Friction in Line Rider
Is there friction in Line Rider? Does it function as physics would expect? To test this, I set up a simple track:
![Page 6 1](http://blog.dotphys.net/wp-content/uploads/2008/09/page-6-1.jpg)
Basically, a slope with a flat part to start with and to end with. Let me show you something simple before further analysis:
![Page 6 2](http://blog.dotphys.net/wp-content/uploads/2008/09/page-6-2.jpg)
This is the x-position vs. time for the line rider on the first horizontal portion of the track (before he or she goes down the incline). This shows the rider traveling at a constant speed of 0.71 m/s. If friction were present, the rider would slow down. If you do not believe me (and why should you?) try creating your own line rider track with a long horizontal section. The rider will not stop, but continue on at a constant speed.
Ok, so no friction on the horizontal line. This makes a little bit of gaming sense. Who would want a rider to stop in the middle of the track and be stuck? That wouldn’t be fun. But, is there friction on non-horizontal portions? To test this, I will use the work-energy principle.
Work – Energy
Here is a crash course in the work-energy theorem. Basically, the work done on an object changes its energy. (see, that wasn’t complicated). Where work is defined as:
![Page 6 3](http://blog.dotphys.net/wp-content/uploads/2008/09/page-6-3.jpg)
Where F is the force acting on the object and delta r is displacement. Since these are both vector quantities, you can not simply multiply them. In this case the dot product (or scalar product) is used. If you don’t like that, you can use the following instead:
![Page 6 4](http://blog.dotphys.net/wp-content/uploads/2008/09/page-6-4.jpg)
Where F and delta r are now the scalar magnitudes of the vectors and theta is the angle between F and delta r.
For the line rider track, there are only two forces (assuming no or negligible air resistance) acting on the line rider. There is gravitational force and there is the force the track exerts on the rider. The force the track exerts on the rider can be broken into a component perpendicular to the track (called the normal force) and a component parallel to the track – friction.
Below is a diagram (a free body diagram) representing the forces on the rider as he (or she) is going down the incline.
![Page 6 5](http://blog.dotphys.net/wp-content/uploads/2008/09/page-6-5.jpg)
To calculate the work, all forces will need to be included. The work can be calculated one of following ways:
![Page 6 6](http://blog.dotphys.net/wp-content/uploads/2008/09/page-6-6.jpg)
Where the thetas are for the angles between the displacement and each force.
For this case, I will calculate the work for each individual force. First, let us look at the work done by the normal force. The rider is moving down the incline, and the normal force is perpendicular to the incline, so work would be:
![Page 6 7](http://blog.dotphys.net/wp-content/uploads/2008/09/page-6-7.jpg)
Now work done by friction:
![Page 6 8](http://blog.dotphys.net/wp-content/uploads/2008/09/page-6-8.jpg)
Where there is a relationship between the friction force and the normal force (in this model). The harder two surfaces are pushed together, the greater the frictional force. This gives the following relationship between the magnitude of the normal force and the friction force:
![Page 6 9](http://blog.dotphys.net/wp-content/uploads/2008/09/page-6-9.jpg)
Where ? is the coefficient of kinetic friction between the two surfaces (in this case the line rider and the track).
The goal is to calculate the ?, so an expression for the normal force is also needed. For this case, the line rider stays on the track. This means that his velocity perpendicular to the track is zero and stays at zero. If his perpendicular velocity stays zero, his (or her) acceleration must be zero perpendicular to the track (notice that the acceleration is zero because the velocity STAYS zero, not because the velocity is zero. MANY MANY MANY people mess that part up). Anyway, if the acceleration perpendicular to the track is zero, the forces perpendicular to the track must add up to zero (vector sum).
The normal force is already perpendicular to the track. The friction force is not, but the gravitational force has some component in the direction perpendicular to the track
![Page 6 10](http://blog.dotphys.net/wp-content/uploads/2008/09/page-6-10.jpg)
Where the yellow vector represents the component of gravity in the direction perpendicular to the track. Since this creates a right-anlge triangle, the magnitude of this component will be
![Page 6 10](http://blog.dotphys.net/wp-content/uploads/2008/09/page-6-101.jpg)
In this case, the magnitude of the gravitational force is mass of the object times the local gravitational field (approximately 9.8 Newtons per kg). This means that the work done by friction can be expressed as:
![Page 6 12](http://blog.dotphys.net/wp-content/uploads/2008/09/page-6-12.jpg)
Where ?r is the distance along the track, ? is the angle of the incline of the track.
Finally, the work done by gravity. The angle between gravity and ?r is ?c (90 degrees – ?).
