Spinning a Bucket to measure gravity with the parabolic surface of the water

[Samir_Khalilullah]

Homework Statement

When a bucket full of water is being rotated, the water surface makes a parabolic shape. If we take a cross section of the surface, its slope will be related to gravitational acceleration, g. That means that we can calculate the values of g by taking the slope measurement at various points.

(see figure in the attachments...)

(A) Suppose the bucket has ω angular velocity. Find the slope of the cross-section when the distance from the center of the bucket is x. Consider the horizontal and vertical components of the forces acting on a small packet of water, a distance "x" from the center of the bucket.

The height of the water surface will remain constant when x = R √2 , where R is the radius of the bucket. This is the condition of maintaining a constant column. This provides us with a stable point to measure slope.

Now, consider a laser placed at h height directly above the point is emitting a beam vertically downwards. After reflection, it reaches the height of the laser with d horizontal distance from the source. We can get the slope from the values of d and h. And ω can be measured by finding the time it takes to rotate 20 times. These data are tabulated below, here h = 13 cm , R = 7.25 cm.

__________________________________________________________________________________________
| 20T (s) | 42.68 | 31.6 | 28.44 | 23.98 | 23.48 | 20.9 | 19.8 | 18.8 | 18.16 | 16.78 |
__________________________________________________________________________________________
| d (mm) | 11 | 20 | 26 | 30 | 40 | 51 | 56 | 65 | 70 | 85 |
__________________________________________________________________________________________

(B) Using these data, draw an appropriate linear graph and calculate the value of g.
(C) Find out the relative error of your result using the reference value.

Relevant Equations

none

I did (A) by balancing force on point p (in the figure). I found slope , tanθ = (w^2)x/g. I dont know what to do next. Pls help

 

[BvU]

hello @Samir_Khalilullah ,

$\qquad$ !​

Homework Statement: When a bucket full of water is being rotated, the water surface makes a parabolic shape. If we take a cross section of the surface, its slope will be related to gravitational acceleration, g. That means that we can calculate the values of g by taking the slope measurement at various points.

(see figure in the attachments...)

(A) Suppose the bucket has ω angular velocity. Find the slope of the cross-section when the distance from the center of the bucket is x. Consider the horizontal and vertical components of the forces acting on a small packet of water, a distance "x" from the center of the bucket.

The height of the water surface will remain constant when x = R √2 , where R is the radius of the bucket. This is the condition of maintaining a constant column. This provides us with a stable point to measure slope.

Now, consider a laser placed at h height directly above the point is emitting a beam vertically downwards. After reflection, it reaches the height of the laser with d horizontal distance from the source. We can get the slope from the values of d and h. And ω can be measured by finding the time it takes to rotate 20 times. These data are tabulated below, here h = 13 cm , R = 7.25 cm.

__________________________________________________________________________________________
| 20T (s) | 42.68 | 31.6 | 28.44 | 23.98 | 23.48 | 20.9 | 19.8 | 18.8 | 18.16 | 16.78 |
__________________________________________________________________________________________
| d (mm) | 11 | 20 | 26 | 30 | 40 | 51 | 56 | 65 | 70 | 85 |
__________________________________________________________________________________________

(B) Using these data, draw an appropriate linear graph and calculate the value of g.
(C) Find out the relative error of your result using the reference value.
Relevant Equations: none

I did (A) by balancing force on point p (in the figure). I found slope , tanθ = (w^2)x/g. I dont know what to do next. Pls help

Well, perhaps you want to continue with part B ? In that case you will need a relationship between slope $\theta$ and distance $d$.

A little geometry around point $P$ should do the trick !


PS I must say I don't understand "The height of the water surface will remain constant when x = R √2". Can you explain it ?

$\$

 

[Chestermiller]

Let's see your solution for part A.

 

[Samir_Khalilullah]

hello @Samir_Khalilullah ,

$\qquad$ !​

Well, perhaps you want to continue with part B ? In that case you will need a relationship between slope $\theta$ and distance $d$.

View attachment 340646​

A little geometry around point $P$ should do the trick !


PS I must say I don't understand "The height of the water surface will remain constant when x = R √2". Can you explain it ?

