Summary of uniformly accelerated motion in terms of definitions and graphs:

All of the theory of uniformly accelerated motion comes from two basic definitions  and one fact about the v vs. t graph:

• Average velocity is average rate of change of position with respect to clock time, so vAve = `ds / `dt.
   
• Average acceleration is average rate of change of velocity with respect to clock time, so aAve = `dv / `dt.
   
• If the v vs. t graph is a straight line, then the average of v over a given interval occurs at the midpoint of that interval (from which it follows, among other things, that vAve = (vf + v0) / 2).

In terms of graphs, we first observe the following:

• The change in position over a given time interval is the 'rise' of the s vs. t graph over that interval, from which it follows that slope = rise / run for a given time interval is `ds / `dt, or average velocity for that interval.
• The change in velocity over a given time interval is the 'rise' of the v vs. t graph over that interval, from which it follows that slope = rise / run for a given time interval is `dv / `dt, or average acceleration for that interval.

It follows from what has gone before that:

• The change in position over a given time interval is equal to the corresponding area beneath the v vs. t graph.

Very similar reasoning tells us that

• The change in velocity over a given time interval is equal to the corresponding area beneath the acceleration vs. t graph.

To solve a uniform acceleration problem:

• List the symbols v0, vf, `ds, `dt, a as well as `dv and vAve
• Determine which of these quantities you are given or can immediately determine from the given information.
• List these 'known' quantities at the top of a 'flow diagram'.
• Use the definitions of ave vel and ave accel, along with a v vs. t graph, to figure out everything you can. 

(The v vs. t graph is linear and the average velocity occurs at the middle of the time interval, with v0 and vf at the beginning and the end of the time interval.)

(Use units for every quantity).

• List the equations of motion.
• Circle all the 'known' quantities.
• Find the equations that can be solved for 'unknown' quantities it terms of 'known' quantities. 
• Solve those equations and check that the units work out.
• Substitute the 'known' values (including units) and carefully calculate the 'unknown' values and their units. 
• Make sure the units work.
• Make sure your solutions are consistent.

To analyze today's experiment:

• Analyze the motion from A to B.
• Analyze the motion from B to C.
• Analyze the motion from C back to B.

To 'analyze' means to find the values of all the variables v0, vf, `dt, `ds, a as well as vAve and `dv, using the above procedures.

To find the components of the gravitational force vector for an object on an incline:

• Sketch the incline and the object on the incline.
• Sketch a vector representing the weight of the object.  The weight is vertically downward (in the direction toward the center of the Earth).
• Sketch a set of right-handed x-y coordinate axes with origin at the center of the object and with the x axis parallel to the incline, oriented either up and to the right or down and to the right (depending on whether the incline is sketched ascending from left to right or descending from left to right).
• Project the vector onto the x and y axes and sketch the resulting components.  Each component runs along its axis from the origin to the point of projection.
• Estimate the magnitudes of the parallel and perpendicular components.

• Determine the angle theta of the weight vector, measured in the counterclockwise direction from the positive x axis.  The x and y components will be, respectively, weight * cos(theta) and weight * sin(theta).  The resulting sign of each component indicates whether it is in the direction of the corresponding axis, or in the opposite direction.  The opposite direction is indicated by a negative result.
• Compare your results with the estimates you obtained from your sketches.  If the directions indicated by your results do not make sense, or if the relative magnitudes of the components do not agree with the relative magnitudes indicated by your sketch, then 'debug' your solution by making sure your sketch is reasonably accurate, that you have correctly determined the angle theta, that you have entered the numbers accurately into your calculator, and that the mode of your calculator agrees with the units you used for the angle (e.g., if the angle is in degrees your calculator need to be in degree mode).

To understand the forces on an object moving on a rigid incline under the influence of gravity and friction:

• The gravitational component parallel to the incline contributes directly to the net force on the object.
• The gravitational component perpendicular to the incline has no direct effect on the motion of the object.  This component slightly bends or compresses the material of the incline, and the elastic response of the incline is precisely equal and opposite to the 'perpendicular' gravitational component.
• The frictional force arises in response to the force which 'presses' the object to the incline and the elastic response which 'pushes back'.  In this case the 'pressing' force is the perpendicular component of the gravitational force.  The amount of the frictional force can be regarded as a certain percent of the 'pressing' force, which we call the 'normal force'.  The percent depends on the nature of the two materials and also on the nature of the motion of the object across the surface.  For example, rolling friction of a typical rubber tire on asphalt is about 2% of the normal force, whereas the sliding friction of dry rubber on dry asphalt is about 80% of the normal force.  We use the word 'coefficient of friction' to express this percent, and we usually express the percent as a decimal, so that the coefficient of rolling friction of the tire would be .02 and the coefficient of friction between dry rubber and asphalt would be .80.
• On inclines less than about 6 degrees the normal force will be greater than 99.5% of the weight of the object, and for many experiments the normal force can therefore be taken to be equal to the weight of the object.

