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course PHY 231
9/21 10:05 pm
#$&*Brief at-home experiment
Is the magnitude of a ball's acceleration up an incline the same as the magnitude of its acceleration down the incline?
Test this.
One possible suggested method: Use the short ramp to get the ball started, and let it roll from the short ramp onto the longer ramp, with the longer ramp inclined so the ball rolls up, rather than down. Get the ball started on the short ramp, either by inclining it toward the long ramp or giving it a push (before it reaches the long ramp). Click the TIMER at the instant the ball hits the 'bump' between the two ramps, again when the ball comes to rest for an instant before accelerating back down the long ramp, and once more when the ball again hits the 'bump'.
Everyone will tend to anticipate their 'clicks', and to try to compensate for their anticipation. If the effect of anticipation is the same for all three timed events, then uncertainties in your results will all be due to the TIMER. If the degree of anticipation differs, as it inevitably will, then the uncertainties are compounded. If the degree of anticipation (and/or compensation) tends to be either greater or less for the second event (the ball stopping for an instant) than for the first and third (the 'clicks' made when the ball hits the 'bump'), then a systematic error is introduced (you will have a tendency to 'short' one interval while extending the other).
If there exists a difference between the times up and down the ramp, and if the difference is great enough to show through the uncertainties and systematic errors, then you might get a useful result (e.g., either the actual times are the same, or they differ).
Give a brief report of your data and your results:
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Data is up ramp time then down ramp time.
0.99, 0.982
1.091, 1.085
0.95, 1.006
1.141, 1.106
1.063, 1.311
0.928, 0.791
0.881, 0.903
0.752, 0.744
0.9511.016
These were done with varying v0, i.e. farther up or down the start ramp. It seems that the time closest to what would suggest equal acceleration were done at about halfway up the start ramp. The higher or lower than that that I went the times seem to have more disparity. Probably somewhat related to the anticipation you mentioned. To me I think that this is like a projectile problem. Something thrown up will land with equal, but opposite in direction, velocity. Here the acceleration the ball experiences is directly due to gravity just it also contributes 'forward', v_x, acceleration due to the incline of the ramp. So how much of the effect of gravity the ball experiences is directly relate to the incline of the ramp. The more acute the incline angle the less the normal reactive force is 'pushing' in the horizontal direction. Because of this I think the acceleration will be the same and the ball will 'leave' the ramp with the same velocity it 'entered' with like a projectile.
Good.
Think about this:
Because of air resistance the ball loses energy going up and coming down, so it in fact has less velocity when it returns to its original point than when it started out.
How does this translate into a statement involving forces? What does this mean for accelerations?
Air resistance is negligible for the ball in this experiment. Are there any forces present that have an effect similar to that of air resistance?
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Displacement and force vector for a rubber band
The points (2, 1) and (4, 6) have the following characteristics:
• From the first point to the second the 'run' is 4 - 2 = 2 and the 'rise' is 6 - 1 = 5.
• The slope of the line segment connecting the points is therefore 5 / 2 = 2.5.
• A right triangle can be constructed whose hypotenuse is the line segment from (2, 1) to (4, 6), and whose legs are parallel to the coordinate axes. The leg parallel to the x axis has length 2, and the leg parallel to the y axis has length 5. The hypotenuse therefore has length sqrt(2^2 + 5^2) = sqrt(29) = 5.4, approx..
• The area of the trapezoid formed by projecting the two points onto the x axis, whose sides are the projection lines, the segment [2, 4] of the x axis and the line segment between the two points, has 'graph altitudes' 1 and 5, giving it average 'graph altitude' (1 + 6) / 2 = 3.5, and width 5 - 1 = 4, so its area is 14.
• The vector from the first point to the second has x component 2 and y component 5. Its length is sqrt(20) and its direction is specified by the ratio of the y to the x component.
Any time you see two points on a graph you should be aware that you can easily construct the right triangle and use it to calculate the slope of the segment joining them, the distance between the points, the area of the associated trapezoid and the components of the vector. Depending on the nature of the graph, some of these quantities will sense and be have useful interpretations, while others probably will not.
