PHYSICS LAB NOTES

FOR

MECHANICS

AND

HEAT

EXPERIMENTS

PHYSICS 37

Los Angeles Harbor College

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TABLE OF CONTENTS

1. Measurement .............................. 5 2. One-dimensional Non-uniform Motion ....... 13 3. One-dimensional Free-fall Motion ......... 17 4. Vector Addition on the Force Table ....... 19 5. Projectile Motion ........................ 23 6. Newton's Second Law ...................... 27 7. Simple Machines .......................... 31 8. The Conservation of Mechanical Energy .... 35 9. The Ballistic Pendulum ................... 3910. Elastic and Inelastic Collisions ......... 4311. Center of Mass ........................... 4712. Torque and Equilibrium ................... 4913. Centripetal Force ........................ 5514. Moment of Inertia ........................ 5715. The Gyroscope ............................ 5916. Archimedes' Principle .................... 6317. The Coefficient of Linear Expansion ...... 6918. Latent Heats of Fusion and Vaporization .. 7319. The Mechanical Equivalent of Heat ........ 79

The Statistics of MeasurementThe Least-Squares Fit to Data

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1. MEASUREMENTPURPOSE:

The purpose of this experiment is to introduce:

a) the practice of making careful measurements with devices of varying precision,

b) the role of significant figures in the collection of raw data (measurements),

c) the role of significant figures in computation, and the rationale for rounding off computed results to the appropriate number of significant digits, and

d) terms encountered in data analysis.

APPARATUS:

MetronomeTape measureTimer, electronicRulerVernier caliperMicrometerCopper disks (50)Meter stick

For the entire class: C-clamp, large Balance, electronic (2) Vernier caliper demo Micrometer demo

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INTRODUCTION:

The Meter Stick and Metric Ruler -

These devices can measure to an accuracy of a millimeter, and all measurements obtained with them should be quoted to this degree of accuracy. The physical size of an object is a fundamental property, and the height, width and depth of an object is measured by its length along each dimension.

Examine the meter stick in front of you. The meter stickis slightly longer than a yardstick, and is divided into 100 centimeters. Each centimeter is further divided into 10 millimeters, so there are 1000 millimeters in a meter. The millimeter can be abbreviated as mm, and the centimeter can be abbreviated as cm. Every measurement of length must be followed by its unit. For example, the length of one-third of a meter stick can be expressed as 333 millimeters, 33.3 centimeters or 0.333 meters.

The left-hand edge of a meter stick or metric ruler is supposed to be at 0.000 meters, but its edge may be damaged or worn. The correct way to get an accurate measurement is to place the object being measured near the center of the ruler, read the position of the left edge of the object, read the position of the right edge of the object, and subtract one number from the other. For example, if the left edge is at 321 mm and the right edge is at 674 mm, the length of the object is 674 - 321 = 353 mm.

The Vernier Caliper -This device can measure to an accuracy

of one-tenth of a millimeter, and all measurements obtained with this device should be quoted to this degree of accuracy. To use this device, the length to be measured is placed in the jaws of the caliper, and the jaws are then closed by pushing on the wheel to the right of the movable window. The measurement is found by reading the ruler visible inside the window of the movable jaw.

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This reading is a two-step process:

- First, look just below the window of the movable jaw. You should see eleven vernier lines, and for this step you use only the leftmost of these eleven vernier lines. That leftmost line points in between two ruler lines directly above it, seen inside the window. Write down the value of the ruler line on the left. The measurement will begin with this value. - Second, which of the eleven vernier lines has a ruler line directly above it? These vernier lines are numbered 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. This number is placed after the value found in the first step. The vernier number 10 is never used.

For example, suppose the vernier caliper has been set as shown:

window

ruler

| | | | | | | | vernier

Performing the first step, the leftmost vernier line (thick arrow) points between the ruler lines 1.1 cm and 1.2 cm. The ruler line on the left is 1.1 cm, so you would write down the value “1.1 cm”. Performing the second step, the vernier line 3 (thin arrow) has a line directly above it, and so a 3 is placed after the previous value. The answer is 1.13 cm.

1 2

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As a second example, suppose the vernier caliper has been set as shown:

3 4

| | | | | | | | | |

Performing the first step, the leftmost vernier line points between the ruler lines 2.5 cm and 2.6 cm. The ruler line on the left is 2.5 cm, so you would write down “2.5 cm”. Performing the second step, the vernier line 8 has a line directly above it, and so an 8 is placed after the previous value. The answer is 2.58 cm.

The Micrometer -This device can measure to an accuracy of

one-hundredth of a millimeter, and all measurements obtained with this device should be quoted to this degree of accuracy. To use this device, the length to be measured is placed in the jaws of the micrometer, and the barrel is rotated until the jaws have closed.

Warning! A large amount of force should not be applied when rotating the barrel, as this will strip the screw inside the barrel and ruin the micrometer. A micrometer is not a C-clamp! Close the jaws so that they are ‘finger-tight’. That means that when your fingers rotate the barrel with only light pressure applied, your fingers slide along the barrel and it doesn’t rotate any more. Some of the micrometers have a small friction barrel on the right-hand side of the barrel, which will prevent the barrel from rotating when too much force is used. If your micrometer has a friction barrel, use it to close the jaws.

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This reading is a two-step process:

- First, look at the ruler between the jaws and the barrel. The left edge of the barrel cuts across the ruler, which is measured in millimeters, and shows both millimeter lines (on top) and half-millimeter lines (on bottom). Write down the value of the ruler line that is visible to the left of the edge of the barrel.

- Second, which of the barrel lines is closest to the horizontal line on the ruler? The two-digit number of that barrel line (from .00 mm to .49 mm) is added to the value found in the first step.

If this is confusing, open the jaws slightly until the barrel value is .00 mm, and the measurement is obvious. Then, slowly close the jaws until the barrel has returned to its former position. The correct value will be slightly smaller than the value obtained when the jaws were slightly open.

For example, suppose the micrometer has been set as shown:

0 5 30

rulerbarrel

horizontal line

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Performing the first step, the ruler line 7.00 is the line that is visible to the left of the barrel, so you would write down the value “7.00 mm”. Performing the second step, 26 is the barrel line closest to the horizontal line, and so 0.26 mm is added to the previous value. 7.00 + 0.26 = 7.26, so the answer is

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7.26 mm.

As a second example, suppose the micrometer has been set as shown:

0 5

40

Performing the first step, the ruler line 6.50 is the line that is visible to the left of the barrel, so you would write down the value “6.50 mm”. Performing the second step, 40 is the barrel line closest to the horizontal line, and so 0.40 mm is added to the previous value. 6.50 + 0.40 = 6.90, so the answer is 6.90 mm. Notice that the answer contains a ‘0’ in the last place, to signify that the length has been measured to this level of accuracy.

PROCEDURE:

A. LENGTH

1. Determine the area of the classroom floor in square meters. Make three separate measurements of both the length and the width. Calculate the average deviation, by consulting “The Statistics of Measurement” sheet at the end of this lab book. Measurements taken with the tape measures and metric rulers should be to the nearest millimeter. Your three separate calculations of area areArea = Length Width.

2. Measure the diameter d and thickness t of a copper disk

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using the vernier caliper. Calculate its volume = πd2·t/4. Measurements taken with a vernier caliper should be to the nearest tenth of a millimeter. You can calculate the volume to one significant figure.3. Measure the diameter and thickness of a copper disk using the micrometer. Calculate its volume = πd2·t/4. Measurements taken with a micrometer should be to the nearest hundredth of a millimeter. You can calculate the volume to two significant figures.

4. Convert the volume into units of cm3.

1 mm3 = 1mm × 1 mm × 1 mm = 0.1 cm × 0.1 cm × 0.1 cm = 0.001 cm3, so the conversion is accomplished by multiplying by 0.001.

B. MASS

1. Determine the mass of a group of 10 copper disks on the electronic balance and record your result on the data sheet. Repeat the measurement on four separate groups of 10 disks. Divide each measurement by 10 to find the mass of one copper disk. The average of these 5 numbers is the mean mass, recorded at the bottom of the data sheet.

2. Calculate the deviation from the mean of each of the 10 masses from the mean mass. Calculate the standard deviation σ of the mass of one copper disk, from these 10 deviations. Consult ‘The Statistics of Measurement’ sheet for the equation.

3. Calculate the standard error α, and use it to determine how much you should round off your value of the mean mass.

C. TIME

1. With the electronic timer, time 50 beats of the metronome set at 120. Set the metronome at 120 by sliding the pendulum weight so its top clicks into position just beneath '120'. Divide 50 by this time to calculate the beats per second, and convert to beats per minute. Repeat three times and calculate the average value.

