Lab 08 - Coefficient of Restitution PDF

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Activity 8 Coefficient of restitution Objective: To investigate the loss of energy of a bouncing ball with each bounce; additionally, to see a practical application of the free-fall kinematic equations Equipment: for this activity you will need • A smartphone. • The phyphox app, available for free for both Apple and Android phones, from their respective online stores. • A ball that bounces well (you need to be able to get at least four bounces out of it). Other than that, size does not matter: it could be a marble or ping-pong ball bouncing on a counter, or a basketball bouncing on a concrete floor. You may be able to get for cheap a small, bouncy rubber ball in the toy section of Walmart or Target, or even a pharmacy. Activity 8.1. Multiple bounces of a dropped ball 8.1.1. Watch the video Before you start on this activity, please watch the video at https://phyphox.org/experiment/inelastic-collision/ which will show you the basic idea. 8.1.2. Trial runs Install the phyphox app on your phone, and do a couple of trial runs dropping your ball and having the “(In)elastic collision” module of the app tell you what the initial height and the height of the subsequent bounces were. You should use these results for calibration: make sure that you set up your experiment in such a way that phyphox correctly determines (within a small error) the initial height, which you should measure with a ruler or meter stick. Things to watch out for are: • Make sure you set the threshold for sound detection of the app appropriately, as shown in the video. • Make sure that the ball is falling straight down, without any sideways velocity. • Make sure you drop the ball with initial zero velocity; that is to say, that you do not accidentally give it a small push up or down upon releasing it. • The app has no way of knowing how big your ball is, so you should measure heights from the bottom of the ball to the floor. This just gives all your potential energies (which typically would be measured from the center of the ball) a constant offset that does not affect the calculations that follow.

When you are satisfied that your setup is giving you reliable results, you can move on to the actual data gathering phase. But first, since the video went over this a bit too fast, here is a little theory. 8.1.3. How the app works things out

Suppose you drop a ball from a height ℎ0 above the ground. How long does it take to hit the ground?

You can set up the equation of motion, 𝑦 = 𝑦𝑖 + 𝑣𝑖 𝑡 − 2 𝑔𝑡 2 , with 𝑦𝑖 = ℎ0, and the initial velocity 𝑣𝑖 = 0 (since you are dropping, not throwing). Then set the final height 𝑦 = 0, and solve for 𝑡: 1 (1) 0 = ℎ0 − 𝑔𝑡 2 2 1

𝑡=√

2ℎ0 𝑔

Your phone does not know when you dropped the ball, so it cannot measure this time directly, but it listens for the first bounce, and then for the second, and so it knows the time in between those two. Let us call that time interval ∆𝑡12 (“from 1 to 2”). The ball spends half of that time going up, and the other half coming down. To find how high it rises, we can just apply this modified version of Eq. (1) (2) 1 ∆𝑡12 2 ) 0 = ℎ1 − 𝑔 ( 2 2

to the falling part of the motion, and solve this for ℎ1 = 𝑔(∆𝑡12 )2 /8, which is the maximum height the ball rises after the first bounce.

We can continue this way, and measure the time between the second and third bounces, ∆𝑡23. From this we get the next maximum height, ℎ2 = 𝑔(∆𝑡23 )2 /8.

Let’s stop here for a minute, and look at the system’s energy. Let the system be the Earth and the ball. This has kinetic, gravitational, elastic and thermal energies (and possibly other forms of energy not directly relevant now). At the top of each bounce, though, we only have gravitational energy, 𝑈 𝐺 = 𝑚𝑔ℎ, plus some unknown amount of thermal energy. So, from the first drop, and through the first two bounces, the mechanical energy of the system drops in this sequence: 𝐸𝑚𝑒𝑐ℎ = 𝑚𝑔ℎ0 → 𝑚𝑔ℎ1 → 𝑚𝑔ℎ2

Assume that the energy drops by a constant factor (or percentage) every bounce. For instance, say that every bounce you lose 25% of your energy. That means that ℎ2 should be 25% smaller

than ℎ1 , that is, you should have ℎ2 = 0.75ℎ1 . Phyphox then applies this logic to the first bounce, that it, it assumes ℎ1 = 0.75ℎ0 , and uses this equation to estimate ℎ0.