Force and acceleration

• If net force F_net acts on an object of mass m the acceleration of the object will be a = F_net / m.
• A simple rearrangement of this equation is F_net = m a.  The units of m are kg, the units of a are m/s^2 so the units of F_net are kg * m/s^2, also called Newtons, abbreviated N.
• When a force in Newtons is divided by a mass in kg the calculation a = F_net / m yields units of N / kg, which is the same as kg m/s^2 / kg, which simplifies to just m/s^2 (just as we would hope, since we are calculating acceleration a).

The idea of work/energy in terms of a 2-ramp system

• When an object speeds up, it gains what we call kinetic energy (KE).  So when a ball rolls from rest down an incline from A to B, then up another incline from B to C, its kinetic energy increases between A and B, then decreases from B to C.  If the object comes to rest at C, its kinetic energy at C will be 0, equal to its kinetic energy when it was at rest at A.
• When an object in the vicinity of the Earth moves closer to the center of the Earth, it loses gravitational potential energy ('potential energy' is abbreviated 'PE'), and when it moves further from the center of the Earth it gains gravitational potential energy.  When the ball rolls from A to B it moves 'downward', i.e., closer to the center of the Earth, so it loses gravitational PE.  When it rolls from B to C it moves 'upward' and hence gains gravitational PE.
• The object on the incline from A to B therefore loses gravitational PE while gaining KE, and from B to C gains gravitational PE while losing KE.
• A steel ball on a grooved metal incline will end up almost as far from the center of the earth as it started; i.e., its final 'altitude' at C will be almost as great as its initial 'altitude' at A.  However, it won't be quite as 'high' (i.e., not quite as far from the center of the Earth), so it will have lost a bit of gravitational PE in the process.
• The KE gained by the ball from A to B will be almost equal to the PE lost from A to B, and the PE gained from B to C will be almost equal to the KE lost from B to C.
• The total mechanical energy of the ball is equal to the sum of its KE and its PE.  From A to B, it loses a little more PE than it gains in KE, so the total mechanical energy of the ball decreases.  From B to C, the ball loses a little more KE than it gains in PE, so its total mechanical energy again decreases. 
• From A to C, the KE does not change (it's zero at both points) but there is a bit of a loss of gravitational PE, so again we see that the total mechanical energy decreases.
• These decreases in total mechanical energy are mostly the result of frictional losses, which end up being dissipated as the random thermal energy of the particles in the ball and the ramp, and ultimately in the surrounding air.  Losses also occur to vibrational motion (sound waves), which also ultimately end up in the form of random thermal energy.
• Thermal energy is the kinetic and potential energy of the particles that make up the solids, liquids and gases in a system. 
• Mechanical energy can be converted from one form to another (e.g., in the present system from gravitational PE to KE and vice versa), and mechanical energy can be converted to thermal energy.
• Thermal energy can be partially converted to mechanical energy, provided we have two objects at different temperatures so that thermal energy can 'flow' from one object to the other.  The greater the temperature difference, the more of the existing thermal energy can be converted, but it is impossible to convert it all, and the proportion of the thermal energy that can be recovered is strictly limited by the two temperatures.

Analysis of Data for Energy Conservation Experiment 1:

You should have data for the experiment in which you observed a ball descending one ramp from rest starting as point A, rolling directly at point B to a second ascending ramp then rolling to rest at point C on the second ramp.  In this experiment you will have measured the vertical and horizontal positions of the ball, information from which you can determine the slope of each ramp, and will have information from which you can determine the clock times at points A, B and C.

You are given the following information:

