See also conservation of energy on an incline, short version, which is appropriate for all Principles of Physics students, and for General College Physics students whose lab goal is to just pass the required lab portion of the course.

Note that the data program is in a continual state of revision and should be downloaded with every lab.

Hypothesis: The total of the potential and kinetic energies is nearly constant as a ball rolls down a grooved track, except for a small friction loss.

In this experiment we test the hypothesis that, for a ball rolling without slipping on a constant incline, the increase in the kinetic energy of the ball is very nearly equal to the decrease in its potential energy, and that the difference is within experimental error equal to the work done against friction.

See CD EPS01 for Lab Kit Experiment 16.  The video file shows a toy car rolling down an incline; in this experiment we will use a ball on an incline.

The kinetic energy in this experiment includes both the translational kinetic energy, which is the kinetic energy 1/2 m v^2 associated with the motion at velocity v of the center of mass, and the rotational kinetic energy, the kinetic energy associated with the 'spinning' motion of the ball about its center of mass.

As in previous experiments, we will determine the velocity of the ball by using its projectile behavior as it rolls down then off of the end of the incline before falling to the floor.  You will be able to use the data analysis program to determine the translational velocity attained by the ball.

Once the translational velocity and hence the translational kinetic energy is determined, the dimensions of the ball and the ramp will determine the ratio of rotational to translational kinetic energy. 

Using a 'constant-velocity ramp' you will determine the slope necessary to overcome friction, which provide a good indication of the energy lost to friction.

Once you know the translational and rotational kinetic energies gained by rolling down the incline, and the energy required to overcome friction, you will be able to compare the total of these energies with the potential energy loss of the ball as it rolls down the incline.

Begin by setting up the ramp and determining the slope at which the frictional force and the component of the gravitational force directed down the plane are in equilibrium. From this we will determine the frictional force for small slopes.

The key is to use a ramp that is just barely steep enough that the ball will travel up then 'turn around'.

Note:  If you have only the 15 cm ramp and a large marble, rather than the 30 cm ramp and steel ball, it is still possible to get good results for this experiment.  Use dimes instead of quarters in order to get a ramp just barely steep enough to allow the marble to turn around and come back down (other combinations of coins may also be useful; for example a quarter on one end and a dime on the other will make the ramp even less steep).  It isn't hard to get a 15-cm ramp on which the ball takes at least a couple of seconds to travel up, and a couple more seconds to travel back down.

 Use the larger of the steel balls provided with your lab kit; if you don't have the steel balls use the larger marble..

• Place the incline on a reasonably level tabletop or countertop.  Place a quarter under the right end of the incline.  Place the ball at the 'bottom' of the incline and give it a nudge, sufficient to carry it at least halfway up the incline.  If the ball rolls up the incline, then turns around and rolls back down, then the quarter 'works' for right end.
• Then place the quarter under the left end of the incline, and repeat the trial.  If the ball rolls up the incline, then without any interference reverses direction and rolls back down, then the quarter 'works' for the left end and you can proceed.
• If the quarter didn't 'work' both ways, then try again using two quarters.  Keep increasing the number of quarters one at a time, until the system 'works' both ways.
• Make a note of the number of quarters you end up using.

• Place your quarter(s) under the right edge of the incline, place the ball a couple of centimeters above the other end and give the ball a quick nudge.  Time the ball as it travels up the incline, then as it travels back down.  Also note the point at which the ball turns around (this can be done by placing a marker at the 'highest' point of the ball's path).
• Record the distance up and the distance down, as well as the time up and the time down.
• Repeat until you have 5 trials.  The distances don't have to be the same every time, but the times up and the time back down should in each case (each) be at least 2 seconds.

In the box below report in the first line the distance up, the time required to travel up, the distance down and the time to travel down the incline on the first trial.  In the second thru the fifth lines, report the same information for your subsequent trials.  Starting in the next line, indicate the meanings of the numbers you have reported.

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In the box below report, one trial to a line in comma-delimited format, the acceleration up and the acceleration down the incline, and the difference in these accelerations.  Beginning in the sixth line explain how you determined your accelerations from the given data.  Be sure your explanation connects your determination of acceleration result to the initial data, and be sure to include an explanation of the algebra of the units.

