Trajectory of a mass element in a string overhanging a pulley

[burian]

Homework Statement

The title, this question I made up myself

Relevant Equations

$$ v= \frac{u}{\cos \theta} $$

Hello everyone, this is a thought experiment I made, it involves two ropes being pulled of a pulley similar to an atwood machine, and a block attached to the two pulleys at the other end. The ropes are being pulled at a constant velocity $ U$ from the end away from the block and the block rises up in the middle. Now, I know the velocity of block,

$$v = \frac{u}{ \cos \theta} $$

but, how would I use this to derive the equation of motion of a mass segment on the string? (supposing the string has mass)

This expression seems quite hard to integrate because $\theta$ is a function of time, (another difficulty)

Any help would be appreciated :)

 

[BvU]

hello and !

a thought experiment
...
(supposing the string has mass)

Doesn't really mattter, because you impose speed $u$.

That is the speed for the rope once over the pulley, so I suppose you are interested in the section that stays between the pulleys ...

You want to make a better sketch (rope segments move away from the symmetry axis, not towards it) and claerly define a few things ($\theta$, for example)

By the way, 'equation of motion' is something else than a trajectory description...

 

[burian]

@BvU Yes I'm interested in any small mass segment belong to the large mass segment hanging over the pulley, the $\theta$ that I refer to is the angle which the string is inclined from the vertical line passing through the block

 

[burian]

See, the trajectory I'm only theorizing what could happen, I am asking for help on deriving a quantitative description of the motion

 

[BvU]

The problem with thought experiments is that you have to think of everything,,,

the θ that I refer to is the angle

I know (from the expresssion). So the rope breaks before $\theta = {\pi\over 2}$, right ?

theorizing what could happen...
a quantitative description of the motion

Ah, so not the equation of motion ?

make a better sketch ...
and clearly define a few things

Well, with the better sketch
$$
v = \frac{u}{ \cos \theta}
$$
and the newly defined 'things' you can write $\cos\theta$ as a function of time, right ?

 

[burian]

1. Yes it breaks there
2. Sorry I misspoke, I meant I want an equation giving position as a function of time
3. I don't understand how I could write theta as a function of time by a better sketch

 

[BvU]

$\qquad\quad$

 

[burian]

Thank you, I'll try doing with this new information and inform if I have any troubles

 

[burian]

so this is what I got,

v= \frac{u}{ \cos \theta}

by using trigonometry on your diagram,

v= \frac{2u (L_o - ut) }{ \sqrt{ (2L_o -ut) -D^2} }

Integrating this I got

y(t) = - \frac{D}{6} [ (\frac{2}{D})^2 (L_o - ut)^2 -1 ]^{\frac32} +C

But this is motion of block.. it seems rather complex? I"m not sure but I did not think a simple string pulley motion would turn this ugly looking

Anyways, my original goal was finding position of a mass segment.. I"m still not sure how to write a position function of it, but what I did is,

\sin \theta = \frac{D}{2(L_o -ut)}
\dot{\theta} \cos \theta =\frac{D}{2} \frac{u}{ (L_o -ut)^2}

rerranging and integrating,

\theta(t) = \int_{t=0}^{t=t_f} \frac{Du dt}{\sqrt{4 (L_o -ut)^2 -D^2} ( L_o -ut)^2}

I'm really confused, is the motion this complex?further, how would I find absolute position of a mass segment piece? I think I could write write the coordinatesin polar form and get a simple expression somehow but I"m not sure

 

[haruspex]

by using trigonometry on your diagram,

v= \frac{2u (L_o - ut) }{ \sqrt{ (2L_o -ut) -D^2} }

Something wrong with the 2s in the denominator.
But why not use trig to go directly to y(t)? You don't need to do any integrals.

 

[burian]

oops, I forgot the square in the denominator. What do you mean? how I can get y(t) without any integrals??

 

[haruspex]

oops, I forgot the square in the denominator. What do you mean? how I can get y(t) without any integrals??

From the diagram in post #7.

 

[burian]

From the diagram in post #7.

I'm not sure how that would give me position as function of time because the diagram is like a snap shot of the motion. How do I determine the position at all times from just a single snap shot of the motion?

 

[haruspex]

I'm not sure how that would give me position as function of time because the diagram is like a snap shot of the motion. How do I determine the position at all times from just a single snap shot of the motion?

One of the distances in the diagram is a function of time.

 

[burian]

Ok, so I thought of it and I could write polar coordinate from near the top of the pulley. As,

$$ x= (L-vt) \cos \theta$$

$$ y = (L-vt) \sin \theta$$
But, I don't understand how I would write the angle as a function of time hmm

 

[haruspex]

Ok, so I thought of it and I could write polar coordinate from near the top of the pulley. As,

$$ x= (L-vt) \cos \theta$$

$$ y = (L-vt) \sin \theta$$
But, I don't understand how I would write the angle as a function of time hmm

Don't you have that $$ D/2 = (L-vt) \sin \theta$$? Doesn’t that give you the angle as a function of time? D, L and v are constants.

 

[burian]

if I were to evaluate it at a point then I would know theta at that point. So, without calculating theta by freeze framing everytime, I want to write very specifically the argument of sine as a function of time

 

[haruspex]

if I were to evaluate it at a point then I would know theta at that point. So, without calculating theta by freeze framing everytime, I want to write very specifically the argument of sine as a function of time

There is no distinction. The equation it gives you is that of theta as a function of time.

 

[BvU]

Dear @burian,

if I were to evaluate it at a point then I would know theta at that point. So, without calculating theta by freeze framing everytime, I want to write very specifically the argument of sine as a function of time

You are making it too difficult for yourself. There is no freeze framing. Suspension point $P$ is moving straight up towards $M$.

Length $PM$ is known as a function of time$$PM = \sqrt{(L_0-ut)^2 - (D/2)^2}$$

Angle $\theta$ is known as a function of time too $$\theta = \arcsin \left (L_0-ut\over D/2\right )$$

The segment that doesn't go over the pulley undergoes a motion that is the sum of the upward motion and the tilting: red, purple, blue in the sketch

The motion of a point on that segment at distance $r$ from $P$ is then easily described as the sum of the motion of $P$ plus $(r\sin\theta, r\cos\theta)$ in a cartesian coordinate system with origin $M$:$$(x,y) = \Bigl (r\sin\theta,\; r\cos\theta- \sqrt{(L_0-ut)^2 - (D/2)^2\;}\ \Bigr )$$

There is no reason to expect a simpler expresssion

 

[burian]

Dear @burian,

You are making it too difficult for yourself. There is no freeze framing. Suspension point $P$ is moving straight up towards $M$.

Length $PM$ is known as a function of time$$PM = \sqrt{(L_0-ut)^2 - (D/2)^2}$$

Angle $\theta$ is known as a function of time too $$\theta = \arcsin \left (L_0-ut\over D/2\right )$$

The segment that doesn't go over the pulley undergoes a motion that is the sum of the upward motion and the tilting: red, purple, blue in the sketch

View attachment 267565

The motion of a point on that segment at distance $r$ from $P$ is then easily described as the sum of the motion of $P$ plus $(r\sin\theta, r\cos\theta)$ in a cartesian coordinate system with origin $M$:$$(x,y) = \Bigl (r\sin\theta,\; r\cos\theta- \sqrt{(L_0-ut)^2 - (D/2)^2\;}\ \Bigr )$$

There is no reason to expect a simpler expresssion

OH I get it now, it clicked from when you wrote it in terms of arcsin. Thank you for bearing with me.