………………………………………………………………………………….20 minutes

4

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Background

Mechanics is the branch of physics that that studies the motion of objects and the

forces and energies that affect those motions. Classical Mechanics refers to the motion

of objects that are large compared to subatomic particles and slow compared to the

speed of light. The effects of quantum mechanics and relativity are negligible in

classical mechanics. Most objects and forces encountered in daily life can be

described by classical mechanics, such as the motion of a baseball, a train, or even a

bullet or the planets. Engineers and other scientists apply the principles of physics in

many scenarios. Physicists and engineers often collect data about an object and use

graphs of the data to describe the motion of objects.

Kinematics is a specific branch of mechanics that describes the motion of objects

without reference to the forces causing the motion. Examples of kinematics include

describing the motion of a race car moving on a track or an apple falling from a tree,

but only in terms of the object’s position, velocity, acceleration, and time without

describing the force from the engine of the car, the friction between the tires and the

track, or the gravity pulling the apple. For example, it is possible to predict the time it

would take for an object dropped from the roof of a building to fall to the ground using

the following kinematics equation:

𝒔 = 1

2 𝒂 𝑡2

Where s is the displacement from the starting position at a given time, a is the

acceleration of the object, and t is the time after the object is dropped. The equation

does not include any variables for the forces acting on the object or the mass or energy

of the object. As long as the some initial conditions are known, such an object’s

position, acceleration, and velocity at a given time, the motion or position of the object

at any future or previous time can be calculated by applying kinematics. This method

has many useful applications. One could calculate the path of a projectile such as a

golf ball or artillery shell, the time or distance for a decelerating object to come to rest,

or the speed an object would be traveling after falling a given distance.

Early scientists such as Galileo Galilee (1564-1642), Isaac Newton (1642-1746) and

Johannes Kepler (1571-1630) studied the motion of objects and developed

mathematical laws to describe and predict their motion. Until the late sixteenth

century, the idea that heavier objects fell faster than lighter objects was widely

accepted. This idea had been proposed by the Greek philosopher Aristotle, who lived

around the third century B.C. Because the idea seemed to be supported by

experience, it was generally accepted. A person watching a feather and a hammer

dropped simultaneously from the same height would certainly observe the hammer

falling faster than the feather. According to legend, Galileo Galilee, an Italian physicist

and mathematician, disproved this idea in a dramatic demonstration by dropping

objects of different mass from the tower of Pisa to demonstrate that they fell at the

same rate. In later experiments, Galileo rolled spheres down inclined planes to slow

5

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down the motion and get more accurate data. By analyzing the ordinary motion of

objects and graphing the results, it is possible to derive some simple equations that

predict their motion.

To study the motion of objects, a few definitions should be established. A vector refers

to a number with a direction and magnitude (or size). Numbers that have a magnitude

but not a direction are referred to as a scalar. In kinematics, vectors are important,

because the goal is to calculate the location and direction of movement of the object

at any time in the future or past. For example if an object is described as being 100

miles from a given position traveling at a speed of 50 miles per hour, that could mean

the object will reach the position in 2 hours. It could also mean the object could be

located up to 100 miles farther away in 1 hour, or somewhere between 100 and 200

miles away depending on the direction. The quantity speed, which refers to the rate of

change in position of an object, is a scalar quantity because no direction of travel is

defined. The quantity velocity, which refers to both the speed and direction of an

object, is a vector quantity.

Distance, or the amount of space between two objects, is a scalar quantity.

Displacement, which is distance in a given direction, is a vector quantity. If a bus

travels from Washington D.C. to New York City, the distance the bus traveled is

approximately 230 miles. The displacement of the bus is (roughly) 230 miles North-East.

If the bus travels from D.C to New York and back, the distance traveled is roughly 460

miles, but the displacement is zero because the bus begins and ends at the same point.

It is important to define the units of scalar and vector quantities when studying

mechanics. A person giving directions from Washington D.C. to New York might

describe the distance as being approximately 4 hours. This may be close to the actual

travel time, but this does not indicate actual distance.

To illustrate the difference between distance and displacement, consider the following

diagrams in Figures 1-3.

Consider the number line in Figure 1. The displacement from zero represented by the

arrowhead on the number line is -3, indicating both direction and magnitude. The

distance from zero indicated by the point on the number line equals three, which is the

magnitude of the displacement. For motion in one dimension, the + or‒ sign is sufficient

to represent the direction of the vector.

Figure 1.