$\$

my bad, it was supposed to be R/sqrt(2). well, that means, even if i change the angular velocity, the height of the water column would remain constant

 

[Samir_Khalilullah]

hello @Samir_Khalilullah ,

$\qquad$ !​

Well, perhaps you want to continue with part B ? In that case you will need a relationship between slope $\theta$ and distance $d$.

View attachment 340646​

A little geometry around point $P$ should do the trick !


PS I must say I don't understand "The height of the water surface will remain constant when x = R √2". Can you explain it ?

$\$

also, after a bit of geometry, i did find the angle adjacent to h (at point P) to be 2 times theta. but i dont know how to proceed from there

 

[Samir_Khalilullah]

Let's see your solution for part A.

(I'm new here so i cant really write with latex, sorry)

So I balanced the forces at point P.(otherwise the height of water column would not remain same). we have centripetal acceleration (w^2)*x towards +ve x axis and gravity g at -ve y axis ( taking point P as centre). So the two forces acting along the tangent to P are ((w^2)*x)cos(theta) and gsin(theta). these forces have to be of equal magnitude. thus i find tangent , tan(theta)=(w^2)x/g

 

[BvU]

also, after a bit of geometry, i did find the angle adjacent to h (at point P) to be 2 times theta. but i dont know how to proceed from there

So you want to expresss $d$ in terms of $\theta$ and $h$. How about a tangent?

$\LaTeX$ tutorial button under lower left of edit panel. Worth the investment, and fun!

$\$

 

[Samir_Khalilullah]

So you want to expresss $d$ in terms of $\theta$ and $h$. How about a tangent?

$\LaTeX$ tutorial button under lower left of edit panel. Worth the investment, and fun!

$\$

yeah i realized that. $\tan 2 \theta = \frac d h$. And i also know, $\tan \theta = \frac {\omega^2 R} {g \sqrt{2}}$ . I can relate these two with $$ \tan 2 \theta = \frac {2 \tan \theta} {1 - {\tan^2 \theta}} $$ . But that would give me a complex equation. i need to get a linear equation so that i can plot this in a graph and find its slope and then find the value of g. So i need help . ( ALSO THANK YOU VERY MUCH FOR THE LATEX GUIDE)

 

[kuruman]

But that would give me a complex equation.

Nothing that you can't handle. It's a quadratic in $g$ of the form $Ag^2+Bg+C=0$. You know that the solution of that is $$g=\frac{-B\pm\sqrt{B^2-4AC}}{2A}.$$Throw out the solution that does not make physical sense and plot $g$ vs. order of measurement. You should get a straight horizontal line.

 

[Samir_Khalilullah]

Nothing that you can't handle. It's a quadratic in $g$ of the form $Ag^2+Bg+C=0$. You know that the solution of that is $$g=\frac{-B\pm\sqrt{B^2-4AC}}{2A}.$$Throw out the solution that does not make physical sense and plot $g$ vs. order of measurement. You should get a straight horizontal line.

ok i did what you told .i got , $$ g = \frac {\sqrt{2} \omega^2 R \left( h + \sqrt{h^2 + d^2} \right)} {2 d} $$

now what do i do? They have given me data of "d" and "T"(from which i can get $\omega$). i need to plot a linear equation in the graph and estimate its slope. pls help.

 

[kuruman]

Use the equation you got to find 10 values of g. If you did everything right, a plot of the value you got vs. the order of the measurement (1, 2, 3,..., 10) should give you a straight horizontal line. If you don't get that, you did something wrong. Go back and check your work.

 

[Samir_Khalilullah]

Use the equation you got to find 10 values of g. If you did everything right, a plot of the value you got vs. the order of the measurement (1, 2, 3,..., 10) should give you a straight horizontal line. If you don't get that, you did something wrong. Go back and check your work.

Thanks, I think I'm getting somewhere. here's the graph i got.

Though I'm a bit doubtful about the result of data number 4, g can't be that high. Do i have to draw a best fit line?? everything else seems to be almost in a straight line. Or do i just have to calculate the average?