In the particular example of a rubber band stretched between the two points, with the coordinates in centimeters, the most important quantities are the components and length of the vector. The vector from the first point to the second is in this case a displacement vector. The displacement vector has x component 2 cm, y component 5 cm and length sqrt( 2 cm)^2 + (5 cm)^2 ) = sqrt( 29 cm^2) = sqrt(29) cm, about 5.4 cm.
• We denote the displacement vector from (2 cm, 1 cm) to (4 cm, 6 cm) as <2 cm, 5 cm>
If we divide the this vector 2 we get a vector of length 2.15 cm. The x component will be 1 cm (half of the original 2 cm) and the y component 5 cm /2 = 2.5 cm (half of the original 5 cm). The ratio of the components is the same, so this vector will be in the same direction as the original vector. If we multiply this vector by 2 we get a vector of length 8.6 cm. Its components will be double those of the original vector, and it will also be in the same direction as the original vector.
If we divide our vector by 5.4 cm, then the resulting vector has length 5.3 cm / (5.3 cm) = 1. The x and y components will be 2 cm / (5.4 cm) = .37 and 5 cm / (5.4 cm) = .92. As before, the ratio of the components is the same as for the original vector (accurate to 2 significant figures),
The vector <.37, .92> has magnitude 1 (calculated to 2 significant figure), as you can easily verify using the Pythagorean Theorem. Its direction is the same as that of the displacement vector <2 cm, 5 cm>.
Now we invoke a rule to find the tension in the rubber band. Assume that for this rubber band the tension is .5 Newtons for each centimeter in excess of its 'barely-exerting-force' length of 4.8 cm.
• The 5.4 cm length of the stretched rubber band is therefore .6 cm in excess of the 'barely-exerting-force' length of 4.8 cm.
• According to the rule, we conclude that the tension is therefore .6 cm * .5 N / cm = .3 N.
The vector <.37, .92> has magnitude 1, and its direction is the same as that of the displacement vector. A rubber band can exert a force only in the direction of its displacement vector.
• If we multiply our vector <.37, .92> , which we again mention has length 1, by .3 N, we will obtain a vector whose magnitude is .3 N. (whatever quantity we multiply by a vector of length 1, we end up with a vector whose length is equal to that quantity, or more correctly whose length represents that quantity)
• The direction of this vector will be the same as that of our original vector.
• <.37, .92> * .3 N = <0.11 N, 0.28 N>
• This is the tension vector. It tells us that the tension of this rubber band has x component 0.11 N and y component 0.28 N.
Using your data from the rubber band experiment, check your previous calculations of the displacement vector, the unit vector and the force vector for each rubber band. If your original work on this experiment was not done correctly, please correct it and resubmit. Please acknowledge that you have read this instruction and understand it.
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I feel confident that my results were obtained correctly and all of these ideas are firmly implanted in my mind.
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Linear force function and work\energy
If the graph below indicates the tension in Newtons vs. length in cm for a rubber band:
At what length does the rubber band begin exerting a force?
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5 cm
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What is the average force exerted between the lengths x = 5 cm and x = 7 cm?
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(0+1)/2 = 1/2 N
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As the rubber band is stretched from length 5 cm to length 7 cm, through what distance is the force exerted?
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7-5= 2 cm
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How much work is therefore done in stretching the rubber band from 5 cm to 7 cm length?
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1/2 * 2= 1 N*cm
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How much work is done in stretching the rubber band from the 5 cm length to the 8 cm length?
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F_Ave= (0+1.5)/2 = 0.75 N, ∆x= 8-5= 3 cm-----> W=0.75 N * 3 cm= 2.25 N*cm
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How much work is done in stretching the rubber band from the 6 cm length to the 8 cm length?
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F_Ave= (0.5+1.5)/2 = 1 N, ∆x= 8-6= 2 cm ----> W= 1 N *2 cm= 2 N*cm
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What are the areas beneath the graph between each of the following pairs of lengths:
• x = 5 cm and x = 7 cm
• x = 5 cm and x = 8 cm
• x = 6 cm and x = 8 cm
• x = 7 cm and x = 8 cm
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(5,7)----> A= 1/2 (1 *2) = 1/2
(5,8) ----> A= 1/2(3*1.5)= 2.25
(6,8)----> A = 1/2 (1/2 +1.5)*2 = 2
(7,8)----> A= 1/2(1+1.5)*1= 1.25
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Verify that the equation of the straight line in the given graph is F(x) = (x - 5) * .5, where F(x) is force in Newtons when x is position in centimeters.