The main thing to take away from this, from a practical perspective, is that the only things phyphox measures directly are the time intervals ∆𝑡12 , ∆𝑡23 , … in between bounces. (These are referred to as “Time 1,” “Time 2”, etc. in the app; that is to say, “Time 1” is what I have called here ∆𝑡12 , etc.) The respective maximum heights after each bounce, ℎ1 , ℎ2 , … are then calculated from these time intervals using the kinematic equations, and the initial height ℎ0 is inferred from this plus the additional assumption that energy decays by a constant factor each bounce. It’s a neat trick, you have to admit, but how accurate is it? 8.1 “Trust but verify,” part 1 Run your experiment, making sure you measure the height ℎ0 from which you drop the ball, and compare it to the result inferred by phyphox: ℎ0 (measured) = 6.25 inch (15.875 cm) ℎ0 (phyphox-inferred) = 16.91 cm (6.65 inch) Percentage error = 6.4%

8.2 “Trust but verify,” part 2 Now to make sure that my description above is trustworthy, take the “Time 1” and ℎ1 that phyphox gives you; assume that “Time 1” equals the ∆𝑡12 I introduced above, and that ∆𝑡12 and ℎ1 are related as in Eq. (2), and answer this question: • What is the value of 𝑔 the phyphox designers used?

Explain how you got this: 982 cm/s2 I did 44.07 = would be 9.82

m/s2

𝑔(0.599)^2 8

and if you convert it to meters, it

8.1.4. The coefficient of restitution How does the coefficient of restitution come into all this? Recall that the coefficient of restitution, 𝑒, could be used to figure out what part of the “convertible energy” was in fact “lost”, that is, converted to internal forms of energy, in a collision between two objects. Specifically, the energy lost was equal to 𝑒 2 𝐸𝑐𝑜𝑛𝑣,𝑖 , and 1 2 𝐸𝑐𝑜𝑛𝑣,𝑖 = 𝜇𝑣𝑟𝑒𝑙,𝑖 2 where 𝜇 is the reduced mass of the system, and 𝑣𝑟𝑒𝑙,𝑖 is the initial value of the relative velocity of the two objects. In this case, since the mass of the Earth is so much greater, the reduced mass 𝜇 is simply the mass of the ball, 𝑚, and since the Earth does not budge, the relative velocity is just the ball’s velocity just before the collision. So, the initial convertible energy just before each bounce is just the whole kinetic energy of the ball, and, since at that point the

gravitational potential energy of the system is zero (the ball is just about to touch the ground), this is also the total mechanical energy of the system: 1 2 𝐸𝑐𝑜𝑛𝑣,𝑖 = 2 𝑚𝑣𝑖 = 𝐸𝑚𝑒𝑐ℎ,𝑖 Here, the subscript 𝑖 means “just before the collision.” In terms of the coefficient of restitution, 𝑒, then, the energy dissipated in each bounce is 𝐸𝑙𝑜𝑠𝑡 = (1 − 𝑒 2 )𝐸𝑐𝑜𝑛𝑣,𝑖 = (1 − 𝑒 2 )𝐸𝑚𝑒𝑐ℎ,𝑖 However, the mechanical energy just before, say, the first bounce is just 𝑚𝑔ℎ0 , that is, the initial gravitational potential energy (that has been totally converted, without losses as long as air resistance is negligible, into kinetic energy); and the mechanical energy just after is also numerically equal to 𝑚𝑔ℎ1 , as I noted above, so the “lost” energy equals 𝑚𝑔ℎ0 − 𝑚𝑔 ℎ1 , and we can write 𝑚𝑔ℎ0 − 𝑚𝑔 ℎ1 = 𝑒 2 𝑚𝑔ℎ0 Or, simplifying this, ℎ1 = 𝑒 2 ℎ0 If the coefficient of restitution does not change from collision to collision, we have here an equation that we can apply to each successive bounce. If we number the successive maximum heights with a subscript 𝑛, with 𝑛 = 0,1,2, …, then we expect ℎ𝑛+1 = 𝑒 2 ℎ𝑛

