Thermal expansion apparatus

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Homework Statement


Im hoping that this experimental setup is common enough in 1st year physics that someone could at least point me in right direction because I may be misunderstanding how the arm and pointer connect together.

I need to come up with an equation that relates how far the metal arm gets pushed out as the metal rod inside the chamber expands under heat, to the amount that the pointer rises on the ruler.
As the arm is pushed to the right (in the sketch) the pointer rises vertically along the ruler. I measured the length of the whole pointer to be 30cm, and also the length of the little arm that is pushed out (4cm).
I know my basic trig and I recognize that the angles change as the pointer rises. But what should the equation look like? Its a pretty bad sketch but hopefully it makes sense enough to comment, thanks for any help
sketch.jpg


Homework Equations




The Attempt at a Solution

 

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  • #2
haruspex
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Its a pretty bad sketch
Too true.
Where exactly is the pivot?
What does the triangle represent, and how is it connected to the lines that touch it?
Which part expands?
Which points are fixed?
What is this screw doing?
You need to make clear which sets of lines constitute rigid assemblies, and how these different assemblies are connected to each other.
 
  • #3
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image2.jpg
image2.jpg
Thanks for answering haruspex , I appreciate it.

Where exactly is the pivot?
It is the arm that pivots, The arm is touching the end of the right end of steel so that that when the rod expands it pushes the arm to the right which then raises the pointer arm vertically alongside the ruler.

What does the triangle represent, and how is it connected to the lines that touch it?
The triangle represents the arm that is in contact with the right end of steel rod.

Which part expands?
the right end of the rod because the left side is held in place by the lockable screw. (linear expansion)

Which points are fixed?
unsure how to answer this question?

What is this screw doing?
the screw (on the left in sketch) prevents the steel rod from expanding in the left direction.

You need to make clear which sets of lines constitute rigid assemblies, and how these different assemblies are connected to each other.

I attached an actual pic so it should make alot more sense, because my sketch is pretty horrible. Thanks in advance for any more help.
 

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  • #4
haruspex
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It is the arm that pivots,
That's not what I asked, but I think I can see from the photograph.
The pointer and short arm form an L-shaped rigid unit. The pivot is where they meet.
If the rod expands by Δx, how much does this unit rotate by? What will that do to the height of the pointer tip?
Edit: note that it is important to measure lengths from the pivot, which is a bit less than the overall length of the arm.
 
  • #5
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If the rod expands by Δx then the whole unit rotates by Δx? I know that height of pointer tip increases but would it be by Δx?
 
  • #6
haruspex
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If the rod expands by Δx then the whole unit rotates by Δx?
Rotation would be an angle, not a distance.
 
  • #7
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Sorry Im not seeing which angle I should be considering. Am I right in saying that as Δx increases the angle decreases right? So hypotenuse is the 30cm pointer? thanks
 
  • #8
haruspex
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Sorry Im not seeing which angle I should be considering. Am I right in saying that as Δx increases the angle decreases right? So hypotenuse is the 30cm pointer? thanks
The short arm is pivoted at its base. When the heated rod expands, it pushes the top of that arm a bit to the right. That will rotate the arm about the pivot.
For a small displacement Δx at the tip of the arm, what angle will it rotate through?
 
  • #9
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haruspex are you still available for this? I still dont see what angle it will rotate through. I am still confused please help
 
  • #10
haruspex
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haruspex are you still available for this? I still dont see what angle it will rotate through. I am still confused please help
The angle will be quite small, so we can use a small angle approximation.
If the short arm rotates a small angle θ to the right in the diagram, how far will its tip move to the right, approximately?
 
  • #11
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the same amount approx?
 
  • #12
haruspex
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the same amount approx?
The rotation is an angle, the distance moved is a distance. A distance cannot equal an angle.

If a car moves distance s and its wheels are radius r, through what angle have the wheels rotated?
(Do you know what a radian is?)
 
  • #13
Tom.G
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If you remove the steam tube and its internal metal rod from the apparatus you will find that the Pointer can freely move up and down (depending on the construction, the pointer may already freely move).

As you move the pointer you will find that it has a pivot point somewhere near its right end. There will also be a part of the apparatus that moves when the pointer moves. Assume that these two moving parts are one piece with a pivot point.

What do you call a rigid piece which pivots around a point?

Hope this helps in your mental image of how this experiment works.