2. With the electronic timer, time 50 beats of your own or your lab partner's pulse. Calculate the beats per second and the pulse rate (beats per minute). Repeat two times for the same person and determine an average value.

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2. ONE-DIMENSIONALNON-UNIFORM MOTIONPURPOSE:

This experiment illustrates the definition of position, displacement, velocity and acceleration of an object in non-uniform horizontal motion.

INTRODUCTION:

In the study of kinematics, an understanding of the terms that describe motion is of utmost importance. Position describes the location of an object relative to a reference point. Displacement is a change in position. Average velocity is the displacement divided by the amount of time it took the displacement to occur. Instantaneous velocity is the velocity at a very small (infinitesimal) time interval. Average acceleration is the change of velocity divided by the time interval. Instantaneous acceleration is the change in velocity over a very small (infinitesimal) time interval. As you observe and record the motion of an object in horizontal motion in this lab experiment, you will perhaps appreciate the physical meaning of these terms and be able to describe motion more accurately.

APPARATUS:

Graph paperTimer, electronicStraightedgeFrench curveTwo-meter stickMasking tapeChalk

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PROCEDURE:

1. Form a team of seven students.

2. With a piece of chalk, draw markers at one-meter intervals along the length of the blackboard. Let the edge of the blackboard be x0, the starting point. Two more sets of intervals can be marked out with pieces of masking tape at the back of the room.

3. Let one member of the team volunteer to be the traveler. The traveler will start the trip from position x0 = 0.00 m.

4. Let each of the other members of the team reset a timer to zero and locate themselves across from the x1 = 1.00 m, x2 = 2.00 m, x3 = 5.00 m, x4 = 5.00 m, x5 = 4.00 m, and x6 = 3.00 m positions. You may use your cell-phone timer if you wish.

5. At the shout of “GO”, the traveler will walk forward at a non-uniform rate, and all the timers will be started.

6. On the trip forward, students with timers t1, t2, and t3 at x1, x2, and x3 will each stop the timer as the traveler walks past that position.

7. Past position x3 = 5.00 m, the traveler will pause for a short while. On the trip back, students with timers t4, t5, and t6 at x4, x5, and x6 will each stop the timer as the traveler walks past that position.

8. Record the data from each timer at each position in Table 1.

9. Plot a graph of position (on the vertical axis) vs. time (on the horizontal axis). Label both axes, including units in parentheses, and supply a title to the graph. Use the French curve to draw a smooth curve through all these points.

10. Use the straightedge to draw tangent lines to this curve at

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t1, t2, t3, t4, and t5, and calculate the slopes of these five straight lines. These are the instantaneous velocities at these five times. Record these results in Table 1.

11. Plot a graph of instantaneous velocity vs. time, labeling both axes and include a title for the graph. Draw a smooth curve through these five points, draw tangent lines at t2, t3, and t4, and calculate the slopes of these three straight lines. These are the instantaneous accelerations at these three times. Record in Table 1.

12. Calculate the average velocity ∆x/∆t for each time interval between the observed times, and place them in Table 2. Note that these are similar to, but not equal to, the instantaneous velocities.

13. Calculate the average acceleration ∆v/∆t for each time interval between t1 and t5 by using instantaneous velocities in Table 1. Record the results in Table 3. Note that these are similar to, but not equal to, the instantaneous accelerations.

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3. ONE-DIMENSIONALFREE-FALL MOTIONPURPOSE:

In this experiment, the numerical value of the acceleration due to gravity will be determined by a graphical technique, and the kinematic equations will be applied in the study of an object in free-fall.

INTRODUCTION:

If the effect of air friction is neglected, objects relatively close to the Earth's surface undergo uniformly-accelerated motion. For our purposes, we will take this value of acceleration to be 9.80 m/s2.

In this experiment, the data are obtained by analyzing a wax paper tape which has a series of spark holes. The apparatus which produces the tape, sparks every 1/60 of a second as the free-fall body descends. Thus a time-distance record of the object in free-fall is produced and the acceleration due to gravity can be calculated.

By definition, acceleration is the time rate of change of velocity, so a graphical plot of the instantaneous velocity vs. time should yield a straight line, the slope of which is the acceleration. For each spark interval, the average velocity is readily calculated, being the distance the object falls in the interval divided by time it takes to fall that interval distance. Use the fact that the average velocity is equal to the instantaneous velocity at the midpoint in time of the interval.

APPARATUS: y1 1 t1

Graph paper ∆y ∆t = 1/30 sElectro Release y3 3 t3

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Spark generator, 1/60 sSpark tapeRulerStraightedgeFree-fall apparatus 5Masking tape

PROCEDURE:

1. Obtain a spark tape, and secure it to the table with masking tape. Starting with holes that are about one centimeter apart, draw a straight line perpendicular to the length of the tape through every other hole, as shown in the diagram.

2. Measure and record the interval distance between adjacent lines.

3. Calculate the average velocity for each of your intervals by dividing the interval distance ∆y by the elapsed time for each interval. The elapsed time is 1/30 second.

4. When constant acceleration occurs, the average velocity of each time interval is also the instantaneous velocity at the midpoint in time of that interval. For example, the average velocity between holes 1 and 3 is the instantaneous velocity of the free-fall object when it is at hole 2. Plot a graph of the instantaneous velocities of the midpoint in time of each interval (that is, at holes 2, 4, 6, 8 etc.) on the vertical axis versus time (1/60, 3/60, 5/60 etc.) on the horizontal axis. Draw the best straight line for the data points by fitting the line so that the line is the closest it can be to all the data points.

5. Calculate the slope of your straight line. This is the rate at which the velocity changes as the free-fall object accelerates downwards due to gravity. Divide the rise (∆v) by the run (∆t) for any two points on the line, not necessarily data points. Choose the two points so that they are widely separated.

Slope = ∆v/∆t = a = g (measured value)

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6. The local acceleration of g has been well-determined to be 9.80 m/s2, and this given value may be compared to your measured value by calculating the percent difference:

observed value - theoretical valuePercent difference = 100% .

theoretical value

4. VECTOR ADDITIONON THE FORCE TABLEPURPOSE:

To establish the condition 1.0for equilibrium of a suspendedmetal object. 0.6

37° 0.8APPARATUS:

Force tablePulley (3)Level, bubble Mass holder, 50-gram (3)Masses, slotted (3 sets)Metal cubes (brass and iron)PaperElectronic balanceRulerProtractor

INTRODUCTION:

A vector is a quantity that has both magnitude and direction. The most obvious example is displacement; if you walk 0.8 meters east and 0.6 meters north, the result of these two displacements is the same as if you had walked only 1.0 meters at an angle of 37° north of east, as shown in the diagram above. This is very different from the behavior of a scalar quantity, for which 0.8 + 0.6 = 1.4 . Displacement, velocity, acceleration and force are examples of vector quantities, while mass and

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temperature are examples of scalar quantities, which have magnitude but not direction. When vectors are added together, the total is called the resultant. When scalars are added together, the total is called the sum.

The mass (= m) of an object, measured in units of kilograms, is the scalar quantity that measures the object’s inertia. The inertia of an object is its resistance to changes in motion. If you pick up the largest slotted mass and wave it back and forth, you will feel this resistance. The slotted mass will have this same resistance to changes in motion anywhere in the Universe.

The weight (= W) of an object, measured in units of newtons, is the vector quantity that measured the object’s force of gravity downwards towards the Earth. If you pick up the slotted mass and hold it motionless in your hand, you will feel this force. The slotted mass will have this weight only near the surface of the Earth. The weight is calculated as W = mg, where g is the local acceleration of gravity. Near the surface of the Earth, g = 9.80 m/s2.

Equilibrium occurs when two or more forces balance, adding as vectors to zero. Each pulley on a force table rotates the tension on its string, generated by the weight of the object at the end of the string, from a vertical direction to a horizontal direction without changing its magnitude. These forces add together at the ring in the center of the force table, and when the pin holding the ring in place is removed, the ring will stay in place only if the forces are in equilibrium. This arrangement will allow you to experiment in obtaining equilibrium.

PROCEDURE:

1. Place the bubble level on the force table, and adjust the force table’s legs until it is level.

2. Attach the three pulleys to the edge of the force table at the angles 37°, 323° (= -37°) and 180°, with a string from the central ring running over each pulley. Hang a mass of 500 grams (this includes the mass of the massholder) from the pulleys at ±37°, and a mass slightly less than 1000 grams from the pulley at 180°. Remove the pin from the ring, and tap the ring gently. If it is not at the center of the table, add or subtract slotted masses to the pulley at 180° and at -37° until the ring remains exactly centered when gently tapped.