(3)

We will be using this equation later to determine 𝑒 from your data. 8.3 Comprehension check: In the example I gave above, where I supposed 25% of the energy was lost in each bounce, what would be the coefficient of restitution? 0 < e < 1 because 75% of the energy was recovered with the first bounce, 50% with the second, 25% in the third, and 0 in the last, where it came to rest Explain how you got this value. Since some of it was recovered, that means the its elastic and the coefficient is greater than 0, but since not all of the energy was recovered in the first bounce, that means it wasn’t completely elastic, but still had some recovery, so its greater than zero but less than one

8.1.5. Finally, the experiment The main goal of this experiment is to determine the coefficient of restitution for the ball and the surface you have chosen, and to see whether it remains approximately constant from one collision to the next. Set up your experiment, measure the initial height from which you will be dropping the ball, start phyphox running, and drop the ball. Try to get at least four good bounces. Write down the heights you got on this table. Use your measured value (not phyphox’s inferred value) for ℎ0 .

8.4 Height vs. bounce table 𝑛

ℎ𝑛 (cm)

0

68

1

44

2

28

3

22

4

18

5

12

8.5. Energy versus bounce graph

To begin with, plot the successive heights versus the bounce number 𝑛, beginning at 𝑛 = 0. Since the total energy is directly proportional to the height, this shows you visually how the energy drops with every bounce. If the coefficient of restitution 𝑒 is constant, this graph should be an exponential decay. The values of ℎ𝑛 should also be an example of what is called a geometric series. Note that this is not actually an energy vs. time graph, since the actual time between bounces changes, becoming shorter and shorter. 80 70 60 50 40 30 20 10 0 0

1

2

3

n

4

5

6

8.6 Height vs. previous height graph. Now plot the value of each height versus the previous one: that is to say, ℎ1 versus ℎ0 , ℎ2 versus ℎ1 , … Another way to put it, plot the points (ℎ𝑛+1 , ℎ𝑛 ), with 𝑛 = 0,1,2, …. Fit a trendline to the points,

display the equation of the trendline on the graph, and paste it below:

80 y = 1.6555x - 5.0563

70 60

h(n)

50 40 30 20 10 0 0

10

20

30

40

50

h(n+1)

8.7 Coefficient of restitution Examine again Equation (3) above and answer: how is the coefficient of restitution related to the slope of the line you just fitted to your data? They’re related because the coefficient squared is equal to the slope, hence why the graph looks almost exponential What value do you get for 𝑒 based on this fit? 𝑒 = 0.8

8.1.6 Analysis Now let’s see what we can make from these data. Answer the following questions:

(8.8) Does the value of 𝑒 appear to change much from one collision to the next? Justify your answer. Yes slightly. It ranges from 0.8-0.9

(8.9) Based on the above, does it appear as if 𝑒 depends much (or not) on the relative speed of the objects colliding? How can you tell? (Think carefully about what’s going on in the experiment!) Not necessarily on speed, but more so on the type of collision, the weight of the object, and how far it bounces back up

(8.10) Based on this, if you dropped the ball from, say, double the height, would you expect to get a very different value for 𝑒 ? Again, explain your reasoning. The e would stay the same because if you just doubled the values of hn, the value of e would stayed the same. The coefficient of reinstitution is not dependent on height.

(8.11) Do you expect 𝑒 to change very much if you change the surface on which you bounce the ball? Explain. (Some quick experiments here would be useful. Note that you don’t need to do all the plotting and tables again—you can use phyphox’s estimate of the rate of energy loss as a proxy for the coefficient of restitution.) Yes. Lets say you bounce the ball on carpet and on hard wood floors. Its obviously going to bounce back up multiple times on hard floors versus carpet, therefore giving different e.

Finally, tell us what you used for the ball and the surface in the experiment whose data you reported here.

I used a rubber bouncy ball, wood floors, tape measure, and my phone....