Cambridge Physics Problems 9
Finally, I'm back after a 3-month hiatus!
This is a part of the "Cambridge Physics Problems" thread, an exercise which I have taken up myself. Please read here for a short introduction on the types of questions featured in threads of the same name. Please search for "Cambridge Physics Problems" for threads featuring past questions.
I appreciate all of your help on this. If possible, please show me the mathematical workings of the question (since this is what the question emphasises).
Thank you!
Question:
A light horizontal wire is attached to two supports 1m apart, so that it is just taut but has negligible tension. The diameter of the wire is 1.0mm and the Young modulus for its material is 2.0 x 1011 Pa. A mass of 100g is attached to the centre of the wire and allowed to fall. Calculate the distance through which it falls before beginning to rise again. Assume that this distance is small compared with the length of the wire.
The answer given is actually 1.46 x 10-2m. I think what I did up there was a bunch of crap. Oh fellow physics deities, please enlighten me!
This is a part of the "Cambridge Physics Problems" thread, an exercise which I have taken up myself. Please read here for a short introduction on the types of questions featured in threads of the same name. Please search for "Cambridge Physics Problems" for threads featuring past questions.
I appreciate all of your help on this. If possible, please show me the mathematical workings of the question (since this is what the question emphasises).
Thank you!
Question:
A light horizontal wire is attached to two supports 1m apart, so that it is just taut but has negligible tension. The diameter of the wire is 1.0mm and the Young modulus for its material is 2.0 x 1011 Pa. A mass of 100g is attached to the centre of the wire and allowed to fall. Calculate the distance through which it falls before beginning to rise again. Assume that this distance is small compared with the length of the wire.
my working
The answer given is actually 1.46 x 10-2m. I think what I did up there was a bunch of crap. Oh fellow physics deities, please enlighten me!
(edited 12 years ago)
Your core physics is correct.
The problem is solved by considering the loss in PE of the mass, falling through a distance h, and equating it to the PE stored in the wire when stretched.
This gives mgh = (T/2).2dL ---(1) (as you have in your working)
Pythagoras on the triangle, assuming dL2 is negligible, gives dL = h2/L --- (2) (as in your working)
Also correct is that the Young Modulus is stress/strain and this is used to eliminate T from the formula. (1)
So somewhere after that point you have made an error in the algebra.
Can I suggest you follow these steps.
In the modulus formula (3) rearrange to get T from
E = (T/A)/(2dL/L) ---- (3) stress/strain
Substitute the expression for T in (1)
Substitute for dL = h2/L
You should arrive at the end with
mgh = 2AEh4/L3
This will give the correct value of h
The problem is solved by considering the loss in PE of the mass, falling through a distance h, and equating it to the PE stored in the wire when stretched.
This gives mgh = (T/2).2dL ---(1) (as you have in your working)
Pythagoras on the triangle, assuming dL2 is negligible, gives dL = h2/L --- (2) (as in your working)
Also correct is that the Young Modulus is stress/strain and this is used to eliminate T from the formula. (1)
So somewhere after that point you have made an error in the algebra.
Can I suggest you follow these steps.
In the modulus formula (3) rearrange to get T from
E = (T/A)/(2dL/L) ---- (3) stress/strain
Substitute the expression for T in (1)
Substitute for dL = h2/L
You should arrive at the end with
mgh = 2AEh4/L3
This will give the correct value of h
Original post by Stonebridge
...
Thanks! A further question: how do we justify us neglecting the elastic potential energy contained in the wire? Do we use the same justification "dl2 is approximated to zero" as in the Pythagoras' theorem? In what cases do you think we should take into account the elastic potential energy?
Original post by johnconnor92
Thanks! A further question: how do we justify us neglecting the elastic potential energy contained in the wire? Do we use the same justification "dl2 is approximated to zero" as in the Pythagoras' theorem? In what cases do you think we should take into account the elastic potential energy?
I'm not sure what you mean here. We have taken account of the elastic PE in the wire.
That was the conservation of energy statement at the start where we equated mgh = (T/2).2dL
The RHS is mean force times extension and is the work done extending the wire, and hence the elastic PE stored in it.
Original post by Stonebridge
I'm not sure what you mean here. We have taken account of the elastic PE in the wire.
That was the conservation of energy statement at the start where we equated mgh = (T/2).2dL
The RHS is mean force times extension and is the work done extending the wire, and hence the elastic PE stored in it.
That was the conservation of energy statement at the start where we equated mgh = (T/2).2dL
The RHS is mean force times extension and is the work done extending the wire, and hence the elastic PE stored in it.
But using 0.5 k dl2 (elastic potential energy) would render thee energy conservation to be approximated to zero right? Why so?
Original post by johnconnor92
But using 0.5 k dl2 (elastic potential energy) would render thee energy conservation to be approximated to zero right? Why so?
We put dl2 as approx zero is the pythagoras formula because, compared with L, it is very small. dL is then found in terms of L and h to a very close approximation.
In the 0.5kdL2 expression it is not negligible, and you can use the value obtained in terms of L and h to get the same answer for h, using that formula, as by the other method.
The answer is that dL is not equal to zero. What you are doing with the approximation is comparing it with the other terms to see if it can be ignored in that formula. This doesn't mean it can be ignored in a different formula.
(edited 12 years ago)
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