The three forces on the ring are now in equilibrium. Fill in the first two blank rows on the data sheet, below the row that

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has been filled in for you. The angles 37° and 53° have been chosen for this lab so that you may use the values ±0.6 and ±0.8 for their sines and cosines, as shown in the diagram on the first page.

y3. A graphical method for analyzingthe addition of vectors is to draw themproportionally (twice the magnitude, twice the length of the arrow) and xoriented according to the measured angles, as shown in the diagram.

Draw the three calculated force vectors on a blank sheet of paper, stating the scale you are using (2 cm/N may be appropriate). Each student should draw his or her own diagram, and should leave room for two more diagrams on the same sheet. Theoretically, the head of the last vector should meet the tail of the first vector for equilibrium, but experimental error may create a small gap.

y4. An algebraic method for analyzingthe addition of vectors is to resolve them into their horizontal and vertical F·cos(θ) F·cos(θ) x components. This is done by multiplyingthe magnitude of the vector by the cosine and sine respectively of the vector’s position angle. Each component is now a scalar, not a vector, and can be added using the ordinary rules of scalar addition to get the magnitude each component of the resultant. This is shown in the diagram for the x component of two of the three vectors.

Add together the three numbers in the last two columns in your data sheet. The sum for each component should be close to zero, but experimental error may create a small excess or deficit.

5. Repeat steps 2 through 4 for angles of 53° and 80°. Notice that the third pulley requires a decreasing mass as the x components decrease, illustrating the vector nature of forces.

6. Use the electronic balance to measure the masses of two different metal cubes. Hang the metal cubes from pulleys at two

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different angles, and experiment with different angles and masses for the third pulley supporting slotted masses until equilibrium is reached. On the reverse side of your blank piece of paper, perform a graphical analysis and an algebraic analysis.

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5. PROJECTILE MOTIONPURPOSE:

The object of this experiment is to determine the initial velocity of a projectile from the position measurements. Also, the equations of motion will be used to predict the point of impact of a projectile.

INTRODUCTION:

A projectile is any object which has its weight as the only significant force acting upon it. A soccer ball that has been kicked into the air is an example of a projectile, if air resistance is ignored.

In order to determine the initial velocity of a projectile fired horizontally, one first makes use of the equation y = at2 to calculate the time of flight as t = √2y/a. Then, from a measurement of the horizontal displacement, the initial velocity vo can be determined from the equation s = vot.

For a projectile fired at an angle, the horizontal displacement of a projectile can be determined if the angle of elevation, the initial velocity and the initial height of the projectile above the landing point are known.

APPARATUS:

C-clampBallistic pendulumClamp, tableClamp for wooden boardRod for wooden boardCatch box

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Plumb bobWooden boardPaperPaper, carbonCardboardRulerInclinometerMeter stickMeter stick, 2mSupport rod, shortMasking tape

y

Initial position of vi projectile xf x

yf Final position of projectile

Fig. 1

PROCEDURE:

A. INITIAL VELOCITY

1. Be extremely careful not to hit anybody with a projectile during this experiment!

Rotate the pendulum arm up so that it rests on the ratchet, out of the way of the projectile. Clamp the gun (not too tightly) near one of its four feet to the table, at least two meters from the catch box on the floor. Use the inclinometer to orient the gun to fire horizontally by rotating the gun’s feet, and take a trial shot. Tape a piece of cardboard to the floor centered on the spot where the projectile landed. On top of the cardboard, tape a piece of carbon paper face-down above a piece of plain paper to record the point of impact. Place the wooden catch box just beyond the carbon paper, to catch the projectile.

2. Measure the height from the floor to the bottom of the ball to the nearest millimeter. This is y (a negative number) and is

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the vertical displacement of the projectile. Use a plumb bob to get the exact vertical direction. Calculate the time of flight t from this measurement, with a = -9.80 m/s2 in the coordinate system shown in Figure 1.

3. Take six shots. Measure xf (the horizontal distance traveled) of each shot to the nearest millimeter. To accomplish this, use a plumb bob to get the exact vertical drop from the end of the gun to the floor.

4. Calculate the average value of xf, and use this number to calculate the initial velocity vi.

B. PREDICTION OF THE HORIZONTAL DISPLACEMENT

y vi

xf x

yf

Fig. 2

1. Clamp the spring gun to a board at an angle θi between 10° and 20°. Measure this angle precisely with an inclinometer.

2. Measure the distance of fall yf (a negative number).

3. Calculate the initial vertical component of velocityvyi = vi·sin(θi) from your value of θi, and your value of vi from part A. Calculate the horizontal component of velocityvxi = vo·cos(θi). Calculate the time of flight t, by solving the quadratic equation yf = vyit + at2, with a = -9.80 m/s2 in the coordinate system shown in Figure 2.

4. Fire the projectile and measure the horizontal distance traveled xf. Fire the projectile five more times and determine an average value. Again, use the cardboard and the catch box.

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5. Calculate the theoretical value of xf, from xf = vxi·t. Calculate the percent difference between this theoretical value, and the average of the measured values of xf.

Remove the masking tape from the carbon paper, the cardboard and the floor.

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6. NEWTON'S SECOND LAW

INTRODUCTION:

The acceleration of an object is directly proportional to the resultant force acting on it and inversely proportional to the mass being accelerated. Furthermore, the direction of the acceleration is in the direction of the resultant force, so

∑ F = ma (Newton's Second Law).

Using an air track, the acceleration of masses due to an unbalanced applied force will be determined, and compared with the acceleration calculated from the equation of motion for a uniformly accelerated object.

From Newton's Second Law: m1

m2g = (m1 + m2)a .

m2g a m2Solving, a = .

m1 + m2 W = m2g

From the equation of motion, s = vot + at2.

2s

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With vo = 0, a = . t2

APPARATUS:

Air track accessory kit Air track gliderPhotogate power adapterPhotogate timer with memoryPhotogate accessory for photogate timerAir trackScissorsString, thinBalance, electronicMass holder, 5-gramMasses, slotted, five-gram (4)Air track air supplyAir hoseAir track power cordPROCEDURE:

1. Connect the air supply to the air track and turn it on. Level the air track by adjusting the leveling feet so the glider is motionless at the center of the air track. Do not lean on the air track or the table (use another table for writing) during the experiment.

2. Place the metal flag on top of the glider, and tie the string to a small attachment from the accessory kit, plugged into the front of the glider. Determine the mass m1 of the glider, flag, attachment and string on a balance and convert this measurement to kilograms. Attach the pulley to the end of the air track opposite the air hose, and run the string from the glider over the pulley.

3. Place the photogate timer near the position 70 cm and the accessory photogate near the position 140 cm. Slowly slide the glider towards the photogate timer until the LED on top lights up because the flag breaks the light beam. Back up the glider and move it forward several times until you can determine the position x1 of the front of the flag, to the nearest millimeter. Do the same with the accessory photogate to get x2, and calculate s = x2 - x1.

4. Place a 5-gram mass holder at the end of the string running over the pulley. Add a 5-gram mass onto the mass holder, so m2 = 10.0 grams = 0.010 kg.

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5. Set the photogate timer to the "pulse" mode, and the resolution scale to 1ms.

6. Delicately hold the glider as close to the light beam of the photogate timer as possible (just before the LED on top lights up). Push the “reset” button, then release the glider (do not push or perturb the glider) and record the displayed time.

7. Repeat the procedure two more times. Average the three values and record in the table.

8. Add a 5-gram mass onto the mass holder. Repeat steps 6 and 7. Remember that m2 equals the mass of the holder plus the mass on the holder, so the total mass for this step is 15 g.

9. Repeat steps 6 and 7 for m2 = 20 grams (including hanger) and m2 = 25 grams (including hanger).

10. Calculate the acceleration of the masses, kinematically, by using the equation of motion s = vot + at2. With vo = 0, this equation becomes s = at2, so

a = 2s/t2 .

11. Calculate the acceleration of the masses, dynamically, by using Newton’s Second Law F = ma. With F = m2g and m = m1 + m2, this equation becomes m2g = (m1 + m2)a, so

a = m2g/(m1 + m2) .

12. Compute the percent difference between them, assuming the kinematic value of the acceleration to be the theoretical value.

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7. SIMPLE MACHINESPURPOSE:

In this experiment the principle of work is studied, using the inclined plane and pulleys as examples of simple machines.