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Now place the quarters under the other end and repeat to obtain at least 5 trials.

• For each trial determine the acceleration of the ball as it travels up the incline, and as it travels back down.

In the box below report in the first line the distance up, the time required to travel up, the distance down and the time to travel down the incline on the first trial.  In the second thru the fifth lines, report the same information for your subsequent trials. 

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In the box below report, one trial to a line in comma-delimited format, the acceleration up and the acceleration down the incline, and the difference in these accelerations.  Beginning in the sixth line explain how you determined your accelerations.

Accelerations are calculated in the usual manner, assuming uniform acceleration.

• Acceleration can be determined either by direct reasoning from definitions (using the displacement and time interval you can find the average velocity; if the initial or final velocity is zero you can infer the final or initial velocity; from the change in velocity and time interval you can find the acceleration).

• Acceleration can be determined by using the equations of uniformly acceleration motion. 

Acceleration up and acceleration down have the same direction; in both cases the acceleration is down the ramp. Depending on which direction you choose as positive, both accelerations will be either positive or negative.

• If the direction up the incline is regarded as positive, then the motion up the incline is characterized by positive `ds and v0, with vf = 0, so that that change in velocity and therefore the acceleration are negative.  The motion down the incline will be characterized by negative `ds, v0 = 0 and negative vf, so that the acceleration would also be negative.

• If the direction down the incline is regarded as negative, then the motion up the incline is characterized by negative `ds and v0, with vf = 0, so that that change in velocity and therefore the acceleration are positive.  The motion down the incline will be characterized by positive `ds, v0 = 0 and positive vf, so that the acceleration would also be positive.

It would not be appropriate to choose different directions as positive when the object undergoes continuous motion. Even if the motion wasn't continuous, there would have to be a good reason before we would choose different positive directions for different trials.

STUDENT QUESTION:

Shouldn't g, the acceleration of gravity, be involved here?

To calculate these accelerations I used `ds = (v0 + vf)/2 * dt and a = (vf – v0)/`dt with vf = 0 cm/s on the way up and v0 = 0 cm/s on the way down. Is this the correct way we are supposed to calculate it? I thought I should be using a formula with g but it doesn’t account for `ds. Also using these calculations, if I get the percentage of gravitation force that the acceleration is, I come up with a very small percentage.

INSTRUCTOR RESPONSE:  You should indeed come up with a very small percentage of g.

The ball isn't undergoing free fall, so there's no reason to assume acceleration g.   We are in any case measuring the acceleration from actual data, as opposed to making a theoretical prediction based on accepted values.

The ball is accelerating due to the component of the gravitational force parallel to the incline. Since the incline is nearly horizontal, this component will be very small compared to the gravitational force, and the acceleration will therefore be a very small percentage of the gravitational acceleration.

Determine the mean difference in accelerations for the first set of 5 trials and the standard deviation of these differences.  Give your results as 2 number is the first line, delimited by commas.  Give the mean and standard deviation of the acceleration differences for the second set of 5 trials in the second line.

In the box below, explain why you think the magnitude of the acceleration up the incline is greater than the the magnitude of the acceleration down, and in the process consider the following questions:

• What force is it that is responsible for the difference? 
• Is the work done on the ball by this force positive or negative as the ball travels up the incline? 
• Is the work done on the ball by this force positive or negative as the ball travels down the incline? 
• Does your answer depend on whether the incline is sloped to the right or to the left?

The rolling friction of the ball on the incline is responsible for the difference in the accelerations.  Rolling friction enhances the acceleration of gravity when the ball is traveling up, and works against the acceleration of gravity when the ball is traveling down the incline.

Since rolling friction acts opposite the direction of motion, and since motion up the incline is in the direction opposite the (parallel component of the) acceleration of gravity, friction increases the magnitude of the acceleration when the ball travels up the ramp.  When rolling down the ramp the frictional force, being directed opposite the motion, is opposed to the gravitational force and hence reduces the magnitude of the acceleration.  So the magnitude of the acceleration down is less than the magnitude of the acceleration up, and the time down the ramp will be greater than the time up.

