Mechanics - Elastic springs and strings

[mr bob]

A particle P of mass m is attached to one end of a light elastic string of natural length L whose other end is attached to a point A on a ceiling. When P hangs in equilibrium AP has length \frac{5l}{3}. Show that if P is projected vertically downwards from A with speed \sqrt(\frac{3gl}{2}), P will come to instantaneous rest after moving a distance \frac{10l}{3}.

I thought about working all this out by finding the energies before and after the projection.

Before:-
KE =\frac{3gl}{4}
GPE = 0
EPE = 0

After:-
KE = 0
GPE = -(y - 5/3L)g where y is the full length of stretched string

However i can't figure out how to work out the EPE after the projection as i don't have the modulus of elasticity of the string.

Any help would be greatly appreciated.

 

[andrewchang]

well, if the mass is hanging off the spring, you know the force it exerts. and also how far the spring stretches.

 

[mr bob]

Sorry, i don't understand.

 

[andrewchang]

If the mass pulls down with the force of mg, then the spring must pull up with a force equal to that. (it hangs without moving at 5L/3)

Therefore, you can apply Hooke's Law to get the spring constant, which is what you were asking for.

 

[mr bob]

Of course! Thank you.

 

[mr bob]

Argh, I've tried the question again and again and can't get the answer. Are the above values for energy correct?

Thank you.

 

[Doc Al]

Don't worry about getting the answer until you've figured out the spring constant of the elastic string; you need it to find the final EPE. Use Hooke's law as andrewchang suggested.

Your energies are not quite correct. For one thing, you left out the mass when calculating KE and GPE. And when computing the final GPE, be sure to measure from the same reference point as you did for the initial GPE: point A.

 

[mr bob]

I get the Modulus of elasicity to be
mg = \frac{\lambda\frac{2}{3}L}{L}

Therefore:-

\lambda = \frac{3mg}{2}

Am i correct?

Thanks.

 

[Doc Al]

Not exactly. (But you are close.) You need to calculate the spring constant (k), not Young's elastic modulus.

From Hooke's law:

F = k \Delta x

mg = k (2L/3)

 

[mr bob]

Ok, We've only been taught the elastic modulus, so i assume the question should be able to be done using that. Do i use K and i would use lambda in the EPE equation?

Thanks once again,
Bob

 

[Doc Al]

Interesting that you've been taught Young's modulus but not Hooke's law in its simplest form. You can express the EPE in terms of k or \lambda, whichever you have learned. If you use \lambda, be sure to define it correctly:

F/A = \lambda \Delta L / L

 

[mr bob]

In my book, Lambda is defined as \lambda = \frac{mgl}{x} where x is the compression/extension. As the equilibrium length is \frac{5l}{3} the extension is \frac{2l}{3}.
Subbing that into the above gives:

\lambda = \frac{3mg}{2}

Why is this wrong?

 

[Doc Al]

There's nothing wrong with using such a definition, as long as you stick to it. (It's just not the standard definition of Young's elastic modulus, which is what I thought you were using.)

 

[mr bob]

I'm calling the entire length of the string (at maximum stretch after projection) Y. The energies are measured from the equilibrium point.

Before projection:-
KE = \frax{3mgl}{4}

EPE = \frac{3mg(\frac{2}{3}l)^2}{4l}

PE = 0

After projection:-

KE = 0

EPE = \frac{3mg(y - l)^2}{4l}

PE = -(Y - l)mg

Are these values correct, as from this:-

\frac{3mg(\frac{2}{3}l)^2}{4l} + \frac{3mgl}{4} = \frac{3mg(y - l)^2}{4l} - (Y - l)mg

 

[mr bob]

Hmm. Still cannot get the correct answer. Please help, this is turning me mad.

Thank you for all the help i have received so far.

 

[Doc Al]

Before projection:-
KE = \frac{3mgl}{4}

You had a typo in your Latex.

EPE = \frac{3mg(\frac{2}{3}l)^2}{4l}

I don't understand this. Initially, the string is unstretched.

PE = 0

OK, you're measuring GPE from point A.

After projection:-

KE = 0

OK.

EPE = \frac{3mg(y - l)^2}{4l}

OK.

PE = -(Y - l)mg

OK.

You're getting close!

 

[mr bob]

EPE = \frac{3mg(\frac{2}{3}l)^2}{4l}

The string is initially stretched by 2/3L as natural length is L and stretched length is 5/3L.

I really appreciate your help Doc Al, this question has been giving me trouble for the last couple of days.

Thank's once again.

 

[Doc Al]

The string is initially unstretched--no tension at all, just hanging loose. The mass only begins to stretch the string when the string has reached its natural length L--that means that until the mass has fallen a distance L from point A, the string hasn't begun to stretch at all. No tension = no stored energy.

(Note: An elastic string is not the same as a rigid spring, which can be compressed as well as stretched.)

 

[mr bob]

Hmm, I've canceled it all down to get

\frac{3mgl}{4} - \frac{2mgl}{3} + \frac{mgl}{3} = \frac{3mg(y - l)^2}{4l} - (Y - l)mg

Which goes to:-

3(y - l)^2 - 4l(y- l) - \frac{20l^2}{12}

Which after applying the quaratic formula, equates y - l = \frac{10l}{6}. (y - l) is the distance i want (distance past 5/3l stretch).

Im not sure why I am getting this, when the answer is apparently \frac{10l}{3}

 

[Doc Al]

After projection:-

KE = 0

EPE = \frac{3mg(y - l)^2}{4l}

PE = -(Y - l)mg

Oops... I just realized that your final gravitational PE is incorrect. Remember that you are measuring the postion with respect to point A. So:
PE = -Ymg

Also, I believe what you should be solving for is Y. Y is the distance that the mass moves after being thrown down at point A.