INTRODUCTION:

A machine is any device used to do work by changing the magnitude or direction of a force. A simple mechanical machine exerts an output force which is greater than the applied force. Work or energy is not multiplied by a machine, force is multiplied. In practical applications, the work output is always less than the energy input.

The efficiency of a machine is defined as:

work output Fout·dout Fout doutEfficiency = = = · , and work input Fin·din Fin din

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θ

FoutActual Mechanical Advantage = AMA = ______ Fin

dinIdeal Mechanical Advantage = IMA = , so dout

AMAEfficiency = ______ IMA

APPARATUS:

Clamp, table for support rodClamp for wooden boardRod for wooden boardBoardHall's carriage (cart)Balance, double-pan Scissors massString hangerInclinometerMass hanger Pendulum clamp dout dinClamp-on pulleyPulley, double tandemPulley, single Fin = m1gMasses, slotted m2Meter stick (2)Support rod, short θ

Load = Fout = m2gFig. 1.

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PROCEDURE:

A. Inclined Plane

1. Set up the apparatus as in Figure 1, with θ = 30° and with an empty mass hanger strung over the pulley, attached to the cart (mass = m2) by the string.

2. Add slotted masses to the mass hanger until the cart movesup the incline at a constant speed. Calculate Fin = m1g and Fout = m2g. Measure din, and calculate dout = din·sin(θ).

3. Calculate AMA, IMA and the efficiency.

4. Repeat steps 1 through 3 for a smaller angle.

B. Pulley Systems

*

* *

cart *

mass hanger

System A System B System C System D System E

Fig. 2.

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1. Set up System A as in Figure 2, with a single string running from the mass hanger to the cart. The top of the pulley can be attached to the horizontal pendulum bar with a small loop of string.

2. Add slotted masses to the mass hanger to move the cart up at a constant speed.

3. Measure m1 (the mass of the mass hanger and slotted masses) and m2 (the mass of the cart and any pulleys that move up or down). System A has no moving pulleys. Calculate Fin and Fout. Measure din (the distance the mass hanger travels) and dout (the distance the cart travels). Make your measurements of din and dout as accurate as possible, using two meter sticks.

4. Calculate AMA, IMA and the efficiency.

5. Repeat the procedure for System B, with a single string running from the mass hanger to the asterisk (*). The lower pulley is a moving pulley.

6. Repeat for Systems C, D and E.

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8. THE CONSERVATION OFMECHANICAL ENERGYINTRODUCTION:

Though conservation of energy is one of the most powerful laws of physics, it is not an easy principle to verify. If a boulder is rolling down a hill, for example, it is constantly converting gravitational potential energy into kinetic energy (linear and rotational), and into heat energy due to the friction between it and the hillside. It also loses energy as it strikes other objects along the way, imparting to them a certain portion of its kinetic energy. Measuring all these energy changes is no simple task.This kind of difficulty exists throughout physics, and physicists meet this problem by creating simplified situations in which they can focus on a particular aspect of the problem. In this experiment you will examine the transformation of energy that occurs as an air track glider moves down an inclined track. Since

35

there are no objects to interfere with the motion and there is minimal friction between the track and glider, the loss in gravitational potential energy ∆(mgh) as the glider moves down the track should be very nearly equal to the gain in kinetic energy ∆KE, with

∆(mgh) = mg·∆h and ∆KE = mv22 - mv12

In these equations, m is the mass of the glider, g is the local acceleration of gravity and ∆h is the change in the vertical position of the glider.

APPARATUS:

Air track gliderPhotogate power adapterPhotogate timer with memoryPhotogate accessory for photogate timerAir track air supplyAir hoseAir track power cordAir trackBalance, electronicVernier caliperShim blocks, 2 (use 100-gram masses)Meter stickPROCEDURE:

PART A:

1. Level the air track as accurately as possible by setting the glider at the middle of the track and adjusting the leveling screws until there is no movement of the glider. Once leveled, do not lean on the table or push down on the glider.

2. Measure D, the distance between the air track support legs. Measure the thickness H of a shim block with a vernier caliper, and place the shim block underneath the single support leg of the track.

3. Set up a photogate timer and an accessory photogate as shown in Figure 1.

d

36

θ

HD

Fig. 1

4. Measure and record d, the distance the glider moves on the air track from where it first triggers the first photogate, to where it triggers the second photogate. You can tell where the photogates are triggered by watching the LED on top of each photogate. When the LED lights up, the photogate has been triggered. Your measurements of air-track positions should be accurate to the nearest millimeter.

5. Measure and record L, the length of the glider. The best technique for this is to move the glider slowly through one of the photogates, and measure the distance it travels from where the LED first lights up to where it just goes off.

6. Measure and record m, the mass of the glider.

D d

∆h H

θ

Fig. 2

7. From the geometry of Figure 2, calculate θ and ∆h. In this figure, θ is exaggerated for clarity.

8. Set the photogate timer to GATE mode, leave the memory function in the "OFF" position, and press the RESET button.

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9. Hold the glider steady near the end of the air track, then release it without pushing, so it glides freely through the photogates. Record t1 (the time during which the glider blocks the first photogate) and t2 (the time during which it blocks the second photogate). Notice that t2 = ttotal - t1. That is, the photogate timer first displays t1, then ttotal = t1 + t2, and does not display t2 by itself.

10. Repeat the measurement three more times. You need not release the glider from the same point on the air track for each trial, but it must be gliding freely and smoothly (minimum wobble) as it passes through the photogates.

11. Calculate v = L/t for position 1 and position 2.

12. Calculate the kinetic energies at position 1 and position 2. Calculate ∆KE and ∆(mgh). Theoretically, these two values should be equal.

PART B:

1. Repeat Part A with a larger value of ∆h, by using two shim blocks.

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9. THE BALLISTIC PENDULUMPURPOSE:

In this experiment we will determine the initial speed of a projectile by using the principles of the conservation of momentum and the conservation of energy.

APPARATUS:

C-clampBallistic pendulum apparatus

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Level, bubbleBalance, electronicRulerShims (use copper disks)

INTRODUCTION:

A device called a ballistic pendulum will be used in this experiment to determine the initial speed of a projectile. The device consists of a spring gun that propels a metal ball of mass m into a pendulum bob of mass M. This pendulum-ball combination then swings up onto a rack with a speed V just after impact. The change in height ∆h through which it rises depends directly on the initial speed vo of the ball.

m vo M V hf hi hi

Just Just Motionless, before after at maximum impact impact height

Fig. 1 Fig. 2 Fig. 3

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From the law of conservation of momentum, with Figure 1 as the initial state and Figure 2 as the final state, mvo = (m + M)V, so

m + Mvo = · V Eq. 1

m

From the law of conservation of energy, with Figure 2 as the initial state and Figure 3 as the final state,

KEi + PEi = KEf + PEf , so

(m + M)V2 + (m + M)ghi = 0 + (m + M)ghf,

____which simplifies to V = √2g∆h, with ∆h = hf - hi.

____Substituting V = √2g∆h into Eq. 1 gives

m + M ____vo = · √2g∆h Eq. 2

m

It is Eq. 2 which permits a calculation of the initial speed as a function of ∆h.

PROCEDURE:

1. Level the apparatus on the lab table using a spirit level. You may need to shim the apparatus. Lightly clamp the apparatus to the table using a C-clamp. Once leveled and clamped, do not lean on the table or otherwise disturb the level of the apparatus.

2. Determine the height (hi) of the center of mass of the freely-hanging pendulum relative to the base plate. The center of mass is indicated by the pointed projection on the side of the pendulum.

3. Determine the mass of the ball and record it on the data sheet.

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4. Fire the ball into the pendulum six times, each time recording the number of the notch on which the pendulum comes to rest.

5. Calculate the average notch number. Place the pendulum at this average position and determine the height (hf) from the base plate to the pendulum center of mass. Calculate ∆h = hf - hi.

6. Calculate the initial speed vo of the ball, from Equation 2. ____7. Calculate the speed V = √2g∆h of the ball and pendulum just after impact.

8. Calculate the momentum mvo before impact, and the momentum (m + M)·V after impact.

9. Calculate the energy lost in joules. The kinetic energy before impact is mvo2, and immediately after impact the kinetic energy is (m + M)V2. What percent of the original kinetic energy was ‘lost’ to other forms of energy? Where did this energy go?

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10. ELASTIC ANDINELASTIC COLLISIONSPURPOSE:

In this experiment we will verify that linear momentum is conserved for both elastic collisions and inelastic collisions.