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Figure 2 Figure 3

The arrows in Figures 2 and 3 represent displacement vectors for an object. The long

lines represent a displacement with a magnitude of five. This displacement vector can

be resolved into two component vectors along the x and y axes. In all four diagrams

the object is moved some distance in either the positive or negative x direction, and

then some distance in the positive y direction; however, the final position of the object

is different in each diagram. The total distance between the object’s initial and final

position in each instance is 5 meters, however to describe the displacement, s, from the

initial position more information is needed.

In Figure 2, the displacement vector can be given by 5 meters (m) at 53.1°. This vector

is found by vector addition of the two component vectors, 3 m at 0° and 4 m at 90°, using conventional polar coordinates that assign 0° to the positive x direction and

progress counterclockwise towards 360°. The displacement in Figure 3 is 5 m at 143.1°.

In each case the magnitude of the vector is length of the arrow, that is, the distance

that the object travels. Most texts will indicate that a variable represents a vector

quantity by placing an arrow over the variable or placing the variable in bold.

To indicate the magnitude of a vector, absolute value bars are used. For example the

magnitude of the displacement vector in each diagram is 5 m. In Figure 2 the

displacement is given by:

s = 5 m at 51.3°

The magnitude of this vector may be written as:

| 𝒔 | = d = 5 m

The displacement vector in Figure 2, s = 5 m at 53.1°, can be resolved into the

component vectors 3 m at 0° and 4 m at 90°.

Two more terms that are critical for the study of kinematics are velocity and

acceleration. Both terms are vector quantities.

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Velocity (v) is defined as the rate of change of the position of an object. For an object

moving in the x direction, the magnitude of the velocity (speed) may be described as:

 

𝒗 = 𝑥2 − 𝑥1

∆𝑡

Where x2 is the position at time t2 and x1 is the position of the object at time t1. The

variable ∆t represents the time interval t2 -t1. The symbol, ∆, is the Greek symbol delta,

and refers to a change or difference. ∆t is read, “delta t”. Time in the following

examples is provided in seconds (s). Please be sure that you do not confuse the “s” unit

for seconds, and the “s “ unit for displacement in these formulas.

For example if an object is located at a position designated x1 = 2 m and moves to

position x2 = 8 m over a time interval ∆t = 2 s, then the average speed could be

calculated: 8𝑚 − 2𝑚

2s = 3𝑚/s

The velocity could for this object could be indicated as:

𝒗 = 3 𝑚/s

Because velocity is a vector quantity, the positive sign indicates that the object was

traveling in the positive x direction, at a speed of 3 m/s.

Acceleration is defined as the rate of change of velocity. The magnitude of

acceleration may be described as:

𝒂 = 𝒗𝟐 − 𝒗1

∆𝑡

For example, an object with an initial velocity v1 = 10 m/s slows to a final velocity of v2 =

1 m/s over an interval of 3 s.

1 𝑚

s⁄ − 10 𝑚 𝑠⁄

3s = −3 𝑚

s s⁄⁄

The object has an average acceleration of ‒3 meters per second per second, which

can also be written as ‒3 meters per second squared, or ‒3 𝑚 s2⁄ .

Because only the initial and final positions or velocities over a given time interval are

used in these equations, the calculated values indicate the average velocity or

acceleration. Calculating the instantaneous velocity or acceleration of an object

requires the application of calculus. Only average velocity and acceleration are

considered in this investigation.

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Materials

Included in the Central Materials kit:

Tape Measure

Rubber Bands

Protractor

Included in the Mechanics Module materials kit

Constant Velocity Vehicle

Steel Sphere

Acrylic Sphere

Angle Bar

Foam Board

Block of Clay

Needed, but not supplied:

Scientific or Graphing Calculator

or Computer with Spreadsheet Software

Permanent Marker

Masking Tape

Stopwatch, or smartphone able to record

video

Reorder Information: Replacement supplies for the Kinematics investigation can be

ordered from Carolina Biological Supply Company, kit 580404 Mechanics Module.

Call 1-800-334-5551 to order.

 

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Safety

Safety goggles should be worn while conducting this investigation.

Read all the instructions for this laboratory activity before beginning. Follow the

instructions closely and observe established laboratory safety practices.

Do not eat, drink, or chew gum while performing this activity. Wash your hands with

soap and water before and after performing the activity. Clean up the work area

with soap and water after completing the investigation. Keep pets and children

away from lab materials and equipment.

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Alternate Methods for Collecting Data using Digital Devices.