 

[kuruman]

I am doubtful about data point 4 too. You can ignore as long as you justify your reasons for doing so. For the remaining points, taking an average would mask a non-zero slope. Do a linear fit and see if the slope you get is within experimental error. If it's not, you have to explain what it could possibly mean.

This looks like a test with made up numbers so there is no actual experiment involved.

 

[BvU]

I am a lazy bugger and prefer to let the computer do the work
So I let it calculate $\theta = \arctan(d/h)/2$ and plot $\tan\theta$ as a function of $\omega^2$:

where I consider point 4 a typo or something definitely wrong.

So I leave that out and then let it plot a trendline. Result $\tan\theta = 0.0054 \,x -0.0065$ (confirmed by a regression -- not that lazy, apparently).

Slope of 0.0054 should be $\displaystyle {R\over g \sqrt 2}$ so that $g = 9.82$ m/s²

Bull's eye !!?! Too good to be true ?

So I wholeheartedly agree with

This looks like a test with made up numbers so there is no actual experiment involved.

And then we can earn a meagre 0.5 points in part C:

A serious error discussion is something else ....

I have no idea what reference value is meant here. Maybe 9.81 m/s² ?
In which case we are less than 0.1% off.

$R = 0.075$ m ; any (relative) error goes straight to $g$ (i.e. is a systematic error in the determination of $g$). An optimistic estimate is, say 1 mm, so 1.3 %
( Not very big, this 'bucket' )

$h = 0.13$ m ; error goes straight into $\theta$; estimate 2 mm means 1.5% for $g$, again a systematic error (i.e. doesn't average out but all work in the same direction in $g$)

$d$ let's assume (optimistically) errors average out...

If the systematic errors are independent, we may add them in quadrature and get 2%.
Meaning $g = 9.8 \pm 0.2$ m/s² is what we should report.

And all that without a decent motivation for throwing out an observation that may or may not be 'in error' ...

$\$

 

[TSny]

So I leave that out and then let it plot a trendline. Result $\tan\theta = 0.0054 \,x -0.0065$ (confirmed by a regression -- not that lazy, apparently).

Slope of 0.0054 should be $\displaystyle {R\over g \sqrt 2}$ so that $g = 9.82$ m/s²

With R = 7.25 cm instead of 7.50 cm, g = 9.49 m/s². Still in the ballpark.

I like this method of getting g from the slope of a linear graph. From the wording of the problem statement, I tend to think this is the approach they want.

 

[kuruman]

I like this method of getting g from the slope of a linear graph. From the wording of the problem statement, I tend to think this is the approach they want.

I am a purist. I believe that plots should only have directly measured quantities on the right hand side.

I have no idea what reference value is meant here. Maybe 9.81 m/s2 ?

For a meagre 0.5 pts, I think that "relative error" here means discrepancy $$\text{Relative error %}=\frac{\text{Plot value}-\text{Accepted value}}{\text{Accepted value}}\times 100.$$

 

[BvU]

I am a purist. I believe that plots should only have directly measured quantities on the right hand side.

Me too ( a purist, that is; always disliked parity plots ). But I wonder what the righthand side of a plot is. And measured/calculated tend to contaminate frequently despite good intentions.

And I do prefer the figure in #12 over the figure in #10.

... there is the smalll matter that it's accidentally the wrong figure (with intercept 0).
The one intended is:

with a 4% different $g$

(more work needed: discarding a point is a no-no; slope is $0.00540\pm 0.00005$, intercept $-0.00650\pm 0.0020$ -- yes/no intercept zero, etc, etc)

$\$

 

[kuruman]

Me too ( a purist, that is; always disliked parity plots ). But I wonder what the righthand side of a plot is. And measured/calculated tend to contaminate frequently despite good intentions.

I composed the message in a hurry. I meant the right hand side of the equation that is being plotted. Sorry for the confusion.

As for discarding a point collected in lab, I too was taught that it is a no-no. However, I was also advised to "plot as you go along" just in case you need to retake a measurement. Look at points 4 and 5.
4: 23.98 s, 30 mm
5: 23.48 s, 40 mm
If I were doing this experiment, I would go back and retake the two measurements just to make sure I have not accidentally discovered a hitherto unknown phase transition of water that shows up only when it is spinning at a certain frequency. (Right!)