INTRODUCTION:

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One of the most widely-used principles in physics is the conservation of linear momentum. The utility of this principle lies in the fact that regardless of the complexity of the actual collision, the total momentum before a collision is equal to the total momentum after a collision:

Momentum before collision = momentum after collision

m1v1i + m2v2i = m1v1f + m2v2f

Two types of two-body collisions will be investigated using the air track. The first is one in which almost no energy is dissipated; this is called an elastic collision. The second is one in which a fraction of the kinetic energy is dissipated; this is called an inelastic collision.

Remember that velocities and momenta are vector quantities and consequently, in one dimension, add algebraically. That is to say, momenta in the right-hand direction are positive and momenta in the left-hand direction are negative.

APPARATUS:

Air track accessory kit Air track glider (2)Photogate power adapter (2)Photogate timer with memory (2)Air trackBalance, electronicAir track air supplyAir hoseAir track power cord

PROCEDURE:

A. ELASTIC COLLISIONS

Photo- Photo-Glider 1 gate 1 gate 2 Glider 2

m1 m2

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Fig. 1

1. Level the air track by placing the glider at the middle of the track and adjusting the leveling screws until the glider is motionless. Place two independent photogate timers near 70 cm and 130 cm.

2. Place a rubber-band bumper at the front of glider 1 and place a counter mass at the back of glider 1. Place a metal flag on the top of each glider. Determine and record the effective length L to the nearest millimeter of each flag by measuring the glider's positions when the LED on top of the photogate turns on, then off. Determine the mass of each glider with attachments and record the mass of each.

3. Set the photogate timers to ‘GATE’ mode and turn on the memory feature. Launch the gliders with moderate speed from opposite ends of the air track. The flags will pass through the photogates and a time will be displayed on each photogate. From these times the initial velocity of each glider can be calculated. After impact, the gliders will pass back through the photogates, adding to the times in memory. Subtract the displayed time from the memory time to determine the time for the final velocities. Be sure to stop the gliders after they have passed through the photogates the second time.

4. Repeat the above procedure two more times so that you will have three trials altogether. Change the mass on the gliders for these two trials by adding masses symmetrically to the sides of one of the gliders.

5. Calculate the total momentum for each trial before and after the collision, in Table C. Use negative signs for the velocities and momenta where appropriate. Is linear momentum conserved? Calculate the total kinetic energy both before and after collision. Is kinetic energy conserved?

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B. INELASTIC COLLISION

Wax Needle plug

Fig. 2

1. Repeat the above procedure for a completely inelastic collision. Use the needle on one glider and the wax plug on the other glider so that they stick together, and place a counter mass at the back of glider 2. After the collision, stop the gliders after one flag has passed through a photogate. Only one timing is necessary, as v1f = v2f.

2. Calculate the momenta both before and after the collision. Is momentum conserved for the inelastic case? Calculate the total kinetic energy both before and after collision. Is kinetic energy conserved?

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11. CENTER OF MASSPURPOSE:

To determine the center of mass (CM) of objects of uniform density and thickness.

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INTRODUCTION:

One way to locate the center of mass of an object of uniform thickness is to balance the object on a pivot. Another way is to drop a plumb line from one corner with the object suspended from that corner, and trace out the plumb line, then repeat this for another corner. The point of intersection of the two lines is the center of mass.

A third way to locate the center of mass is to divide an object into rectangles, and imagine the mass of each rectangle to be concentrated at the rectangle's center. Each rectangle can be weighted by area instead of mass, as the object has uniform surface density and thickness. Then the center of mass of the object is the average of the positions of these centers, weighted according to area.

For example, suppose that the upper bar of Object A in Figure 1 has an area of A1 = 10 cm2 extending from y = 10 cm to y = 14 cm, so y1 = 12 cm, and suppose the vertical stem has an area of A2 = 15 cm2 and extends from y = 0 cm to y = 10 cm, so y2 = 5 cm. Then,

(10 × 12) + (15 × 5) 195yCM = = = 7.8 cm. 10 + 15 25

∑Aixi ∑Aiyi Generally, xCM = and yCM = .

∑Ai ∑AiAPPARATUS:

Plumb bobNail Cardboard sheetScissorsStringRuler

PROCEDURE:

1. Tape a sheet of white paper to Object A, aligned with the x and y axes shown in Figure 1. Puncture one corner of Object A with a nail, let it hang freely, and trace the path of the plumb line across the paper (not the cardboard), starting at the nail. Repeat this from another corner, and find the point where these two lines intersect. Mark this point on the paper, and label it

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as the “experimental” center of mass. The presence of the paper will distort your results slightly.

2. Verify that this is the center of mass by balancing the object with your index finger.

3. Calculate the position of the center of mass for Object A, by imagining Object A as two rectangles of unequal mass, each replaced by point masses at their geometric centers. Mark and label the "calculated" position of the center of mass on the paper.

y y y

x x x Object A Object B Object C

Fig. 1.

4. Compare the "experimental" and the "calculated" center-of-mass positions.

5. Repeat procedure for Object B.

6. Repeat procedure for Object C.

12. TORQUE AND EQUILIBRIUMPURPOSE:

The object of this experiment is to use the method of balancing torques to determine the center of mass of a

49

non-homogeneous meter stick, and to determine the unknown mass of an object.

APPARATUS:

Knife-edge standClamp, knife-edgeMetal cubeScissorsString, thinBalance, electronicMasses, hookedMeter stick, weighted

Fulcrum at midpoint of stick

x d1 d2

m2

F1 = m1g F2 = m2gτ

τCounter-clockwise Clockwisemotion produces a motion produces a positive torque negative torque

Fig. 1

INTRODUCTION:

If a rigid object is in rotational equilibrium, the net torque acting on it, about an axis, is zero. This equilibrium condition can be stated as:

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∑τ = 0

where τ = Fd, F is the applied force, and d is the distance from the horizontal axis of rotation to the point where the downward force is applied. The plus sign {+} corresponds to a counter-clockwise torque and the negative sign {-} corresponds to a clockwise torque.

The center of mass, denoted here by CM, is the point at which the mass of the object can be considered to be concentrated. The position x of the CM of a non-homogeneous meter stick can be determined by balancing the torque of the stick on one side of the fulcrum with the torque of a known mass on the other side of the fulcrum.

Having established the position of the CM and knowing the mass of the stick, the same procedure can be used to determine the unknown mass of another object.

PROCEDURE:

A. CENTER OF MASS OF A NON-UNIFORM METER STICK

1. Record the mass of the non-uniform meter stick m1 indicated on the electronic balance.

2. Set up the apparatus as shown in Fig. 1, making sure that the fulcrum is at the midpoint of the stick, and using one of the hooked standard masses suspended from a loop of string as m2. Slide m2 in along the stick until the stick is in equilibrium. Record m2 and d2. Be sure to include the mass of the string in the mass of m2.

3. Use Eq. 1 to obtain your first estimate of x, the distance of the CM from the weighted end of the stick. This equation was obtained from the equilibrium condition:

τcounter-clockwise + τclockwise = F1d1 + (-F2d2) = 0 , so

F2 m2gF1d1 = F2d2 and d1 = ·d2 = ·d2 .

F1 m1g

m2 d1 = ·d2 = distance of CM from fulcrum Eq. 1

m1

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x = position of CM from weighted end

x = fulcrum position minus d1

4. Move the fulcrum 5.0 cm away from the midpoint, toward the weighted end of the stick as shown in Fig. 2. Slide m2 to establish equilibrium. Record m1 and m2 and the new value of d2. Use Eq. 1 to calculate the new value of d1 and subtract this from the position of the fulcrum to obtain your second estimate of x, the position of the CM.

Fulcrum

Midpoint of stick

x d1 d2

m2

F1 = m1g F2 = m2g

Fig. 2

5. To obtain your third estimate of x, remove m2 and balance the stick on the knife-edge clamp as in Figure 2 below. Record x as the position of the fulcrum.

B. DETERMINATION OF AN UNKNOWN MASS

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1. Having calculated the position of the center of mass on the previous page, set up the apparatus as shown in Fig. 1, only this time m2, a metal cube, will be an unknown mass. Using a loop of string, hang this unknown mass on the stick and slide it along the stick to balance.

2. Record the new value of d2. Use Eq. 2 to obtain your estimate of the unknown mass, m1.

F1 m1g∑τ = 0 , so F1d1 + (-F2d2) = 0 , F2 = ·d1 and m2g = ·d1 .

d2 d2

d1m2 = ·m1 Eq. 2

d2

3. Weigh the metal cube on the electronic balance and find the percent difference between the two measurements of m1.

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C. MULTIPLE-TORQUE SYSTEM: DETERMINATION OF THE MASS OF A METAL CUBE

Fulcrum at midpoint of stick

d2 d4

x d1 d3

m2 m3 m4

F2 F1 F3 F4

Fig. 3

1. The equilibrium condition can be used even when there are several torques involved. Set up the apparatus as shown in Figure 3, using the same m2 as in Part B, and using hooked standard masses suspended from loops of string as m3 and m4.