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F(5)= (5-5)*.5= 0 N, F(6)= (6-5)*.5= 0.5 N. Also, ∫ F(x)= 0.25x^2-2.5x, over the interval of (5,8) is 2.25 and F'(x)= 0.5, our slope of the graph is up 1 over 2, or 1/2.
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University Physics Students: Confirm your results for graph areas by integrating the force function over appropriate intervals. Suggested method: Integrate the function symbolically from x = a to x = b, then use the resulting expression for appropriate values of a and b.
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Well I just did this above before seeing this. I also used the derivative to show the slopes were the same.
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Nonlinear force function and work\energy
The graph given previously was linear. That graph would be realistic for a well-made spring, but not for a rubber band.
The graph given below is more realistic.
Estimate the average force exerted by this rubber band between the x = 5 cm and x = 6 cm lengths. Give your estimate in the first line, and explain how you made the estimate in the second:
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(0+0.75)/2 = 0.375 N is a start but I think the 'true' value is going to be larger than this since the curve spends more 'time' above this value than below so about 0.4 N.
This is just a straight average and depending on the shape of the curve between the two points can give under or over estimates. This is an underestimate.
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Repeat, for the x = 6 cm to x = 7 cm length interval.
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(0.75+0.95)/2 = 9=0.85 N
I think this is close the the actual value since the curve between 6 and 7 is almost linear.
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Repeat, for the x = 5 cm to x = 5.5 cm length interval.
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(0+0.5)/2 = 0.25 N
This is also reasonable since the slope between the two points seems close to linear.
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Repeat, for the x = 5.5 cm to x = 6 cm length interval.
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(0.5+ 0.75)/ 2= 0.625 N
This interval is close enough here to be reasonably close to this value. Of course creating smaller and smaller intervals creates better averages up until you reach a limit and then you have the true value, the integral.
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Estimate the work done in stretching the rubber band for each of these intervals. Give your four estimates in the first line below, separated by commas. Starting in the second line explain how you got your estimates:
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0.4, 0.85, 0.125, 0.3125
Here I used the F_Ave's from above and then multiplied by ∆x.
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How much work do you estimate is done between the x = 5 cm length and the x = 7 cm length? Give your estimate in the first line, your explanation starting in the second.
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Just doing this linearly it would be something like 0.95 N*cm. But looking at the 'sub-intervals' I would say its more like 1.15 N*cm. That's finding the average force of the forces above to obtain a better average force for this interval and then using ∆x=2 cm.
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If a trapezoid was constructed by projecting the x = 5 and x = 7 points of the graph, what would be its area, and would this result be an overestimate or an underestimate of the actual area beneath the graph between these two points?
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The area would actually just be a triangle since F=0 at 5 cm. That is 1/2(0.95)*2= 0.95 N*cm. This is an underestimate and you can see it on the graph by drawing a straight line between the to two altitudes as the curve is more above this line than below. All of this is a lot like the trapezoidal Riemann sums but I don't know that you want to go that way with it since some of the general college students probably don't know what that is. Actually, from what I have gathered some of the other 231/241 students don't either.
You don't need to use calculus rules or formulas to see whether the curve lies above or below the straight-line approximation, so I would expect everyone to be able to make a commonsense comparison of the areas.
I will expect 241 and 231 students tO break it down into Riemann sums and/or trapezoidal approximations, but maybe with a little more help and discussion in and out of class. As I said today, it can take awhile to master this, even after a good calculus course.
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If a trapezoid was constructed by projecting the x = 5 and x = 8 points of the graph, what would be its area, and would this result be an overestimate or an underestimate of the actual area beneath the graph between these two points?
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Here again it's actually just a triangle but the idea is understood about the trapezoidal projections. The area is 1/2*3*1.75 = 2.625. Drawing this line shows that the curve is under the line some and over the line sum and as near as I can see from the picture it's pretty close to even both ways, which means this would be pretty close to the actual value and I can't say with certainty which way it would lean toward, under or over.
We regard a 'graph triangle' as a trapezoid; technically though it's a limiting shape of a trapezoid as one side approaches zero. It doesn't have two parallel sides so it really isn't a trapezoid.