Much of the uncertainty in these experiments arises from human error in measuring the

times of events. Some of the time intervals are very short, which increases the effect of

human error due to reaction time.

Observing the experiment from a good vantage point that removes parallax errors and

recording measurements for multiple trials helps to minimize error, but using a digital

device as an alternate method of data collection may further minimize error.

Many digital devices, smart phones, tablets, etc. have cameras and software that

allow the user to pause or slow down the video.

If you film the experiment against a scale, such as a tape measure, you can use your

video playback program to record position and time data for the carts. This can

provide more accurate data and may eliminate the need for multiple trials.

If the time on your device’s playback program is not sufficiently accurate, some

additional apps may be available for download.

Another option is to upload the video to your computer. Different video playback

programs may come with your operating system or software suite or may be available

for download.

Some apps for mobile devices and computer programs available for download are

listed below, with notes about their features.

Hudl Technique: http://get.hudl.com/products/technique/

 iPhone/iPad and Android

 FREE

 Measures times to the hundredth-second with slow motion features

QuickTime http://www.apple.com/quicktime/download/

 Free

 Install on computer

 30 frames per second

 Has auto scrubbing capability

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Preparation

1. Collect materials needed for this investigation.

2. Locate and clear an area of level floor space in order to conduct the constant

velocity experiment. The space should be free of obstruction and three to four

meters long with a surface which will allow the vehicle to maintain traction but not

impede the vehicle.

Activity 1: Graph and interpret motion data of a moving object

One way to analyze the motion of an object is to graph the position and time data.

The graph of an object’s motion can be interpreted and used to predict the object’s

position at a future time or calculate an object’s position at a previous time.

Table 1 represents the position of a train on a track. The train can only move in one

dimension, either forward (the positive x direction) or in reverse (the negative x

direction).

Table 1

Time (x-axis), seconds Position (y-axis), meters

0 0

5 20

10 40

15 50

20 55

30 60

35 70

40 70

45 70

50 55

1. Plot the data from Table 1 on a graph using the y-axis to represent the displacement

from the starting position (y = 0) and the time coordinate on the x-axis.

2. Connect all the coordinates on the graph with straight lines.

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Activity 2: Calculate the velocity of a moving object

In this activity you will graph the motion of an object moving with a constant velocity.

The speed of the object can be calculated by allowing the Constant Velocity Vehicle

to travel a given distance and measuring the time that it took to move this distance. As

seen in Activity 1, this measurement will only provide the average speed. In this activity,

you will collect time data at several travel distances, plot these data, and analyze the

graph

1. Find and clear a straight path approximately two meters long.

2. Install the batteries and test the vehicle.

3. Use your tape measure or ruler to measure a track two meters long. The track should

be level and smooth with no obstructions. Make sure the surface of the track

provides enough traction for the wheels to turn without slipping.

Place masking tape across the track at 25 cm intervals.

4. Set the car on the floor approximately 5 cm behind the start point of the track.

5. Set the stopwatch to the timing mode and reset the time to zero.

6. Start the car and allow the car to move along the track.

7. Start the stopwatch when the front edge of the car crosses the start point.

8. Stop the stopwatch when the front edge of the car crosses the first 25 cm point.

9. Recover the car, and switch the power off. Record the time and vehicle position on

the data table.

10. Repeat steps #5‒9 for each 25 cm interval marked. Each trial will have a distance

that is 25 cm longer than the previous trial, and the stopwatch will record the time

for the car to travel the individual trial distance.

11. Record the data in Data Table 1.

Note: The vehicle should be able to travel two meters in a generally straight path. If

the vehicle veers significantly to one side, you may need to allow the vehicle to

travel next to a wall. The friction will affect the vehicle’s speed, but the effect will be

uniform for each trial.

Note: Starting the car a short distance before the start point allows the vehicle to

reach its top speed before the time starts and prevents the short period of

acceleration from affecting the data.

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Data Table 1

Time (s) Displacement (m)

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

2.00

12. Graph the time and displacement data points on graph paper.

13. Draw a line of best fit through the data points.

14. Calculate the slope of the line.

15. Make a second data table, indicating the velocity of the car at any time.

Data Table 2

Time (s) Velocity (m/s)

1

2

3

4

5

6

7

8

Note: The points should generally fall in a straight line. If you have access to a

graphing calculator or a computer with spreadsheet software, the calculator or

spreadsheet can be programmed to draw the line of best fit, or trend line.

