Question involving joined particles, string and elastic...

leedavies · **Posted:** Mon, 3 Dec 2007 01:54:37 UTC

AAAGGHHH !!!!

Can anybody help with this question. I've been working on it here and there for the pass week.

Quote:

One end of an elastic string, of natural length

and modulus

, is tied to a particle of mass

lying on a smooth horizontal table, the other end being fastened to a fixed point

on the table. An inelastic string, also fastened to the particle, passes over the edge of the table and carries a mass

hanging vertically.

The particle is held at rest at

, the inelastic string being taut, and is then released. Find the greatest extension of the elastic string and prove that

returns to

after time -

$2(\pi+\sqrt{2}-arctan{\sqrt{2}})\sqrt{[(M+m)a/Mg]}$

Here's my diagram illustrating the problem.

To tackle this problem I've used two models, one for $x\leq a$ and the other for $x\geq a$ .

Using the following:

$T_{i}=\frac{gmM}{(m+M)}$

$T_{e}=\frac{Mg(x-a)}{a}$

$\hline$

Case 1 $x\leq a$

Concentrating on the acceleration on particle

.

$M\ddot{x}=Mg - T_{i}$

$\ddot{x}=g - \frac{T_{i}}{M}$

Integrating produces the velocity

$\dot{x}=gt- \frac{T_{i}}{M}t$

Integrating produces the distance

$x=\frac{gt^{2}}{2}-\frac{T_{i}}{2M}t^2=\frac{gt^{2}}{2}-\frac{gmt^{2}}{2(m+M)}=\frac{gt^{2}}{2}(1+\frac{m}{(m+M)})$

Rearranging for

$t=\sqrt{\frac{2x(m+M)}{g(2m+M)}}$

When

x=a

, $t=\sqrt{\frac{2a(m+M)}{g(2m+M)}}$

$\hline$

Case 2 $x\geq a$

Concentrating on the acceleration on particle

.

$M\ddot{x}=Mg - T_{i} - T_{e}$

$\ddot{x}=g - \frac{gm}{(m+M)} - \frac{g(x-a)}{a}=2g - \frac{gm}{(m+M)} - \frac{gx}{a}$

Rearranging.

$\ddot{x}+\frac{gx}{a} =2g - \frac{gm}{(m+M)}$

This is in the form $\ddot{x}+\alpha^{2}x=\beta$ (SHM equation), with the general solution of $x=Acos{\alpha}t+Bsin{\alpha}t+\frac{\beta}{\alpha^{2}}$

therefore $x = Acos(\sqrt{\frac{g}{a}})t+Bsin(\sqrt{\frac{g}{a}})t+{2a}-\frac{am}{(m+M)}$

To find

we use

,

$a = A+2a-\frac{am}{(m+M)}$

Therefore $A = - \frac{aM}{(m + M)}$

To find

we use velocity at t = 0

,

.

$\dot{x} = -\sqrt{\frac{g}{a}}Asin(\sqrt{\frac{g}{a}})t+\sqrt{\frac{g}{a}}Bcos(\sqrt{\frac{g}{a}})t$

$\dot{x} = \sqrt{\frac{g}{a}}B$

at x = a

, $t=\sqrt{\frac{2a(m+M)}{g(2m+M)}}$ ,

$\dot{x}=gt-\frac{gmMt}{M(m+M)}=g(\sqrt{\frac{2a(m+M)}{g(2m+M)}})-\frac{gmM(\sqrt{\frac{2a(m+M)}{g(2m+M)}})}{M(m+M)}$ YUK!!!

therefore $B = \frac{gt-\frac{gmMt}{M(m+M)}=g(\sqrt{\frac{2a(m+M)}{g(2m+M)}})-\frac{gmM(\sqrt{\frac{2a(m+M)}{g(2m+M)}})}{M(m+M)}}{\sqrt{\frac{g}{a}}}$

There's probably no point in continuing, as you can see this is rapidly running out of control!!!

Can somebody nudge me back onto the path?

Many thanks,

Lee

tashirosgt · **Posted:** Mon, 3 Dec 2007 02:56:01 UTC

Are

and

the tension in the inelastic thread? If so, since the thread is inelastic they should have the same magnitude. It seems to me the simple minded approach (for x < a) is

(Mg - T) = M y" , where y'' is the acceleration of the particle hanging down.

T = mx'' where x'' is the acceleration of the particle on the table.

x'' = y'', in magnitude. Would that be consistent with your result?

leedavies · **Posted:** Mon, 3 Dec 2007 09:33:14 UTC

$T_{i}$ is the tension in the inelastic thread connecting particle

to particle

$T_{e}$ is the tension in the elastic thread connecting particle

to table at point

.

Sorry, I forgot to label these in the diagram.

Lee

tashirosgt · **Posted:** Mon, 3 Dec 2007 16:19:12 UTC

I understand how you got the formula for Ti when x < a. But are you using this same formula when x > a? If Ti is constant, wouldn't the larger mass always move as if under a constant force?

leedavies · **Posted:** Mon, 3 Dec 2007 17:25:30 UTC

Yes I'm using the same formula for $T_{i}$ when $x\geq a$ , however I assumed the movement of larger Mass

isn't a constant since we also have $T_{e}$ of the elastic.