However the difference in the accelerations is small.  Not all students will be successful in timing the ball with sufficient precision to detect the difference in accelerations.  On trials where it takes about 2 seconds for the ball to travel up, and about 2 seconds to travel back down the incline, the difference in times is only around .1 second.  If an individual has a tendency to anticipate or delay the timing at the beginning, at the top and at the end of the incline, then the small difference in times might get lost in the 'noise' of the experimental uncertainty. 

The less the slope of the ramp, the greater the times required to travel up and down, and the greater the difference in the times.  To get good results on this experiment it is best to make the slope as small as possible.

COMMON STUDENT ANSWER:

The magnitude of the acceleration is greater up the incline, because we are giving it the initial push. The push that we give
it helps it overcome the friction. When the ball stops on top of the ramp and turns around it must overcome this on its own.
The friction force will be negative for either direction.  The magnitude of the acceleration down also takes into account the turn-around time at the top of the ramp.

INSTRUCTOR RESPONSE:

The push is over, or should be, at the instant the timing starts. The only effect of the push is to give the ball its initial upward velocity. The push does nothing during the timed intervals.

The point is that friction helps gravity slow the ball on the way up, while it acts against gravity's tendency to speed the ball up as it descends.

The turn-around is pretty much instantaneous. However it's difficult to see exactly when that occurs, so there is significant uncertainty in timing associated with the turn-around.

You may if you wish do the remainder of this experiment using your data from the experiment 'Uniformity of Acceleration for a Ball on a Ramp' (that experiment is in turn a continuation of the experiment Ball and Ramp Projectile Behavior, which preceded it).  In those experiments you set up '2-, 3- and 4-domino ramps' and carefully observed the range of the ball as it rolls down and off the ramp.  Based on the ranges and the vertical drop, you used the data analysis program to determine the speed of the ball at the end of the ramp.

•  If you have only the 15-cm ramp then you would have used half the specified number of dominoes (i.e., 1 domino instead of the specified 2, and 2 dominoes instead of the specified 4), and your 10, 20 and 30 cm rolls were probably 5, 10 and 15 cm).
• The data program requests the number of dominoes, from which it calculates a slope based on a 30-cm ramp.  A 15-cm ramp has only half the 'run' of a 30-cm ramp, so if you used a 15-cm ramp you would enter double the number of dominoes you actually used.  That is, if you use 1 domino on a 15-cm ramp, tell the program you used 2; if you used 2 dominoes tell the program you used 4.  If you happened to use 3 or 4 dominoes, you would tell the program you used 6 or 8, respectively.

If not, you should review the instructions for those experiments, set up ramps with 2 and 4 dominoes (1 and 2 dominoes if using a 15-cm ramp).  Using 5 trials you should obtain horizontal ranges for the ball rolling distances of 15 cm and 30 cm along each ramp (use 7.5 and 15 cm if you have only the 15-cm ramp), and calculate the mean and standard deviation of ranges for each set of 5 trials.  You will have a total of 4 five-trial means and standard deviations to report.

If you use the data from the experiments you will have information for 10, 20 and 30 cm rolls on 2- and 4-domino ramps, for a total of 6 five-trial means and standard deviations, which you will have reported once already (you should report once more).

In the box below report in the first line the number of dominoes, the distance of the roll down the ramp, and the mean and standard deviation of the horizontal range for your first setup.  In the second line report the same information for your second setup, etc., until you have reported your results for all of your setups (4 setups if you obtained new data for the experiment, 6 setups if you used your old data).  In the first subsequent line, give the distance the ball fell after leaving the end of the ramp.

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If you are using data obtained specifically for this experiment, you will need to use your mean horizontal ranges to find the corresponding velocities.  The instructions for doing so were given in the previous experiments, and are repeated here for easy access:

Using the 'Experiment-Specific Calculations' button, select 1, as you did in the preceding experiment, and respond with the information necessary to calculate the speed of the ball at the end of the ramp, based on the mean distance observed for your first set of 5 trials.  (Remember to enter double the actual number of dominoes if you used the 15-cm ramp, a discussed above).

In the box below, report in the first line the number of dominoes, the distance of the roll and the final velocity on the ramp for the first trial.  In subsequent lines report the same information for subsequent trials:

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Measure the height of a stack of 4 dominoes, as accurately as you can.  In the box below give in the first line the height per domino.  In the second line describe how you made your measurement and how you then determined the height per domino.