Note: Based on the equation of a line that cross the y-axis at y = 0, the slope of the

line, m, will be the velocity of the object. 𝑦 = 𝑚𝑥 𝑑 = 𝑣∆𝑡

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16. Graph the data points from the Data Table 2 on a second sheet of graph paper.

Label the y-axis Velocity and the x-axis Time.

17. Draw a vertical line from the x-axis at the point time = 2 seconds so that it intersects

the line representing the velocity of the car.

18. Draw a second vertical line from the x-axis at the point time = 4 seconds so that it

intersects the line representing the velocity of the car.

19. Calculate the area represented by the rectangle enclosed by the two vertical lines

you just drew, the line for the velocity of the car, and the x-axis. An example is shown

as the blue shaded area in Figure 4.

 

Figure 4

Note: Because the object in this example, the battery-powered car, moves with a

constant speed, all the values for the velocity of the car in the second table should

be the same. The value of the velocity for the car should be the slope of the line in

the previous graph.

Note: When the data points from this table are plotted on the second graph, the

motion of the car should generate a horizontal line. On a velocity vs. time graph, an

object moving with a constant speed is represented by a horizontal line.

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Note: In order to calculate the area of this rectangle, you must multiply the value

for the time interval between time t=2 s and time t=4 s, by the velocity of the car.

This area represents the distance traveled by the object during this time interval.

This technique is often referred to as calculating the “area under the curve”. The

graph of velocity vs. time for an object that is traveling with a constant

acceleration will not be a horizontal line, but using the same method of graphing

the velocity vs. time and finding the “area under the curve” in a given time

interval can allow the distance traveled by the object to be calculated.

Distance = velocity × time

In this equation, the time units (s) cancel out when velocity and time are

multiplied, leaving the distance unit in meters.

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Activity 3: Graph the motion of an object traveling under constant

acceleration

Collecting data on freefalling objects requires accurate timing instruments or access to

a building with heights of several meters where objects can safely be dropped over

heights large enough to allow accurate measurement with a stopwatch. To collect

usable data, in this activity you will record the time objects to roll down an incline. This

reduces acceleration to make it easier to record accurate data on the distance that

an object moves.

1. Collect the following materials:

Steel Sphere

Acrylic Sphere

Angle Bar

Clay

Tape Measure

Timing Device

Protractor

2. Use the permanent marker and the tape measure to mark the inside of the angle

bar at 1-cm increments.

3. Use the piece of clay and the protractor to set up the angle bar at an incline

between 5° to 10°. Use the clay to set the higher end of the anglebar and to

stabilize the system. (Figure 5)

Figure 5

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Set up the angle bar so that the lower end terminates against a book or a wall, to stop

the motion of the sphere (Figure 6.)

Figure 6

4. Place the steel sphere 10 cm from the lower end of the track.

5. Release the steel sphere and record the time it takes for the sphere to reach the

end of the track.

6. Repeat steps #4‒5 two more times for a total of three measurements at a starting

point of 10 cm.

7. Repeat steps #4‒6, increasing the distance between the starting point and the end

of the track by 10 cm each time.

8. Record your data in Data Table 3.

Note: You are recording the time it takes for the sphere to accelerate over an

increasing distance. Take three measurements for each distance, and average the

time for that distance. Record the time for each attempt and the average time in

Table 4.

 

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Data Table 3

Time (s) Average time (s) Average Time 2 (s2) Distance (m)

Trial 1 =

0.1 Trial 2 =

Trial 3 =

Trial 1 = 0.2

Trial 2 =

Trial 3 =

Trial 1 =

0.3 Trial 2 =

Trial 3 =

Trial 1 =

0.4 Trial 2 =

Trial 3 =

Trial 1 =

0.5 Trial 2 =

Trial 3 =

Trial 1 =

0.6 Trial 2 =

Trial 3 =

Trial 1 =

0.7 Trial 2 =

Trial 3 =

Trial 1 =

0.8 Trial 2 =

Trial 3 =

9. Calculate the average time for each distance and record this value in Table 4.

10. Create a graph of distance vs. time using the data from Table 4.

11. Complete Table 4 by calculating the square of the average time for each distance.

12. Create a graph of displacement vs. time squared from the data in Table 4.

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Graphing the displacement vs time data from Table 4 will generate a parabola.

When data points generate a parabola, it means the y value is proportional to the

square of the x value, or:

𝒚 ∝ 𝑥2

That means the equation for a line that fits all the data points looks like:

𝑦 = 𝐴𝑥2 + 𝐵𝑥 + 𝐶.

