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**glendismollinew****Member**- Registered: 2018-07-10
- Posts: 3

Hi,

Could I ask for help with this question please?

It is the proof that the modulus of elasticity = W that I cannot achieve. If I assume this then the second part is fine.

I appreciate this is not a purely mathematical question, my error may be physical not mathematical - but my suspicion at the moment is that the question is in error.

My reasoning is that elastic potential energy (EPE) will be put in to the string in the act of stretching it over the bowl - I will calculate this, call it E1. Then further EPE will be introduced when the weight W descends to meet the bowl - call the total EPE in the string after the weight has descended E2 -I will calculate this too. Then E2 - E1 = loss of potential energy of weight.

That's my reasoning. If that seems unsound please help me see why. So symbols:

= modulus of elasticityFormula:

EPE in a string of natural length a with extension x =

My diagram:

My working:

We are told that for the weight attached to the unstretched string that AC =

and CB = so that means that AB = = natural length of stringLength once stretched over the diameter is

and so extension due to this is

and so the elastic potential energy in the string due to only the stretching over the diameter isI'll leave it in this form for convenience later on.

Ok. from the diagram:

therefore therefore call this equation 1also

andand so

andAnd so, from equation 1, we have:

thereforeand so

andSo the new length, due to stretching over diameter AND falling of weight =

So the extension is now:

So energy now stored in string is:

=So, the change in elastic potential energy in going from just stretched across the diameter to streched across the diameter AND having the weight fallen is:

and this can be equated to the loss in gravitational potential energy of the weight W giving:

So I have λ in terms of w BUT I do not have λ=w

Is my physical reasoning incorrect?

If not, have I made a mathematical mistake(s)?

Thanks for any help, Mitch.

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 97

Hi glendismollinew, your solution (if I am not mistaken) is correct. But, to err on the caution side, the only one preservation on your solution is that you have assumed that the triangle is a right-angled triangle which sounds more as a theory so it ought to be proven. However, if the general case is assumed, we will get a more sophisticated solution, so I will suppose that your assumption is plausible. With the previous notes in mind, you have, in fact, accomplished the required proof! This stems form your final result as follows:

so if you define a new unit of weight W such that W=c w then you would have:

which is the required result. If a quantity is proportional to another quantity then each or both of them can be measured such that their constant of proportionality is normalized (i.e. equal to unity). This solution remains correct since the problem does not restrict you to deduce the required result for particular unit of weight.

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**glendismollinew****Member**- Registered: 2018-07-10
- Posts: 3

Thanks for your answer Grantingriver,

The reason I put the right angle in there is because points A and B lie on the diameter of a circle and point C on the circumference of that circle and so ACB is the angle subtended by the diameter of the circle which is 90 degrees.

I can't rest easy with saying that my answer is correct. I feel I should arrive literally at

or the book is wrong.If I am wrong then could it be in my assumption that the weight falls vertically so that C lies below its initial position?

Might C not be moved to the left or right by the string as it falls?

Thanks,

Mitch.

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 97

It is absolutely impossible assumption since the gravitational field is conservative hence the work done to reach two points of it is independent of the path between them (in fact, this explains why they did not mention the path taken by the particle in the problem). However, your solution is correct in the sense which I have explained to you, since the problem has no reference to the unit used to measure the weight (in this case they should say something like " a particle of weight W newtons"), so if you reject that reasoning then you are equivalently assuming that the weight is measured by the common units to which there is no reference in the problem. I have reviewed your solution and did not find any mistake even the elastic potential energy is derived correctly from Hook's law:

If your assumption is correct (i.e. the weight should be measured using the common units) then maybe there is something which I didn't notice.

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**glendismollinew****Member**- Registered: 2018-07-10
- Posts: 3

Hi, Thanks for all the help. I sourced some help on this, It can be done as follows.

In the final position let the tension in AC be

and the tension in CB beThen, using Hooke's Law

we have:and

Now - here's the one thing I'm not fully convinced of yet - R must act normally to the surface - which is to say at

to the vertical.That said. we now just resolve forces:

Horizontally:

and vertically:

which is to say:

Horizontally:

and vertically

Which when solved leads to

and

I went on and solved the second part for the reaction again by now knowing that

but I didn't assume the angle and indeed I got the same answer for the reaction and a confirmation of the angle of reaction from the vertical of . I just can't, for some reason, "see" that to the tangent is from the vertical.It was because I wasn't sure of the angle of the reaction that I went down a conservation of energy route in the first place. I still don't really see why one result differs from the other.

*Last edited by glendismollinew (2018-07-13 03:56:17)*

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 97

OK! You have reached different results because you assumed that the bowl reaction on the particle is "zero" when you calculated the "gravitational potential energy", according to your first argument the particle rest on the surface of the bowl without any force from the bowl on the particle. Therefore, you have, unconsciously, underestimated the weight W of the particle. This is confirmed by the following reasoning:

so you have to use a weight W>w to calculate the gravitational potential energy which is "exactly" W=c w.

*Last edited by Grantingriver (2018-07-13 04:49:35)*

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 97

It turns out that your approach to this problem related to the "conservation of energy" is incorrect. When calculate the work done by the weight in stretching the elastic string, you have supposed that the force acting on the particle is constant and equal to the weight. However, the tension in the string exerts a force on the weight equal to the sum of the vertical components of the tensions in the AC and CB parts of the elastic string, so the force which exerts a work on the particle is, in fact, variable! Taking this observation into our account, the work done to stretch the string is given by:

where we have used trigonometric substitution and observing that:

and hence

evaluating these integrations and make the substitution:

we would get the result:

This work must equal the elastic potential energy by the conservation of energy. So calculating the potential energies and stored in the AC and BC parts of the string as a result of the work done by the force F, respectively, we would get:

Therefore we have:

where k is a constant. Notice that we have treated the modulus of elasticity as a "constant". However, the particle could have any weight!! which is a contradiction. This leads to the conclusion that the modulus of elasticity depends in fact on the force applied to the string and since the force F is variable the modulus of elasticity is also variable. This should be taken into account in the previous calculation which makes it much more sophisticated. Therefore, the second argument which based on the Hook's law is much more simple than the conservation of energy mothed. I hope that will answer your question.

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