Do you think the initial formula for acceleration after $x\geq a$ is incorrect?

$M\ddot{x}=Mg-T_{i}-T_{e}$

Cheers,

Lee

tashirosgt · **Posted:** Mon, 3 Dec 2007 18:00:46 UTC

Let T (without subscript) be the magnitude of the tension in the inelastic thread.
For x > a, I think of the situation as:

eq 1: T - (x-a)k = mx'' (As I read the problem, k = Mg)
eq 2: T - Mg = My''
eq 3: y'' = -x'' (Taking positive y to point upwards and positive x to point to the right.)

So I would solve for x'' and then for T. That might agree with what you get, but it isn't obvious to me that T is the sum of the force from elastic thread and some constant.

leedavies · **Posted:** Tue, 4 Dec 2007 02:47:16 UTC

Mmmm...

using $\ddot{x}=-\ddot{y}$

$T=\frac{gM(Mx-Ma+ma)}{a(m+M)}$

(Notice when x=a

at

in the second model, $T=\frac{gmM}{(m+M)}$ , as in the original formula)

Do I substitute this into $T - (x-a)k=m\ddot{x}$ , producing...

$\frac{gM(Mx-Ma+ma)}{a(m+M)} - \frac{gM(x-a)}{a} = m \ddot x$

Cheers,

Lee

tashirosgt · **Posted:** Tue, 4 Dec 2007 03:39:47 UTC

You like to manipulate symbols more than I do. I would do this:

T = Mg + My'' (from eq 2)
= Mg - Mx''

Substitute for T in eq 1:

Mg - Mx'' - (x-a)k = mx''

Mg - (x-a)k = Mx'' + mx''

Mg = (x-a)k + (M+m)x''

Mg +ak = kx + (M+m)x''

I would solve the differential equation. Then if I was interested in T for its own sake, I would find it from T = Mg - Mx''

leedavies · **Posted:** Thu, 6 Dec 2007 00:43:50 UTC

Many thanks tashirosgt.

Using your equations, I have almost derived the answer given.

$\hline$

Case 1 $x \leq a$

$m\ddot{x}=T$ .......(1)

$M\ddot{y}=-Mg+T$ .......(2)

since $\ddot{x}=-\ddot{y}$ , we can rewrite (2) as

$M\ddot{x}=-Mg+T$

Rearranging for T produces

$T=-M\ddot{x}+Mg$ .......(3)

Sub (3) into (1) produces

$\ddot{x}=\frac{Mg}{m+M}$

Integrating produces the velocity

$\dot{x}=\frac{Mgt}{m+M}$ .......(4)

Integrating again produces distance

$x=\frac{Mgt^2}{2(m+M)}$

Rearranging for t, produces

$t=\sqrt{\frac{2x(m+M)}{Mg}}$

At x=a

, $t=\sqrt{\frac{2a(m+M)}{Mg}}$ .......(5)

$\hline$

Case 2 $x \geq a$

$Mg + ak = kx + (M+m)\ddot{x}$ , where $k=\frac{Mg}{a}$

$2Mg = \frac{Mgx}{a} + (M+m)\ddot{x}$

Using the general formula for SHM, produces

$x=sin(\sqrt{\frac{Mg}{a(M+m)}}t)\sqrt{2}{a}-cos(\sqrt{\frac{Mg}{a(M+m)}}t)a+2a$

Differentiate for velocity. Equating velocity to zero will yield the time at which particle 'm' reaches its maximum distance x, before changing direction.

$0=\frac{sin(\sqrt{\frac{Mg}{a(M+m)}}t)\sqrt{2aMg}-cos(\sqrt{\frac{Mg}{a(M+m)}}t)\sqrt{aMg}}{\sqrt{M+m}}$

Rearranging for t produces

$-arctan(\sqrt{2})\sqrt{\frac{a(m+M)}{Mg}}$ .......(6)

Adding (5) and (6) together and doubling the result produces the total travel time from 'A and back again.

$t_{total}=2(\sqrt{2}-arctan(\sqrt{2})) \sqrt{\frac{a(m+M)}{Mg}}$

I must have missed something, since in the question there's a $\pi$ term, which I don't have?

Cheers,

Lee

tashirosgt · **Posted:** Thu, 6 Dec 2007 16:04:11 UTC

How are the initial conditions used when you derived the solution to the equation for simple harmonic motion? What is "time zero" for that equation. You equation shows that at time 0 the x(0) = a, but the first time x(t) = a is sometime later than when the experiment starts (the release of the body at point A).

leedavies · **Posted:** Fri, 7 Dec 2007 00:57:57 UTC

I treated the second case ( $x\geq a$ ), as an independent model from the first with initial conditions

$x=a, t=0, \dot{x}=\sqrt{\frac{2a(m+M)}{Mg}}$

I then summed the 'travel time' for both models to produce the total travel time. Is this approach incorrect?

Cheers,

Lee

PS thanks for being so patient

tashirosgt · **Posted:** Fri, 7 Dec 2007 17:11:03 UTC

I believe it is OK to use two models. If I take a legalistic reading of the problem, it says to prove that the particle returns to A at a certain time, not that this is the first time that it gets there. I am curious whether the time in answer works given the models we have. (not a particularly easy question to answer)

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