Determine the acceleration of a puck on an inclined air hockey table

[duchuy]

Homework Statement

Inclined hockey table

Relevant Equations

a = g sin alpha

Hi,
I'm struggling to start this exercice, where I have an inclined air hockey table with an angle alpha.
They gave me this chronophotography (this is online) saying that the ratio is 1:1

I really don't know how to proceed since this is online and I'm thinking if I measure this with a ruler, the ration may not be adapted anymore (right?)
I'm tried to calculate the instantaneous speed then deriving it to get to the accelaration but I don't have the speed function so that didnt really work out. Excuse me if I have misused any vocabulary, I translated this from French.
Thank you so much for your help!

 

[berkeman]

So an inclined air hockey table is just an approximation for a frictionless inclined plane or ramp. Yes, use a ruler to determine each displacement from the left starting point, and make a chart of those measurements versus the $\delta t$ time ticks. Do you know the length of the time ticks, or are you just supposed to keep that as a variable?

You then just calculate the velocity and acceleration based on the deltas. You can make another two columns in your table where you calculate the velocity as the change in position for each time tick, and then use the velocity v(t) column to likewise calculate the acceleration numbers. Does that help? (Excel is a good tool to use for making this table...)

 

[duchuy]

Yes, it's equal to 40 ms
I don't really know how to calculate the velocity and acceleration based on the deltas.. If I have the speed, how do I proceed to get the acceleration if I don't have the speed function?
I'll try with excel but, if I zoom on the image, will it falsify the ratio? And for a 1:1 ratio, does it mean that if I measure 1cm on the pc screen, I'll use that measurement directly for my calculations?
Thank you for your help sir!

 

[kuruman]

I assume you are familiar with the basic kinematic equations under constant acceleration. If not, you can find them here. Have you made the necessary measurements to get the position of the puck at different times?

I see that @berkeman preempted me so I will continue with the alternative suggestion of linearizing the data. First you measure all the distance $x_i$ from the first dot, which I call $x_0$. Make a table with an $x_n$ column, and a second $t_n$ column with the times at which the puck is at $x_n$. Now make a third column of the ratio $x_n/t_n$. Use Excel to make a plot of the $x/t$ versus the time $t$ points. You should get a straight line.
1. What is the significance of the slope (gradient) of the straight line?
2. What is the significance of the zero time intercept?

Hint: $x(t)-x_0=v_0t+\dfrac{1}{2}gt^2.$

 

[berkeman]

Using Excel, make a first column for time. Start with the first entry as 0s, the 2nd row entry is 0.040s, 3rd is 0.080s, and so on. I typically enter 0 for the first row entry, and then make the 2nd row entry = to the previous entry + 0.040. Do you know how to do that in Excel by entering a simple formula?

Then fill that first column down with enough time ticks to account for all of the ruler measurements you got (like 10 rows if you have 10 x(t) datapoints, etc.). In the 2nd column, enter your x(t) data in each row. Then in the 3rd column, you will use a formula to calculate the average velocity between each set of two x(t) values. What is the formula for the average velocity given the distance and time? Note that you will start the v(t) entries in the 2nd row, because it has to be the difference between the 1st and 2nd x(t) values. Does that make sense?

And then make the a(t) column using the same indea, using the difference between v(t) values and the 40ms delta t...

If you want to post your table for us to check, just take a screenshot of it or turn it into a PDF. It's not a good idea to post Word or Excel files directly in a forum, since they can contain macros and other problematic things.

 

[duchuy]

I managed to get this graph, and I tried to set Acceleration = Speed(n+1) - Speed n / Time (n+1) - Time (n)
Does this mean that the accelaration is equal to 0.625 m/s^2?
Thank you!

 

[berkeman]

Looks like you are getting close, but where is your position data? Or are the column labels shifted left somehow? The 4 columns should be Time, Position, Velocity, Acceleration. And the position at t=0 should be x=0.

 

[duchuy]

Yes sorry it was shifted to the left. But I tried to use the value of the slope to determine the acceleration but it was incorrect. And if speed is a straight line, it means that the acceleration is constant no?
Thank you!

 

[duchuy]

1. What is the significance of the slope (gradient) of the straight line?
2. What is the significance of the zero time intercept?

Hint: $x(t)-x_0=v_0t+\dfrac{1}{2}gt^2.$
[/QUOTE]
I was thinking that the value of the gradient would be the acceleration but apparently it wasnt the correct answer. Does it mean that I made an unit conversion error or there is no correspondance?
I'm not sure I understand what zero time intercept means I'm sorry..
But are you suggesting that I could use the equation you gave, since x(0) = 0 and initial velocity = 0 so :
x(t) = 0,5at¨^2?
So it means I should take the value of the distance at a given time and be able to determine the acceleration?
Thank you!

 

[kuruman]

I was thinking that the value of the gradient would be the acceleration but apparently it wasnt the correct answer. Does it mean that I made an unit conversion error or there is no correspondance?
I'm not sure I understand what zero time intercept means I'm sorry..
But are you suggesting that I could use the equation you gave, since x(0) = 0 and initial velocity = 0 so :
x(t) = 0,5at¨^2?
So it means I should take the value of the distance at a given time and be able to determine the acceleration?
Thank you!
[/QUOTE]
If $x(t)-x_0=v_0t+\dfrac{1}{2}at^2$, then $\dfrac{x(t)-x_0}{t}=y=v_0+\dfrac{1}{2}at.$ When you plot $y$ vs. $t$, the intercept (if any) is the initial velocity and the slope is $\frac{1}{2}a$.