2. Use Eq. 3 to obtain another estimate of the unknown mass m2.

∑τ = 0 , so F1d1 + F2d2 + (-F3d3) + (-F4d4) = 0 ,

giving m1gd1 + m2gd2 - m3gd3 - m3gd4 = 0 .

m3d3 + m4d4 - m1d1m2 = Eq. 3

d2

3. Find the percent difference between this measurement and the value obtained directly from the electronic balance.

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55

13. CENTRIPETAL FORCE

INTRODUCTION:

When an object rotates with uniform speed v around a circle of radius r, its inward (centripetal) acceleration has a magnitude of a = v2/r. The inward (centripetal) force that keeps it in its circular path is found by substituting this magnitude into Newton's Second Law F = ma to obtain

F = mv2/r . (Eq.1)

Also, v = 2πr/T where T = period, or time for one revolution,because speed = distance/time.

APPARATUS:

Centripetal force apparatusRubber stopper, #5ScissorsStringBalance, electronicTimer, electronicMasses, hooked: 50g, 100g & 200gMarker penThistle tubeMasses, slotted rMeter stick

Revolving Mass,

m Thistle tube

Mark

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Hanging mass, M

Fig. 1PROCEDURE:

1. Determine the mass m of a stopper. Tie a 1.5 m length of string to the stopper, then thread it through the thistle tube as shown in Figure 1. Tie a hooked mass with M = 0.100 kg to the other end of the string. The weight W = Mg = 0.100 × 9.80 = 0.980 newtons of this hooked mass creates the tension in the string that provides the centripetal force on the stopper.

2. To help you maintain the radial distance, use a mark made by a marker pen at the top edge of contact with the thistle tube.

3. Using the stop clock, measure the total time for 25 revolutions for r = 0.500 m. Maintain a steady horizontal swing. Make three measurements, and average them. The time T for one revolution is this average time divided by 25.

4. The speed of the revolving mass is given by the equation

distance circumference 2πrv = = = , time time for one revolution T

where r is the radius of the circular path and T is the time for one revolution.

5. Calculate the percent difference between the theoretical value F = mv2/r and the measured value F = Mg.

6. Repeat the above procedure for five other values of r and M.

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14. MOMENT OF INERTIAAPPARATUS:

C-clampClamp, tableClamp, 90-degree twist (2)Moment of inertia massesMoment of inertia apparatusBubble levelTimer, electronicVernier caliperFoam padMass holderPulley, rod (2)Masses, slotted2-meter stickMasking tapeSupport rod

PROCEDURE:

1. Set up the equipment as shown in the diagram. Make sure the string runs horizontally from the platform, and the pulleys are aligned with the string. The falling weight should travel about 1.5 meters before striking the foam pad on the floor. Use the vernier caliper to measure the diameter (= 2·r) of the platform’s axle, where the string will be. Make sure the string is long enough to exert torque on the wheel for the full length of travel.

2. Use a total mass m = 0.250 kilograms for the falling weight. Measure the distance that the weight has to fall, to the nearest millimeter. Always start the weight from the same position.

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3. Make three timings of how quickly the falling weight travels through your distance d ~ 1.5 meters when the platform is empty, and find the average time t of these three timings. The average velocity of the object is vave = d/t. Since the acceleration is uniform, the final velocity of the weight just before it lands on the foam pad is v = 2.vave. Since the initial velocity is zero, the acceleration of the weight is a = v/t.

The linear acceleration of the falling weight equals the linear acceleration of the string wrapped around the axle of the platform, so the angular acceleration of the platform is = a/r. The tension in the string caused by the falling mass m is T = m.(g - a), with g = 9.80 m/s2, and this tension exerts a torque on the axle of = r.T.sin(90°). The moment of inertia is I = /.

4. Repeat procedure 3, but with the ring lying on the platform. After you calculate the moment of inertia of the platform and ring, subtract the moment of inertia of the platform to get the moment of inertia of the ring.

Calculate the theoretical moment of inertia of the ring, and calculate the percent difference between the theoretical and the measured values.

5. Repeat procedure 4, using the solid disk instead of the ring.

6. Repeat procedure 4, using two cylinders taped standing on edge at the rim of the platform, instead of the ring.

7. Remove all masking tape from the equipment.

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15. THE GYROSCOPEPURPOSE:

To determine the spin angular velocity (s) and precessional angular velocity (p) of a gyroscope.

APPARATUS:

Photogate timerGyroscopeTransfer caliper, largeBalance, double-panStringTimer, electronicRulerVernier caliperTorque mass, largeTorque mass, smallMasking tape

INTRODUCTION:

A spinning object such as the rotor of a gyroscope possesses angular momentum (L). Precession occurs when a torque is applied, changing the angular momentum.

SPIN TORQUE

y = r × F

Spin axis L s r r x

z F s = spin angular velocity L = angular momentum

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PRECESSION ∆L

y = –– ∆t

p

L·∆ Lf = –––– = L·p

∆ ∆L x ∆t Li

z p = –

L

∆L r F∆ = ––, so ∆L = L·∆ p = ––––

L Is

rmgp = ––– Eq. 1

Is

p is the precession angular velocity. This angular velocity can also be directly measured by timing the gyroscope as it slowly precesses in the horizontal plane through one complete revolution, over the time Tp, from the equation

2πp = –– Eq. 2

Tp

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PROCEDURE:

1. Balance the rotor statically by screwing the pointer of the representative L vector in or out until the pointer is stationary, when pointed horizontally.

2. Physically measure the rotor angular velocity by the following method:

a) Tape a short piece of string Axis ofto the edge of the rotor as shown rotationin the diagram to the right.

b) Set the photogate to the "pulse" mode. Plug in the gyroscope and hold the arrow in a vertical position. Place the photogate beam in the path rotor stringof the rotating string, and press the reset button on the photogate timer. photogateThis will give you the time for one revolution directly. Warning! Students with long hair should be very careful not to get it tangled up in the gyroscope!

c) Take six consistent time readings by pressing the reset button and calculate the average time for one revolution, Ts. Calculate s = 2π/Ts.

3. Use the stop clock to measure the time for one precessional revolution, using the small torque mass clipped into the notch just behind the pointer, pointed horizontally. Do this six times and calculate an average time. Calculate the precessional frequency, p = 2π/Tp. Measure the length of the moment arm r (the distance from the center of the rotor to the center of the torque mass).

4. Repeat procedure 3 for the large torque mass.

5. Calculate the moment of inertia of the rotor itself by the method of superposition, i.e., the moment of inertia of the whole is equal to the moment of inertia of a solid thick disk, minus the moment of inertia of a disk of smaller radius and thickness that would fill in the empty part of the rotor. Assume that the rotor has a density = 7.0 × 103 kg/m3. Use the large caliper to measure the radii accurately.

6. Calculate the percent difference between your measured and

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theoretical precessional angular velocities.

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16. ARCHIMEDES' PRINCIPLEPURPOSE:

Archimedes' Principle will be used to determine:a) the density of a symmetrically-shaped objectb) the density of an irregularly-shaped objectc) the density gravity of an unknown liquid.

APPARATUS:

HydrometerBeaker, 150-mlBeaker, 600-mlGraduated cylinder, 250-mlClamp, tableMetal cubeOverflow canRock sampleBalance, double-panPaper clip, smallScissorsStringLab jackVernier caliperUnknown fluidSupport rod, short

KEY TO SYMBOLS:

B = buoyancy

mo = mass of object in airmw = mass of water displacedmaw = apparent mass of object in water

Vo = volume of objectVw = volume of water displaced

Wo = weight of object in airWw = weight of water displacedWaw = apparent weight of object in water

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o = density of objectw = density of water = 1000 kg/m3INTRODUCTION:

Archimedes' Principle states that an object that is submerged in a liquid is buoyed up by a force that is equal to the weight of the liquid displaced by the object. This force is called the buoyant force, or the buoyancy. The buoyant force can be determined experimentally with the following setup:

Beam balance Beam balancePaperclip

T1 = Wo T2 = Waw B

m ma mg mg

Lab jack

Fig. 1 Fig. 2

T1 = Wo T2 = Waw(the weight (the apparent weight of object of objectin air) in water)

The force diagrams in Figures 1 and 2 both have the same downward force, which is the weight of the object, Wo = mog. The difference between T1 and T2 measured by the balance is the upward push by the liquid, the buoyant force, B. Since B = T1 - T2, the buoyant force can be determined experimentally by measuring the difference in weight of the object in air and when it is submerged in a liquid such as water.