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For the 2-domino ramp, the ramp descends through a distance equal to the height of two dominoes while rolling a distance of about 30 cm.  In the box below, give in the first line the total descent of the ball for the entire 30 cm of roll.  In the second line give the amount of descent per cm of roll.  Starting in the third line explain how you obtained your two results.

Note that 'descent' is clearly defined in the above paragraph being equal to the height of two dominoes; 'descent' refers to the vertical distance through which the ball descends.  If the ball rolls 30 cm, it does not descend 30 cm.  30 cm would be the distance of roll for the trial in which the domino rolled the entire length of the ramp.  Its descent for the entire 30-cm roll will in this case be about equal to the height of the two dominoes (perhaps a little more, depending on exactly where the dominoes were positioned).

For example if 4 dominoes have a height of 3.6 cm, and support a ramp of length 30 cm, then 3.6 cm of descent corresponds to 30 cm of roll along the ramp.  This means that the rate of descent is 3.6 cm / 30 = .12 cm per cm of roll.

If 1 domino has a height of .9 cm and supports a ramp of length 15 cm, then the rate of descent it .9 cm / 15 = .06 cm of descent per cm of roll.

For the 2-domino ramp, use the preceding results to determine the amount of descent that would correspond to rolls of 10 cm, 15 cm, 20 cm and 30 cm.  Give these results as 4 numbers in the first line, delimited by commas.  In the second line explain how you obtained your results.

A '2-domino ramp' of length 30 cm, with dominoes of height .9 cm, has a rate of descent equal to .06 cm of descent per cm of roll.  Note that this is just another way of saying that the ramp has a slope of .06.  A roll of 20 cm along this ramp would correspond to a descent of .06 (cm / (cm of roll) )* 20 cm of roll = 1.2 cm.

Using a similar strategy find the descent corresponding to rolls of 10 cm, 15 cm, 20 cm and 30 cm on a 4-domino ramp and report in similar syntax in the box below:

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For your remaining calculations, assume that the mass of the ball is 100 grams.  This isn't completely accurate, but the mass of the ball isn't critical and if necessary results could later be adjusted very easily for the accurate mass.

Determine how much work it would take to raise the mass of the ball, against the downward gravitational pull, through each of the distances of descent you have calculated.  In the box below indicate in the first line the first distance of descent, then the work.  In subsequent lines give the remaining distances of descent and corresponding amounts of work.  In the first line following your data lines, specify the units of the quantities you have given, and explain how you obtained your results.

For example if the distance of descent for a certain roll is 1.2 cm, then the magnitude of the PE change would be 1.2 cm * 100 g * 980 cm/s^2 = 120,000 ergs, approx., or .012 m * .1 kg * 9.8 m/s^2 = .012 Joule.

The differences in acceleration previously observed, between the ball traveling up the ramp and the ball traveling down the ramp, are due to the change in the direction of the frictional force. 

• At any point on the ramp, the gravitational force has a component along the ramp, which is the same whether the ball is moving up or down. 
• However the frictional force is always in the direction opposite the motion, so when the ball is moving up the ramp the frictional force is in the same direction as the gravitational force component along the ramp, whereas when the ball is moving down the ramp the frictional force is in the opposite direction.
• It follows that the difference in net force while the ball is traveling up the ramp, and the net force while the ball is traveling down the ramp, is double the force of friction.

Based on the above:

• What was the average of the differences previously observed between the two accelerations?  Answer in the first line of the box below.
• On a 100-gram ball, how much difference in force would be required, to result in this much acceleration?  Answer in the second line of the box below.
• What therefore is the force of friction on the ball?  Answer in the third line of the box below.
• Starting in the fourth line, explain how you obtained your results.

Example:  The difference in the accelerations is usually reported to be around 5 cm/s^2, but there is significant uncertainty in timing which results in significant uncertainty in these results.  An acceleration of 5 cm/s^2 would be associated with a force of 100 grams * 5 cm/s^2 = 500 dynes, or .1 kg * .05 m/s^2 = .005 Newton. 

If the difference between upward and downward forces is .005 Newtons, then

Rolling friction is directed downward the incline as the ball travels up the incline, and up the incline as the ball travels down the incline.  The component of the gravitational force parallel to the incline is the same throughout, so the difference in accelerations is due to the difference in the frictional forces.  Since the frictional force reverses direction when the ball does, the magnitude of the difference in forces is double the magnitude of the frictional force. 