In our experiment, the y-axis is displacement and the x-axis is time-; therefore

displacement is proportional to the time squared:

𝒔 ∝ 𝑡2

So, we can exchange y in the equation with displacement (s), to give a formula that

looks like:

𝒔 = 𝐴𝑡2 + 𝐵𝑡 + 𝐶.

We would know the displacement s, at any time t. We just need to find the

constants, A, B, and C.

The equation that describes the displacement of an object moving

with a constant acceleration is one of the kinematics equations:

𝒔 = 1

2 𝒂∆𝑡2 + 𝒗𝟏∆𝑡

The following section describes how to find this equation using the same method of

finding the “area under the curve” covered in Activity 2.

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Finding an Equation for the Motion of an Object with Constant Acceleration

The general form of a line is: 𝑦 = 𝑚𝑥 + 𝑏

 

Whre m is the slope of the line, and b is the y-intercept, the point where the line

crosses the y-axis. Because the first data point represents time zero and

displacement zero, the y-intercept is zero and the equation for the line simplifies

to:

y = mx

The data collected in Activity 3 showed that:

𝒔 ∝ 𝑡2

This means that the displacement for the object that rolls down an inclined plane

is can be represented mathematically as:

𝒔 = 𝑘𝑡2 + c

 

Wher k is an unknown constant representing the slope of the line, and c is an

unknown constant representing the y-intercept.

The displacement of the sphere as it rolls down the incline can be calculated

using this equation, if the constants k and c can be found.

 

Further experimentation indicates that the constant k for an object in freefall is

one-half the acceleration. If the object is released from rest, the constant c will

be zero.

So for an object that is released from rest, falling under the constant

acceleration due to gravity, the displacement from the point of release is given

by:

𝒔 = 1

2 𝒂𝑡2

Where s is the displacement, t is the time of freefall, and 𝒂 is the acceleration.

For objects in freefall near Earth’s Surface the acceleration due to gravity has a

value of 9.8 𝑚 s2⁄ .

Another way to derive this equation, and find the values for k and c, is to

consider the velocity vs. time graph for an object moving with a constant

acceleration. Remember the velocity vs. time graph for the object moving with

constant velocity from Activity 2. If velocity is constant, the equation of that

graph would be: 𝒗 = 𝑘

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Where v represents the velocity, plotted on the y-axis, and k is the constant

value of the velocity. Plotted against time on the x-axis, this graph is a horizontal

line, as depicted in Figure 7.

Figure 7

By definition, the shaded area is the distance traveled by the object during the

time interval: Δ𝑡 = 𝑡2 − 𝑡1

𝒗 = 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕

𝑡𝑖𝑚𝑒 =

𝒔

∆𝑡

∴ 𝒔 = 𝒗∆𝑡

If an object has a constant acceleration, then by definition:

 

𝒂 = 𝒗𝟐 − 𝒗1

∆𝑡

Or : 𝒗2 = 𝒂∆𝑡 + 𝒗𝟏

 

This is equation is in the general form of a line y = mx + b, with velocity on the y-

axis and time on the x-axis. The graph of this equation would look like the graph

in Figure 8.

 

Figure 8

Similar to how the shaded area A1 in Figure 7 represents the distance traveled by

the object during the time interval Δt = t2 – t1, the shaded area A2 combined with

 

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A1 equals the distance traveled by the object undergoing constant

acceleration

The area A1 can be given by:

𝐴1 = 𝒗𝟏∆𝑡

 

The area A2 can be given by:

𝐴2 = 1

2 (𝒗2 − 𝒗𝟏)∆𝑡

 

Because this is the area of the triangle, where the length of the base is Δt and the

height of the triangle is (𝒗𝟐 − 𝒗𝟏),

 

Adding these two expressions and rearranging:

 

𝒔 = 1

2 (𝒗2 − 𝒗𝟏)∆𝑡

 

And substituting: 𝒗2 = 𝒂∆𝑡 + 𝒗𝟏

Gives this equation:

𝒔 = 1

2 (𝒂Δ𝑡 + 𝑣1 + 𝑣2Δ𝑡 + 𝑣1Δ𝑡)

Simplifying gives:

 

𝒔 = 1

2 𝒂∆𝑡2 + 𝒗𝟏∆𝑡

This equation gives the theoretical displacement for an object undergoing a

constant acceleration, 𝒂, at any time t, where s is the displacement during the

time interval, Δ𝑡, and v1 is the initial velocity.