B = Wo - Waw (Eq. 1)

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THE DENSITY OF AN OBJECT

When an object is totally submerged in water, the volume of water displaced is equal to the volume of the object. That is,

Vw = Vo (Eq. 2)

Density = mass/volume, or = m/V, so V = m/, and Equation 2 becomes

mw/w = mo/o

Solving for o and multiplying mw and mo by g gives

mo·go = ·w , with

mw·g

mo·g = Wo, the weight of the object in air, andmw·g = Ww, the weight of the liquid displaced. According to Archimedes’ Principle, Ww = B, so

Woo = ·w

B

Combining this with Equation 1 give

Woo = ·w

Wo - Waw

Dividing the terms in the fraction by g gives

moo = ·w (Eq. 3)

mo - maw

Equation 3 calculates the density of an object, given the measure quantities mo and maw.

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PROCEDURE:

A. DENSITY OF A METAL CUBE

1. Measure the length of one side of the metal cube. Calculate the volume of the cube Vo.

2. Hang a paper clip and a string from the bottom of the left side of the beam balance mounted on a support rod as in Figure 1. Make sure all sliding weights are set at their zero positions and to zero the balance, rotate the small calibration weight until the needle lines up on the center vertical line. Suspend the cube from the beam balance with the attached string, as shown in Figure 1, and determine its mass, mo.

3. Calculate the cube’s density, as = mo/Vo.

4. Immerse the suspended cube in a beaker of water as in Figure 2. Determine the apparent mass of the cube in water, maw.

5. Determine the density of the cube from Equation 3, based on Archimedes’ Principle, and find the percent difference between this and the value from step 3.

B. DENSITY OF AN IRREGULARLY-SHAPED ROCK

1. Suspend a rock from the beam balance and determine its mass mo and apparent mass maw, as shown in Figures 1 and 2.

2. Determine the density of the rock from Equation 3.

3. Determine the mass mb of an empty 150-ml beaker.

4. Place the displacement can on a level surface near the edge of a sink. Fill it with water, and let the excess drain off into the sink.

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5. Without moving the displacement can, slowly lower the rock into the water, allowing the displaced water to flow into the small beaker. Determine the mass mb+w of the beaker with displaced water inside it.

6. Determine the mass of water displaced, mw = mb+w - mb.

7. Calculate the volume of displaced water, from V = mw/w. This equals the volume of the rock.

8. Determine the density of the rock = mo/V, and find the percent difference between this and the value from step 2.

C. DENSITY OF AN UNKNOWN LIQUID

1. With the same cube used in Part A, determine its mass mo and apparent mass maf when immersed in an unknown liquid, as shown in Figures 1 and 2.

2. The density of the cube o is known from Procedure A, so Equation 3 can be solved for w, and re-written as

mo - mawliquid = ·o (Eq. 4)

mo

Calculate liquid.

3. Fill a tall measuring cylinder with the “unknown” liquid. Use a hydrometer to measure the density of the liquid. Find the percent difference between this and the value from step 2.

4. When finished, return the unknown liquid to its container. Thoroughly rinse and dry all equipment before leaving.

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69

17. THE COEFFICIENT OFLINEAR EXPANSIONPURPOSE:

The purpose of this experiment is to measure the coefficient of linear expansion for various metals and to compare the results with the known values.

INTRODUCTION:

In most cases, when materials are heated or cooled, they undergo expansion or contraction respectively. From the standpoint of materials science, this process must be taken into account when designing structures that are subjected to temperature variations. Otherwise, tensile or compressive stresses might develop which could destroy the structure.

The linear (one-dimensional) coefficient of expansion is defined as the fractional increase in length divided by the temperature change. This coefficient is designated by the Greek letter alpha (), and is found to be almost constant over a wide range in temperature. In equation form, the definition of is:

∆L = ––––– Lo·∆T

where ∆L is the change in length, Lo is the original length, and ∆T is the temperature change in degrees Celsius.

In this experiment, the value of the linear coefficient of expansion of several rods of common metals will be determined. The length of the rod is measured at room temperature, then steam is passed over the rod with the resulting temperature increase

70

causing it to expand. The amount of expansion is measured with a dial indicator. The coefficient is then determined using the data gathered.

APPARATUS:

Thermometer, 100 degree Celsius Steam generator, electricDial indicatorGlycerinMeter stickLinear expansion apparatusRods: Fe, Al, Cu

PROCEDURE:

1. Measure and record the initial length of the rod Lo, to the nearest millimeter. Determine and record the ambient temperature (room temperature).

2. Place the dial indicator on the metric ruler, with its movable probe along the millimeter rulings. Slowly move the probe millimeter by millimeter, and see how the two dials change. Spend a minute or so until you feel confident that you understand how this dial measures very small changes in position, so you can read this to an accuracy of 0.01 millimeters.

Steam inlet tube

Steam Thermometer Generator

Rod Dialindicator

Steam

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outlet tube

Fig. 1. Linear Expansion Apparatus with Associated Equipment

3. Set up the apparatus as shown in Figure 1. Use glycerin as a lubricant if necessary. The steam jacket for the rod has an opening for steam, thermometer, and rod ends, and an outlet for the condensed steam. Fill the steam generator about 2/3 full of water and turn on the generator, but do not connect the generator to the expansion apparatus as yet. Insert the rod in the apparatus until it just makes contact with the dial indicator probe and is in firm contact with the screw at the other end.

4. Make sure that the dial indicator is firmly screwed onto its holder and that the graduated ring is tightened down. Record the initial reading of the dial indicator, to the nearest 0.01 mm (= 0.00001 m).

5. When the generator is generating steam briskly, connect the steam tube to the inlet on the apparatus. Warning! Be careful not to scald yourself.

6. Allow the steam to warm up the rod to a constant maximum temperature, Tmax. When the rod stops expanding, record the final reading of the dial indicator. Calculate ∆T = (Tmax - Tambient).

7. Calculate the change in length, ∆L. This equals the final reading of the dial minus the initial reading of the dial. Notice that you don’t have to measure Lfinal and Linitial to get ∆L.

8. Calculate the coefficient of expansion and record it on the data sheet. Compare your values with the known values of the coefficient of linear expansion by calculating the percent difference.

Aluminum = 2.4 × 10-5 /°C Copper = 1.7 × 10-5 /°CSteel = 1.1 × 10-5 /°C

9. Repeat the above procedure for two other rods. Be carefulnot to burn yourself on the hot metal. When finished, dry the equipment thoroughly.

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18. LATENT HEATS OF FUSIONAND VAPORIZATIONPURPOSE:

The value of the latent heat of fusion for water and the latent heat of vaporization for water will be determined by the method of calorimetry.

INTRODUCTION:

When a substance such as water undergoes a change of state from the solid phase to the liquid phase, not all of the heat energy that is added to the system is reflected in a change of temperature of the substance. Some energy is needed to break the permanent bonds between the molecules of the substance and this energy is called the latent heat of fusion of the substance.

In today's experiment the latent heat of fusion will be determined by the method of mixtures and by applying the principle that the heat lost is equal to the heat gained (conservation of energy).

For the latent heat of fusion, ice cubes are placed into a measured amount of warmed water and is left to melt, cooling the water in the process. By noting the temperatures before and after melting, the latent heat of fusion can then be calculated.

For the latent heat of vaporization, steam is introduced into a measured amount of cooled water warming the water in the process. By noting the temperatures before and after the introduction of the steam, the latent heat of vaporization can be calculated.

APPARATUS:

Thermometer

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Pinch clampSteam generatorWater trapDouble-walled calorimeterBalance, electronicRing StandIce cubesBaster

KEY TO SYMBOLS:

cc = specific heat of aluminum calorimeter = 0.22 cal/g·°C cw = specific heat of water = 1.00 cal/g·°C

mc = mass of calorimeter (g) mi = mass of ice (g) ms = mass of steam (g) mw = mass of water (g)

Ti = initial temperature of water and calorimeter (°C) Tf = final temperature of water,

calorimeter and melted ice (°C) in Part A Tf = final temperature of water,

calorimeter and condensed steam (°C) in Part B

0 °C = initial temperature of ice100 °C = initial temperature of steam

PART A: LATENT HEAT OF FUSION OF WATER: Lf

Heat gained by water = Qw = mwcw(Tf - Ti)water

Heat gained by aluminum calorimeter = Qc = mccc(Tf - Ti)Al

Heat gained by ice = Qi = miLf + micw(Tf - Ti)ice

Qw and Qc will both be negative numbers, because the water and the calorimeter both lose heat as they cool down due to the melting ice. The first term of Qi is positive as the ice melts to become liquid water at 0 °C. The second term of Qi is positive as this liquid water absorbs heat and increases in temperature.