For the present example the difference in the forces is .005 N, so that the magnitude of the frictional force would be half this, or .0025 N.

For each distance of roll, determine, based on the frictional force and the distance of the roll, the work done by friction on the ball.  In the box below report distance of roll and work against friction, in comma-delimited format, reporting the results for one distance per line.  In the first line after your data report, indicate how you obtained your results.

For example if the difference in accelerations is 5 cm/s^2 the frictional force on a 100 gram ball would have magnitude .0025 N, as in the preceding example.  If the distance of roll is, say, 20 cm, then the frictional force does work -.0025 N * 20 cm = -.0025 N * .20 m = -.001 Joule.  (note that the frictional force and the displacement are in opposite directions so the work is negative)

The translational kinetic energy of the ball is 1/2 m v^2, where m is the mass of the ball and v its velocity.  Give for each setup the number of dominoes, the distance of roll, and the translational kinetic energy attained by the ball, one setup per line in comma-delimited format.  Then indicate the units of your results, and how you obtained your results.

For example if the final velocity is 40 cm/s, the KE of a 100 gram ball would be .5 * .1 kg * (.35 m/s)^2 = .006 Joule.

If this occurs on 20 cm roll, on a ramp with 2 dominoes, you would report

2, 20, .008

In your explanation line you would indicate how you obtained the KE and the units of your results.

The potential energy (PE) loss of the ball as it descends on the ramp is equal to the work required to raise it against gravity through the distance of descent.

If the ball was rolling, without slipping, on a smooth and ungrooved ramp its rotational KE would be 2/5 of its translational KE.  Assuming this to be the case (it isn't, but we'll use it as a first approximation), report in the first line of the box below the number of dominoes, distance of roll, PE loss, work done against friction, translational KE and rotational KE, for the first setup.  You will report 6 numbers in comma-delimited format.  In subsequent lines indicate the same information for each remaining setup.  In the first remaining line, indicate the units of PE loss, work against friction, translational KE and rotational KE.

Continuing the example of the 20 cm roll, assuming a 1.2 cm descent and final velocity 35 cm/s:

As seen previously friction does -.001 Joule of work and the PE decreases by .012 J.  The translational KE is .5 m v^2 = .006 Joule.  The additional rotational KE is taken to be 2/5 this great, so we have rotational KE = 2/5 * .006 Joule = .0024 Joule.  Thus, if you report the results in Joules, the numbers you would report would be

2, 20, .012, .001, .006, .0024.

(If you chose to calculate energies in ergs, the numbers would be 2, 20, 120 000, 10 000, 60 000 and 24 000).

We need to compare PE loss, work against friction, translational KE and rotational KE.  Are your units for these quantities all the same? 

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Standard units of work/energy are:

• Joules, which are the same as N * m, or kg * m^2 / s^2, and
• ergs, which are the same as dyne * cm, or g * cm^2 / s^2.

We expect that PE loss should be equal to total KE gain + work done against friction, where total KE gain is translational KE gain + rotational KE gain.

Using the results you previously reported, but when necessary converting these quantities to consistent units so they can be compared, report in the first line of the box below the PE loss for your first setup, KE gain + work done against friction for that setup, and the second quantity as a percent of the first (i.e., KE gain + work done against friction as a percent of PE loss); you will report three numbers in comma-delimited format.  In subsequent lines report the same information for the remaining setups.  Starting in the first remaining line, give the units of your results, and explain how you obtained your results.

Continuing the example of a 20 cm roll down a 2 domino ramp we have PE loss .012 Joule, work .001 Joule against friction, total KE = translational KE + rotational KE = .006 J + .0024 J = .0084 J.  The total KE and the work against friction add up to .0084 J + .001 J = .0094 J, so we would report

.012 J, .0094 J, 78.   

The number 78 represents the second quantity, .0094 J, as a percent of the .012 J (.0094 / .012 = .78 or 78%). 

We expect that the work against friction plus the total KE will be equal to the loss of PE, as predicted by the law of conservation of energy.  We have accounted for 78% of the PE loss.  Given the degree of experimental error, this might or might not be an acceptable result.