Cheers,
Tom
 
  • #14
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I dont know what a rigid piece which pivots around a point is called but are you saying that the arc length is = to the amount that pointer moves? thanks
 
  • #15
haruspex
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I dont know what a rigid piece which pivots around a point is called but are you saying that the arc length is = to the amount that pointer moves? thanks
For a small angle, the arc length is roughly equal to the displacement.
 
  • #16
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OK so to know arc length I have to have the radius and do I know that because I know the length of the arm that moves through the angle?
 
  • #17
haruspex
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OK so to know arc length I have to have the radius and do I know that because I know the length of the arm that moves through the angle?
You gave the arm length as 4cm, but I do not know exactly what that includes.
Project the centre line of the rod to where that line is closest to the pivot axis of the arm. I.e. the point where a line to there from the pivot axis of the am makes a right angle to the line of the rod.
You need the distance from there to the axis of the pivot.
 
  • #18
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thanks, the 4cm is the length of arm from point of contact with expanding rod to the bottom part of right side of pointer arm. I unfortunately dont have access to the apparatus anymore so I cant remeasure.
 
  • #19
haruspex
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thanks, the 4cm is the length of arm from point of contact with expanding rod to the bottom part of right side of pointer arm. I unfortunately dont have access to the apparatus anymore so I cant remeasure.
Taking it from the point of contact is probably ok, but I suspect going to the other extreme of the arm is too much. Can you remember well enough to make a guess at how far past the axis that would be? 0.5cm? Less?
 
  • #20
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Sorry I dont know what you mean about what the axis would be. All I know is that pointer moved about 3mm when the rod expanded. The pointer is about 30cm and the arm is about 4cm. I also get that the arm moves through some angle as the pointer rises.
Thanks for any more help. I have some mental block about this for some reason.
 
  • #21
haruspex
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Sorry I dont know what you mean about what the axis would be. All I know is that pointer moved about 3mm when the rod expanded. The pointer is about 30cm and the arm is about 4cm. I also get that the arm moves through some angle as the pointer rises.
Thanks for any more help. I have some mental block about this for some reason.
I find it hard to believe that you cannot tell me the relationship between how far an arm rotates (an angle) and how far its tip moves (an arc length). This is very basic geometry.
See https://www.math.uh.edu/~irina/MATH1330/1330Notes/1330S42.pdf.

The other issue is that with such a short arm it is important to get its effective length right. The effective length is from the tip to the axis, i.e. to the centre of the pin on which it is mounted. If you measure the whole arm length it will go past that pin. This will be an extra few mm at least.
 
  • #22
Tom.G
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Lets see if we can un-block that.
Take an ordinary pencil and balance it horizontally across your finger. Push down on one end and the other end goes up by tthe same distance.

Now grab the pencil at one end and slide it so this end is sticking out from your balance finger only half as far as the far end is. You still have ahold of the short end. Push down and the far end will travel twice the distance as that short end.

You have just demonstrated a lever and found how the distance each end travels is related to the distance to the pivot point.

What do you have in your apparatus that resembles a lever?
EDIT: (Hint, it doesn't have to be straight.)

Cheers,
Tom

EDIT: p.s. That's the simplified version and probably 'close enough' for this experiment. @haruspex gave a link to the more accurate and general approach.
 
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  • #23
haruspex
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Lets see if we can un-block that.
Take an ordinary pencil and balance it horizontally across your finger. Push down on one end and the other end goes up by tthe same distance.

Now grab the pencil at one end and slide it so this end is sticking out from your balance finger only half as far as the far end is. You still have ahold of the short end. Push down and the far end will travel twice the distance as that short end.

You have just demonstrated a lever and found how the distance each end travels is related to the distance to the pivot point.

What do you have in your apparatus that resembles a lever?

Cheers,
Tom
I hope that succeeds. I didn't go down that route because I thought it would be even harder to get across the idea of a lever that isn't straight.
 
  • #24
Tom.G
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Yeah, here's hoping. See also my 2 edits.
 
  • #25
jbriggs444
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I hope that succeeds. I didn't go down that route because I thought it would be even harder to get across the idea of a lever that isn't straight.
One could try the idea of a wheel where one pushes rearward on a lug nut positioned 3 inches directly above the axle and then measures the resulting rotation of the tire tread positioned some 15 inches in front of the axle. [Five to one leverage]

Then carve away everything but an L shaped lever pivoted at the axle. Short leg goes from axle to lug nut. Long leg goes from axle to tire tread.
 

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