• The mass of the ball may be assumed to be .050 kg.  This is not accurate but it will be very easy to modify the analysis for the accurate mass, in those calculations where the mass matters.  The mass won't matter in some of the calculations.
• The acceleration of an object of mass m subject to net force F_net is a = F_net / m.  This can also be expressed as F_net = m a.  Another expression for F_net is sum(F), where 'sum' is often written as a summation sign.  When we multiply m * a the MKS units are therefore units of mass multiplied by units of acceleration, or kg * m/s^2.  The unit kg * m/s^2 is called a Newton.
• The force gravity exerts on an object will, in the absence of other forces, accelerate that object at 9.8 m/s^2.  This force is called the 'weight' of the object.  Therefore weight = m * g, where g is the acceleration of gravity.
• The work done by a force F acting through displacement `ds parallel to F is `dW = F * `ds.  The boldface indicates that F and `ds are vector quantities, meaning that you have to be careful about signs.  (In particular note that when the force and the displacement are parallel and in the same direction, F and `ds will have the same sign and will therefore give us a positive product; but when the directions are opposite, F and `ds have opposite signs so that the product is negative).
• The unit of work is therefore equal to the unit of force multiplied by the unit of distance, i.e., Newtons * meters.  The unit N * m is called a Joule.  Since the unit N is also expressed as kg m/s^2, the unit of work is Joule = N * m = kg m / s^2 * m = kg m^2 / s^2.  The unit 'Joule' is abbreviated 'J'.
• When gravity does work on an object, the gravitational potential energy of the object changes by an amount equal and opposite to that work.  (So, for example, when gravity does negative work on an object its potential energy increases, and when gravity does positive work the potential energy decreaes).
• It follows that, provided the gravitational acceleration remains constant (which will be the case if the percent change in the distance from the center of the Earth is negligible), the change in gravitational potential energy when an object of mass m changes its vertical position by amount `dy is m g `dy.  Note that the units of this calculation are kg * m/s^2 * m = kg m^2 / s^2, or Joules.
• The kinetic energy of a mass m moving with speed v is KE = 1/2 m v^2.  (This is equal to the work required to accelerate the mass from rest to speed v).
• The angle of the incline is arcTan(incline slope).  The arcTan function is typically found on your calculator as 2d fn   Tan .

Using this information for one run of the experiment:

• Analyze the motion from A to B.
• Analyze the motion from B to C.
• Calculate the slope of each incline.
• Calculate the angle of each incline.
• Calculate the weight of the ball.
• Sketch the ball on each incline using conventions covered during the last week, sketching and estimating the x and y components of the weight of the ball.
• Using the weight and the angle of each incline, calculate the x and y components of the weight of the ball.
• Using the x component of the weight, and the mass of the ball, calculate the acceleration of the ball on each incline.  Assume that the frictional force acting on the ball is negligible.
• Compare with the accelerations you calculated when you analyzed the motion of the ball.
• Calculate the work done by gravity on the ball from A to B, and from B to C, based on the displacement along each incline.  Remember that if you use the displacement along the incline, you can use only the component of the force which runs parallel to the incline.  This is because F and `ds must be in parallel directions.  Be sure to consider the signs of F and `ds.
• Using the velocity at B, as determined from your analysis, find the kinetic energy of the ball at point B.
• Using m g `dy, find the change in potential energy from A to B, and from B to C.  Be careful of the sign of `dy.  Remember that `dy is the change in vertical position.

Be sure you have included units at every step of every calculation, and be sure you have actually done the algebra of the units.

How negligible is the percent change in the distance of the ball from the center of the Earth during this experiment?

Obtaining the formula for KE and, while we're at it, the work-energy theorem:

If an object of mass m and initially moving at velocity v0 is subject to a net force F_net while it moves through displacement `ds, then:

• We can easily determine its acceleration, which is a = F_net / m.
• We then know its initial velocity v0, the displacement `ds and the acceleration a.
• We can apply the fourth equation of uniformly accelerated motion to get the final velocity vf.
• We obtain  vf^2 = v0^2 + 2 * (F_net / m) * `ds.
• We can do a little algebra and get this equation into the form F_net * `ds = 1/2 m vf^2 - 1/2 m v0^2.
• F_net * `ds is the work done by the net force on this interval.
• 1/2 m vf^2 is the kinetic energy of the mass m at velocity vf, and 1/2 m v0^2 is its kinetic energy at velocity v0.  So the right-hand side is 1/2 m vf^2 - 1/2 m v0^2 = KEf - KE0 = `dKE.
• Thus F_net * `ds = `dKE, or using `dW_net for the work done on the interval by the net force,
• `dW_net = `dKE.

Discrepant Event:

Using 1/2 m v^2 as the expression for KE in our analysis of the 2-ramp experiment, we will find that the PE gain of the steel ball on the steel track, between points B and C, is greater than the KE loss.  So with no application of external forces, the system we are observing ends up with more mechanical energy at C than at B.

This result is not consistent with the law of energy conservation.

The problem is that 1/2 m v^2 doesn't include the entire kinetic energy of the ball.  v is only the velocity of the center of mass of the ball.  The ball is also rotating, so each of the particles that make up the ball has an organized, non-random velocity of rotation about the center, and this velocity also contributes to the total kinetic energy of the ball.

In this experiment, the total kinetic energy of the ball is somewhat more than 7/5 as great as the translational KE expression 1/2 m v^2 would indicate.  If the ball was rolling without slipping on an inclined plane, the 7/5 would be exact.  However the ball is rolling, most likely without slipping, in a grooved track, and this causes the relative proportion of rotational KE to be even higher.  The analysis of the rotational motion on an inclined plane will fall within the scope of this course, later in the term.  The analysis for the grooved track will be an optional special topic addressed by only a few students, if any.

However we still need to obtain data accurate enough to verify this discrepant event, and ultimately to verify energy conservation for this system.