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University Physics Students Only:
Note that these questions are easy to answer, if you understand what to do. Understanding what to do is fairly challenging at this point. I expect that some will get this, and some will not, and I can't predict who will fall into which category. Of course I'd love it if you would make it easy on me, and everyone would get these:
Suppose the function for the tension is T(L) = (L-6.5)^3*.15 + .5 + (L-5)*.2. How much work is done between length L = 5 cm and L = 7 cm, according to this function?
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W= F_Ave * ∆x= (T(7)-T(5))/2 * (7-5)= 0.91875≈ 0.919 N*cm.
You would average the two tensions by adding and dividing by 2. However this wouldn't give you the average tension, only a (perhaps reasonable) approximation of the average tension.
You need to integrate the function to get the work (or the average tension).
Think about it. We'll do this one in class Monday.
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How much work is done between lengths L = L_1 and L = L_2? Apply your expression to the work done over each of the following intervals:
• L = 5 cm to L = 6 cm
• L = 6 cm to L = 7 cm
• L = 5.5 cm to L = 6 cm
• L = 5.5 cm to L = 7 cm
Give your four results in the first line below, separated by commas. Starting in the second line explain how you got your results.
0.341, 0.8, 0.283, 1.03
I did jus like above by finding the average force by the displacement. I haven't read it yet but I smell an integral here because these Work values are areas and as ∆x gets small and we take the limit we have the integral.
Directions of force and displacement vectors matter for work\energy
If force F and displacement `ds are both along the x axis, then what is the sign of F * `ds in each of the following cases:
• F is in the positive direction and `ds is in the positive direction.
• F is in the positive direction and `ds is in the negative direction.
• F is in the negative direction and `ds is in the positive direction.
• F is in the negative direction and `ds is in the negative direction.
Give your answers in the given order, in the first line below, separated by commas. Starting in the second line explain how you got your results.
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positive, negative, negative, positive
These are determined by the 'pairing' up of the distance and the direction of the acting force. It follows the rules for positive/negative multiplication.
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If the x axis in the car-and-magnet experiment was your meter stick, then what were the directions of F and `dx, in each of the following situations:
• F is the force exerted on the 'car magnet' by the 'fixed magnet', `dx is the displacement of the car for 10 cm after its release.
• F is the force exerted on the 'car magnet' by the 'fixed magnet', `dx is the displacement of the car as you move it toward the last 10 cm toward the 'fixed magnet'.
• F is the force exerted by the 'car magnet' on the 'fixed magnet', `dx is the displacement of the car for 10 cm after its release.
• F is the force exerted by the 'car magnet' by the 'fixed magnet', `dx is the displacement of the car as you move it toward the last 10 cm toward the 'fixed magnet'.
• F is the force exerted on the car by friction, `dx is the displacement of the car after being released.
• F is the force exerted by the car against friction, `dx is the displacement of the car after being released.
Give your six answers in the first line below, separated by commas. Starting in the second line explain how you reasoned out your answers.
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positive, negative, negative, positive, negative, positive
First, the fixed magnet is pushing the car away from itself and that is the direction the car is travelling in; second, here the fixed magnet is pushing away but you are moving the car toward it; third, the car magnet is pushing against the fixed magnet but the car is moving in the opposite direction; fourth, (I think this is supposed to read: by the 'car magnet' on the 'fixed magnet') here the car is pushing against the fixed magnet and you are moving it toward the fixed magnet also; fifth, this friction is acting away from the direction of travel; sixth, the car pushes against friction and it travels in the opposite direction that friction is acting.
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The figure below represents two plausible force vs. separation models for two of the ceramic magnets used in the experiment. Force is in Newtons while separation is in centimeters.
If the two magnets were touching before release, which model appears to predict the greater kinetic energy at the x = 2 cm position, and which at the x = 9 cm position?
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We know that more force leads to more acceleration, this is true since the mass is constant here, and acceleration leads to greater velocity and greater velocity means greater kinetic energy. So I think that the red curve has more KE for 2 cm, since it has show greater force here, and the green for 9 cm, since it shows more force at this point.
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&#Your work looks good. See my notes. Let me know if you have any questions. &#
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