![Page 6 13](http://blog.dotphys.net/wp-content/uploads/2008/09/page-6-13.jpg)
Looking at the track, the slope is inclined at an angle ? and has a length of ?r. The expression ?r sin(?) is equivalent to the opposite side of the right triangle, in this case that is the change in height of the rider (?y), so the work done by gravity is:
![Page 6 14](http://blog.dotphys.net/wp-content/uploads/2008/09/page-6-14.jpg)
I think we are finished with work. So, the total work done on the rider while going down the incline is:
![Page 6 15](http://blog.dotphys.net/wp-content/uploads/2008/09/page-6-15.jpg)
That is great. But….Work. What is it good for?
So, I have talked about work. The work-energy relationship says that the work done on an object is its change in energy. In this case, the line rider will only have a change in translational kinetic energy. So
![Page 7 1](http://blog.dotphys.net/wp-content/uploads/2008/09/page-7-1.jpg)
So the change in kinetic energy will be from the top of the incline to the bottom. Putting together from the total work:
![Page 7 2](http://blog.dotphys.net/wp-content/uploads/2008/09/page-7-2.jpg)
Notice that the mass cancels (good because I never really knew that the mass was anyway)
![Page 7 3](http://blog.dotphys.net/wp-content/uploads/2008/09/page-7-3.jpg)
In this expression, I can measure ?y, vlower and vupper. Solving this expression for ?:
![Page 7 4](http://blog.dotphys.net/wp-content/uploads/2008/09/page-7-4.jpg)
![Page 7 5](http://blog.dotphys.net/wp-content/uploads/2008/09/page-7-5.jpg)
![Page 7 6](http://blog.dotphys.net/wp-content/uploads/2008/09/page-7-6.jpg)
So, I can measure the upper and lower velocities and I can measure ?y and ?x. From this I can calculate ?. After that, I will change the incline and see if ? changes (it shouldn’t change).
Measuring ?y and ?x
![Page 7 7](http://blog.dotphys.net/wp-content/uploads/2008/09/page-7-7.jpg)
Using tracker video analysis, I found the coordinates (with respect to the red origin as shown) for the beginning and end of the incline. The beginning is at (4.77 m, -1.00 m) and the end of the track is at (15.29 m, -14.44 m). This gives a ?y = 13.44 meters. (a big hill for a 5 year old to go down) and ?x = 10.52 meters
Velocity at the bottom
![Page 7 8](http://blog.dotphys.net/wp-content/uploads/2008/09/page-7-8.jpg)
This linear fit to the last part of the run shows a horizontal velocity of 13.22 m/s.
Velocity at the top
I previously stated the velocity at the top. It is 0.71 m/s
So, plugging stuff in:
![Page 7 9](http://blog.dotphys.net/wp-content/uploads/2008/09/page-7-9.jpg)
Notice that this is a unitless quantity (as it should be).
A different situation
Now, we can look at a different track with the same change in y, but different slope. The final velocity should be less because friction will be greater in magnitude AND exert over a larger distance. This will mean that friction will do more work and thus reduce the energy gained (friction is doing negative work). However, the coefficient of friction should be the same.
Here is a track with a different slope:
![Page 8 1](http://blog.dotphys.net/wp-content/uploads/2008/09/page-8-1.jpg)
From this, the following position-time data is obtained.
![Page 8 2](http://blog.dotphys.net/wp-content/uploads/2008/09/page-8-2.jpg)
In this graph, it can be seen that the speed at the top of the incline is 0.68 m/s This is slightly different than the 0.71 m/s from the last run and shows the error associated with the data collection (but that is a whole different page that I have not written).
Also, the final speed is 16.25 m/s (faster than before) – this is really important.
From the video, ?x and ?y can be obtained. The point at the top of the slope is (4.67, -0.99) and at the bottom it is (35.38, -13.86). This gives ?x = 30.71 m and ?y = -12.87 m.
![Page 8 3](http://blog.dotphys.net/wp-content/uploads/2008/09/page-8-3.jpg)
What? That is odd. A negative coefficient of friction? That would mean that friction is making it speed up. Suppose there were no friction at all. Then the work-energy equation would say:
![Page 8 4](http://blog.dotphys.net/wp-content/uploads/2008/09/page-8-4.jpg)
Solving for the final velocity:
![Page 8 5](http://blog.dotphys.net/wp-content/uploads/2008/09/page-8-5.jpg)
And plugging in the data from above:
![Page 8 6](http://blog.dotphys.net/wp-content/uploads/2008/09/page-8-6.jpg)
This is slower than with friction. Maybe I need another test.
Another friction analysis
My favorite method for looking at friction is the measure the motion of an object sliding both up and down and inclined plane. Here is the line rider track I created to do that.