 

[duchuy]

1. What is the significance of the slope (gradient) of the straight line?
2. What is the significance of the zero time intercept?

Hint: $x(t)-x_0=v_0t+\dfrac{1}{2}gt^2.$

So the initial velocity is equal to 0, and by that you're saying that a = 2 times the value of the slope.
But by doing the excel, I ended up with many slope values.
And when I tried enter the value of a = 0.625 x 2 = 1.25 it was the wrong value.
Do I have to calculate the average of all the slope values? Since these are measurements with a ruler on a pc screen, i don't know how I would be able to get to a precise value..
Thank you sir

 

[kuruman]

Excel will give you only one slope if you ask it to "Add a trendline" and select a linear fit. If you don't know how to do that, ask Excel Help.

Do you know the correct answer? I am asking because your distances may need to be rescaled. If you used a centimeter ruler to measure distances on your screen, you will get different values if your "screen" is a desktop, laptop, notebook tablet, cellphone, etc. Whoever wrote this problem should have provided a length next to the dots saying how long it is whatever units. The 1:1 ratio is meaningless.

 

[duchuy]

Excel will give you only one slope if you ask it to "Add a trendline" and select a linear fit. If you don't know how to do that, ask Excel Help.

Do you know the correct answer? I am asking because your distances may need to be rescaled. If you used a centimeter ruler to measure distances on your screen, you will get different values if your "screen" is a desktop, laptop, notebook tablet, cellphone, etc. Whoever wrote this problem should have provided a length next to the dots saying how long it is whatever units. The 1:1 ratio is meaningless.

No i do not know the correct answer. This is a picture of the exercice.

It's just indicated 1:1 ratio. I have zero idea about what is the equivalence of the distance I measured with a ruler. So I just decided to use the distance I measured with a ruler to put in the excel.
And also I'm supposed to insert the correct acceleration value online, so I don't really know if if can find it..
And sorry but I really thought that the value of the slope corresponds to the acceleration? Or in every single case, the value of acceleration is equal to 2 times the slope?
Thank you.

 

[berkeman]

Maybe work backwards to get an idea of what the distance units are. Assume that the surface is frictionless and do you know the angle $\alpha$ that they mention? If you do, you can calculate the acceleration down the slope due to gravity, and use that to work back to what the distance units must be.

 

[duchuy]

Maybe work backwards to get an idea of what the distance units are. Assume that the surface is frictionless and do you know the angle $\alpha$ that they mention? If you do, you can calculate the acceleration down the slope due to gravity, and use that to work back to what the distance units must be.

No they didn't give out the value of the angle alpha.
That is actually the following question where I have to determine its value, but without acceleration, I can't really finish the exercice.

 

[berkeman]

if I zoom on the image, will it falsify the ratio? And for a 1:1 ratio, does it mean that if I measure 1cm on the pc screen, I'll use that measurement directly for my calculations?

If you used a centimeter ruler to measure distances on your screen, you will get different values if your "screen" is a desktop, laptop, notebook tablet, cellphone, etc. Whoever wrote this problem should have provided a length next to the dots saying how long it is whatever units. The 1:1 ratio is meaningless.

@duchuy what is the standard width of paper in your country? In the US it is 8.5" wide. So if the problem looks like it could be printed on standard paper, that is probably the scale 1:1 they are referring to. If you do a Print Preview on the problem, you could scale your measurements with respect to what you measure for the width of the paper. That's the only thing I can think of, short of asking your TA/professor for clarification.

 

[SammyS]

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Yes sorry it was shifted to the left. But I tried to use the value of the slope to determine the acceleration but it was incorrect. And if speed is a straight line, it means that the acceleration is constant no?
Thank you!

By the way:
How do you know the acceleration is incorrect?

Thanks for including the column showing distance. Now I see what is involved in your calculations.

What you have calculated for velocity (speed) at time, $t$, is $\dfrac{x(t)-x(0)}{t-0}$. This is the average velocity over the time interval from 0 to $t$. If the acceleration is constant, then this velocity also corresponds to the instantaneous velocity at time, $\dfrac{t}{2}$.
Alternatively: It appears from the data that initial velocity was very near zero. In this case, the instantaneous velocity at time $t$ is 2 times the average velocity over the time interval from 0 to $t$. You can simply double each value in the velocity column.

As for the acceleration values you obtained: In post #6 you wrote,

...
I managed to get this graph, and I tried to set Acceleration = Speed(n+1) - Speed(n) / Time (n+1) - Time (n)
...

First of all, I think you mean, Acceleration(n) = (Speed(n+1) - Speed(n)) / (Time (n+1) - Time (n)) .
The extra pairs of parentheses are important.
Using the corrected velocity values, which are twice the values you have in the table, will give acceleration values which are likewise twice the values you have in your table.
But I was curious about how you list an acceleration value for the $t=0.44$s, as that would require using velocity and time values from the line below that line. It looks like the table and graph were generated using a spreadsheet, so I guess the spread sheet used zeros for those values. Indeed, $\dfrac{0-0.2931818...}{0-0.44)} = 0.66632...\$ .