If the object is released from rest, as in our experiment, v1 = 0 and the equation

simplifies to:

𝒔 = 1

2 𝒂∆𝑡2

 

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Activity 4: Predict the time for a steel sphere to roll down an incline

In this activity you will use the kinematics equation:

 

𝒔 = 1

2 𝒂∆𝑡2

This will allow you to predict how long the sphere will take to roll down the

inclined track.

First you must solve the previous equation for time:

 

𝑡 = √ 2𝒔

𝒂

If the object in our experiment was in freefall you would just need to substitute

the distance it was falling for s and substitute the acceleration due to Earth’s

gravity for 𝒂, which is

g = 9.8 m/s2

In this experiment, however the object is not undergoing freefall, it is rolling down

an incline.

The acceleration of an object sliding, without friction down an incline is given by:

 

𝒂 = gSINθ

Where θ is the angle between the horizontal plane (the surface of your table)

and the inclined plane (the track), and g is the acceleration due to Earth’s

gravity.

When a solid sphere is rolling down an incline the acceleration is given by:

 

𝒂 = 0.71 gSINθ

The SIN (trigonometric sine) of an angle can be found by measuring the angle

with a protractor and using the SIN function on your calculator or simply by

dividing the length of the side opposite the angle (the height from which the

sphere starts) by the length of the hypotenuse of the right triangle (the length of

the track). Figure 9 shows the formula for deriving sines from triangles

Note: Read the following section carefully.

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Figure 9

 

sin 𝜃 = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒

ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒

 

Activity 4: Procedure

 

1. Set up the angle bar as a track. Measure the length of the track and the angle of

elevation between the track and the table.

2. Rearrange the kinematics equation to solve for time (second equation), and

substitute the value 0.71 g SINθ for 𝒂 (third equation). Use a distance of 80 cm for s.

 

𝒔 = 1

2 𝒂∆𝑡2

𝑡 = √ 2𝒔

𝒂

𝑡 = √ 2𝒔

(0.71𝐠 SINθ)

 

3. Release the steel sphere from the start point at the elevated end of the track and

measure the time it takes for the sphere to roll from position s = 0 to a final position s

= 80 cm.

4. Compare the measured value with the value predicted in Step 2. Calculate the

percent difference between these two numbers.

 

 

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5. Repeat Activity 4 with the acrylic sphere. What effect does the mass of the sphere

have on the acceleration of the object due to gravity?

 

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Activity 5: Demonstrate that a sphere rolling down the incline is

moving under constant acceleration

 

1. Collect the piece of foam board. Use a ruler and a pencil to draw lines across the

short dimension (width) of the board at 5 cm increments.

2. Collect rubber bands from the central materials kit. Wrap the rubber bands around

the width of the foam board so that the rubber bands line up with the pencil marks

you made at the 5 cm intervals. See Figure 10, left panel.

3. Use a book to prop up the foam board as an inclined plane at an angle from 5° to

10° from the horizontal.

4. Place the steel sphere at the top of the ramp and allow the sphere to roll down the

foam board.

5. Remove the rubber bands from the foam board.

6. On the reverse side of the foam board, use a pencil to mark a line across the short

dimension of the board 2 cm from the end. Label this line zero. Mark lines at the

distances listed in Table 5. Each measurement should be made from the zero line.

(see Figure 10).

Note: The sound as the steel sphere crosses the rubber bands will increase in

frequency as the steel sphere rolls down the ramp, indicating that the sphere is

accelerating. As the sphere continues to roll down the incline, it takes less time to

travel the same distance.

If the steel sphere is moving under a constant acceleration, then the displacement

of the sphere from the initial position, if the sphere is released from rest, is given by:

 

𝒔 = 1

2 𝒂∆𝑡2

 

The displacement at each time t should be proportional to 𝑡2

 

 

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Table 2

Displacement (cm)

1

4

9

16

25

36

49

64

81

7. Place rubber bands on the foam board, covering the pencil lines you just made.

8. Set the foam board up at the same angle as the previous trial.

9. Roll the steel sphere down the foam board.

Note: The sounds made as the sphere crosses the rubber bands on the foam board

in the second trial should be at equal intervals. The sphere is traveling a greater

distance each time it crosses a rubber band, but the time interval remains constant

meaning the sphere is moving with a constant acceleration.

 

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Figure 10

Note: For more information on the Trigonometry, Kinematics Equations, and

Rotational Motion exercises, visit the Carolina Biological Supply website at the

following links:

 

Basic Right Triangle Trigonometry

 

Derivation of the Kinematics Equations

 

The Ring and Disc Demonstration

 

 

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