![Page 9 1](http://blog.dotphys.net/wp-content/uploads/2008/09/page-9-1.jpg)
The up and down is important because on the way up the track, gravity is slowing the rider down and so is friction (because friction is in the opposite direction to the motion). On the way down, gravity pulls down the incline, but friction is acting in the opposite direction. The result is that the acceleration up and down the incline will be a little different (depending on the coefficient of friction).
Going up the incline
![Page 9 2](http://blog.dotphys.net/wp-content/uploads/2008/09/page-9-2.jpg)
Here friction and the gravitational force are both pointing down the incline.
Going down the incline
![Page 9 3](http://blog.dotphys.net/wp-content/uploads/2008/09/page-9-3.jpg)
Now they are “working against each other”. Going down the incline should have a smaller acceleration than going up.
Newton’s Second Law
Newton’s second law is what relates forces, mass and acceleration. It is most often written as:
![Page 9 4](http://blog.dotphys.net/wp-content/uploads/2008/09/page-9-4.jpg)
– but this is bad way of writing it. A better way would be:
![Page 9 5](http://blog.dotphys.net/wp-content/uploads/2008/09/page-9-5.jpg)
There are two main differences in these equations. The key difference is that the second version is a vector equation (relating vectors). The other difference is the inclusion of Fnet. This says that it is the sum of all the forces that relates to the acceleration.
To make this analysis simpler, we can let one of the coordinate axes be parallel to the incline.
![Page 9 6](http://blog.dotphys.net/wp-content/uploads/2008/09/page-9-6.jpg)
This will allow us to write the vector equation as the following two equations:
![Page 9 7](http://blog.dotphys.net/wp-content/uploads/2008/09/page-9-7.jpg)
![Page 9 8](http://blog.dotphys.net/wp-content/uploads/2008/09/page-9-8.jpg)
So the net force in the y direction (perpendicular to the incline) must be zero.
Going up the plane, the x-motion can be described by:
![Page 9 9](http://blog.dotphys.net/wp-content/uploads/2008/09/page-9-9.jpg)
Where the friction force can be modeled by:
![Page 9 10](http://blog.dotphys.net/wp-content/uploads/2008/09/page-9-10.jpg)
so, up the plane:
![Page 9 11](http://blog.dotphys.net/wp-content/uploads/2008/09/page-9-11.jpg)
Solving for the acceleration (the mass cancels)
![Page 9 12](http://blog.dotphys.net/wp-content/uploads/2008/09/page-9-12.jpg)
The only thing different going down the plane is the direction of the frictional force, so the acceleration would be:
![Page 9 13](http://blog.dotphys.net/wp-content/uploads/2008/09/page-9-13.jpg)
Here is a plot of the x-position (in the frame with the x-axis parallel to the incline) versus time
![Page 9 14](http://blog.dotphys.net/wp-content/uploads/2008/09/page-9-14.jpg)
In this graph, I have fit a quadratic equation to the part with the rider going up the track and a different function for going down. If you recall the scale of the line rider experiment, I described how the acceleration can be found from a quadratic fit. In this case, the accelerations are
Up the incline: ax = – 4.00m/s2
Down the incline: ax = – 4.00 m/s2
This suggests that either the frictional force is too small to be measured, or there is no frictional force (since the acceleration is essentially the same on the way up and on the way down. Another possibility is that there is a frictional force, but it can not be seen because of excessive error in the data collection process. Even if my scale was off (from before), the acceleration should still be different on the way up and on the way down.
Comparing angle of incline to the acceleration
If there is no friction, the acceleration should be related to the angle of the incline. If you remove the frictional force from the previous equations, you get:
![Page 9 15](http://blog.dotphys.net/wp-content/uploads/2008/09/page-9-15.jpg)
Solving for theta:
![Page 9 16](http://blog.dotphys.net/wp-content/uploads/2008/09/page-9-16.jpg)
This gives us a calculated angle of the incline
? = 24.1 degrees.
Looking at the video, the measured angle of the incline is 35.1 degrees.
Evidence for friction
There is evidence of some type of frictional loss of energy. In this track, the rider goes up the incline, then down. He then goes back up another incline. Below is a plot of his y-postion (in the non-rotated frame of reference) versus time.
![Page 9 17](http://blog.dotphys.net/wp-content/uploads/2008/09/page-9-171.jpg)
If there was no friction present, the rider would go back to the same height as before (conservation of energy). In this case, the rider lost some energy.
I am not sure how friction is implemented in line rider. When you play the game, the implementation seems plausible (it doesn’t look weird). It is possible that I am encountering significant errors due to the way the data is obtained. This could be errors introduced through dropped frames in the screen capture, different time rates or error in locating the rider in each frame.
I suspect that friction is implemented by just making the acceleration lower that what it should be for planes (but the same acceleration up and down the plane). I am very sure that there is zero friction on the horizontal surfaces.
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