From the law of conservation of energy applied to this isolated

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system,

∑(Qw + Qc + Qi) = 0 , so

mwcw(Tf - Ti)water + mccc(Tf - Ti)Al + miLf + micw(Tf - Ti)ice = 0

Solving,

mwcw(Tf - Ti)water + mccc(Tf - Ti)Al + micw(Tf - 0)iceLf = - –––––––––––––––––––––––––––––––––––––––––––––––––––

mi

PROCEDURE: Part A: Latent Heat of Fusion

1. Determine the mass of the inner cup and stirrer of the calorimeter to the nearest tenth of a gram. Do not include the plastic collar in any measurements of mass.

2. Fill the inner cup of the calorimeter to about 2/3 full with warm water at about 40 °Celsius. Use the baster to transfer hot water from the steam generator to reach this temperature. You need to start the experiment with water hotter than room temperature because the calorimeter is not a perfect insulator. Some heat leaks out at the beginning of the experiment, but leaks back in near the end of the experiment as the temperature drops below room temperature. By starting with heated water, these heat transfers approximately cancel each other out.

3. Determine the mass of the inner cup, stirrer, and water. Calculate the mass of water in the cup.

4. Place the cup, stirrer, collar, and water into the outer calorimeter jacket and record the initial temperature just before the ice cube is placed in the water, to the nearest tenth of a degree (yes, I know it's difficult).

5. Wipe all the water from the surface of one large ice cube or three small ice cubes and place the ice carefully into the

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calorimeter cup.

6. Stir the contents occasionally while constantly observing the ice cubes. As soon as the ice cubes are completely melted, record the final temperature.

7. Determine the mass of the cup, stirrer and contents. The mass of the ice cubes can now be calculated.

8. Calculate the latent heat of fusion of water Lf, and compare this to the accepted value by calculating the percent difference.PART B: LATENT HEAT OF VAPORIZATION OF WATER: Lv

Heat gained by water = Qw = mwcw(Tf - Ti)water

Heat gained by aluminum calorimeter = Qc = mccc(Tf - Ti)Al

Heat gained by steam = Qs = (-msLv) + mscw(Tf - Ti)steam

Qw and Qc will both be positive numbers. The first term of Qsteam is negative as the steam condenses to become liquid water at100 °C. The second term of Qsteam is negative as this liquid water cools down and decreases in temperature.

From the law of conservation of energy applied to this isolated system,

∑(Qw + Qc + Qs) = 0 , so

mwcw(Tf - Ti)water + mccc(Tf - Ti)Al + (-msLv) + mscw(Tf - Ti)steam = 0

Solving,

mwcw(Tf - Ti)water + mccc(Tf - Ti)Al + mscw(Tf - 100)steamLv = –––––––––––––––––––––––––––––––––––––––––––––––––––––––

ms

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water thermometer trap

ring stand

steam calorimeter generator

Fig. 1PROCEDURE: Part B: Latent Heat of Vaporization

1. Set up the apparatus as shown in Figure 1. The hose with the pinch clamp can be drained by momentarily opening the clamp, if the water trap begins to fill with liquid water.

2. Determine the mass of the inner cup and stirrer of the calorimeter.

3. Fill the inner cup to about 2/3 full with cold water at about 15 °C below room temperature.

4. Determine the mass of the inner cup, stirrer and water. Calculate the mass of water in the cup.

5. Place the cup, stirrer, collar and water into the outer calorimeter jacket and record the initial temperature just before steam is introduced into the water.

6. Stir the contents occasionally while constantly observing the rise in temperature. Drain the water from the water trap by occasionally opening the clamp on the hose that drains into the sink. Only steam, and no water, should enter the hose leading to the calorimeter. When the temperature has increased to about 15o above room temperature, remove the steam inlet and record the final temperature.

7. Determine the mass of the cup, stirrer and contents. The mass of the steam which was introduced and condensed into water can now be calculated.

8. Calculate the latent heat of vaporization of water Lv, and compare this to the accepted value by calculating the percent

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difference.

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19. THE MECHANICAL EQUIVALENTOF HEATPURPOSE:

Mechanical work (in joules) can be converted into heat (in calories), permitting a conversion factor to be determined between the two.

INTRODUCTION:

In calorimetry, the calorie is a convenient unit for measuring the flow of energy into substances, often immersed in water. In mechanics, the Joule is the mks unit of energy. The conversion between them can be calculated by heating up an object through the exertion of a force through a distance. The heat may be calculated in calories, the work may be calculated in joules, and the Law of Conservation of Energy applied to equate the two.

APPARATUS:

C-clamp, smallMechanical Equivalent of Heat apparatusCord, strong & flat, 2 meters longTube-O-Lube graphite

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Mass holder, 1-kgMass, slotted, 1-kgMass, slotted, 2-kg (4)Multimeter (ohmmeter)Masking tapeBanana wire (2)

PROCEDURE:

1. Attach the apparatus firmly to the side of a table, using a small C clamp for increased stability. Wrap the cord around the drum several times so the upper end hangs free. Attach 10.0 kg of mass to the lower end, hanging freely, taped securely. Add a small amount of graphite to the drum if the cord seizes while the drum is rotated. Turn the ohmmeter on and plug it into the apparatus. Set the counter to zero by rotating the black dial counter-clockwise.

2. Record the initial reading Ni of the counter and the resistance Ri measured by the ohmmeter. Rapidly rotate the handle through 100 revolutions while the instructor monitors the tension on the free end of the cord. Record the final reading Nf of the counter and the resistance Rf. Another group may begin taking their measurements immediately.

3. On the conversion table on the next page, find the resistance readings that are above and below your value of Ri, and their respective temperatures. Interpolate to find the temperature associated with your value of Ti, and record it to a precision of one-hundredth of a degree. Follow the same procedure to find Tf.

4. The work done may be calculated by noting that the cord applied a force of 98.0 newtons as friction along the surface of the aluminum cylinder. The circumference of the cylinder is C = 0.1495 meters, so the cylinder’s surface traveled a distance (Nf - Ni)C in meters against the motionless cord. The work done by friction is force times distance, so

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W = 98.0·(Nf - Ni)C

in units of joules.

5. The mass of the aluminum cylinder is m = 202.8 grams, and aluminum has a specific heat of cAl = 0.22 calories/gram·°C, so

Q = m·cAl(Tf - Ti)

in units of calories. Note that one food Calorie equals1,000 calories. You did less than one food Calorie of work!

6. Calculate the conversion factor between calories and joules, and calculate its percent difference from the accepted value of 1 calorie = 4.186 joules.

Thermistor Conversion Table

R T R T R T(k) (°C) (k) (°C) (k) (°C)

351.02 0.00 100.000 25.00 33.591 50.00332.64 1.00 95.447 26.00 32.253 51.00315.32 2.00 91.126 27.00 30.976 52.00298.99 3.00 87.022 28.00 29.756 53.00283.60 4.00 83.124 29.00 28.590 54.00269.08 5.00 79.422 30.00 27.475 55.00255.38 6.00 75.903 31.00 26.409 56.00242.46 7.00 72.560 32.00 25.390 57.00230.26 8.00 69.380 33.00 24.415 58.00218.73 9.00 66.356 34.00 23.483 59.00207.85 10.00 63.480 35.00 22.590 60.00197.56 11.00 60.743 36.00 21.736 61.00187.84 12.00 58.138 37.00 20.919 62.00178.65 13.00 55.658 38.00 20.136 63.00169.95 14.00 53.297 39.00 19.386 64.00161.73 15.00 51.048 40.00 18.668 65.00153.95 16.00 48.905 41.00 17.980 66.00146.58 17.00 46.863 42.00 17.321 67.00139.61 18.00 44.917 43.00 16.689 68.00133.00 19.00 43.062 44.00 16.083 69.00126.74 20.00 41.292 45.00 15.502 70.00120.81 21.00 39.605 46.00 14.945 71.00115.19 22.00 37.995 47.00 14.410 72.00109.85 23.00 36.458 48.00 13.897 73.00

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104.80 24.00 34.991 49.00 13.405 74.00

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