Olympiad Problem -- Revisiting the problem with 3 blocks and a pulley

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  • #36
sysprog
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Here a simulation with M/m = 5



Done in Algoodo:
http://www.algodoo.com/

I would like to see the input to the simulation; not just the resulting simulation ##-## please post something that includes the full set of parameters.
 
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  • #37
PeroK
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To simplify this a bit, I will assume that ##\theta## is small, so ##\cos(\theta)=1## and ##\sin(\theta)=\theta##. So the new Lagrangian with the extra degree of freedom is $$L=m r \left(g+\dot \theta \dot x\right)+\frac{1}{2} (2 m+M) \dot x^2+m (\theta+1) \dot r
\dot x+m \dot r^2+\frac{1}{2} m r^2 \dot \theta^2 $$ I would not mind at all if someone could check this. Even with the small angle approximation it is a pretty hairy Lagrangian.

The Euler Lagrange equations are: $$ -(2 m+M) \ddot x-m (\theta +1) \ddot r-2 m \dot r \dot \theta -m r \ddot \theta =0$$ $$ m \left(g-2 \ddot r+r \dot \theta ^2-\theta \ddot x-\ddot x\right)=0 $$ $$ -m r \left(2 \dot r \dot \theta +r(t) \ddot \theta +\ddot x\right)=0 $$
It's probably necessary to delay the small angle approximation until at least we have the E-L equations. This gives:
$$2\ddot r + (1+\sin \theta)\ddot x -r\dot \theta^2 - g\cos \theta = 0$$$$(2m + M)\ddot x + m[(1+\sin \theta)\ddot r + 2\dot r \dot \theta \cos \theta + r \ddot \theta \cos \theta - r\dot \theta^2 \sin \theta] = 0$$$$r \ddot \theta + \ddot x \cos \theta + 2\dot r \dot \theta + g\sin \theta = 0$$
 
  • #38
PeroK
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If we look for an "equilibrium" solution of constant ##\theta## the equations (with ##\mu = \frac M m##) are:
$$2\ddot r + (1+\sin \theta)\ddot x = g\cos \theta$$$$(2 + \mu)\ddot x + (1+\sin \theta)\ddot r = 0$$$$\ddot x \cos \theta = -g\sin \theta$$And the solution is:
$$\ddot x = -(\tan \theta)g$$$$\ddot r = \frac{1 + \sin \theta}{2\cos \theta}g$$Where ##\theta## can be obtained numerically from the equation:$$\mu = \frac{1 - \sin \theta}{2\tan \theta}$$E.g. if ##\mu = 5##, then:
$$\theta \approx 0.09$$The small angle approximation is, of course$$\theta = \frac{1}{2\mu + 1}$$I can't prove, however, that the solution tends asymptotically to that equilibrium angle.
 
  • #39
PeroK
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Here's an idea. If we look at the first E-L equation:
$$2\ddot r + (1+\sin \theta)\ddot x -r\dot \theta^2 - g\cos \theta = 0$$All the terms are finite as ##t \to \infty##, except possibly the term ##r\dot \theta^2##. So, this must be asymptotically finite too. But, ##r## increases without limit, so ##\dot \theta## must tend to zero.
 
  • #40
jbriggs444
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I would like to see the input to the simulation; not just the resulting simulation ##-## please post something that includes the full set of parameters.
It has the right qualitative behavior. The falling block never moves rightward. It does (eventually) move leftward. The big block accelerates away leftward starting immediately.
 
  • #41
A.T.
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I would like to see the input to the simulation; not just the resulting simulation ##-## please post something that includes the full set of parameters.

I attached the Algodoo file. You can rename it to "block_pulley.phz" and open it in the free software. There you can check and change all the parameters:

http://www.algodoo.com/

Here an updated animation with a grid.

 

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  • #42
Baluncore
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The big block accelerates away leftward starting immediately.
I thought of falling m as a pendulum bob, that started swinging, driven by big M moving sideways. As a pendulum, the period of oscillation is increasing as the string lengthens, and it can only swing through half a cycle before falling m collides with big M.
Given sufficient string will that collision ever occur before it runs out of string? Or will the period lengthen at a rate that prevents the collision until after it runs out of string?
 
  • #43
PeroK
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I thought of falling m as a pendulum bob, that started swinging, driven by big M moving sideways. As a pendulum, the period of oscillation is increasing as the string lengthens, and it can only swing through half a cycle before falling m collides with big M.
Given sufficient string will that collision ever occur before it runs out of string? Or will the period lengthen at a rate that prevents the collision until after it runs out of string?
My analysis shows that it tends to a fixed angle determined by ##\frac M m##. It doesn't oscillate. See also AT's simulation.
 
  • #44
sysprog
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A.T. said:
There you can check and change all the parameters.
How do I check the parameters?
 
  • #45
A.T.
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How do I check the parameters?
In Windows you right click on objects. Then under Material you have mass and friction.
 
  • #46
sysprog
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Baluncore said:
I thought of falling m as a pendulum bob, that started swinging, driven by big M moving sideways.
If falling ##m## doesn't move leftward exactly when and as much as the pulley does, but at some ##t>0## time later it does move leftward (unlike as in @A.T.'s simulation, in which it tilts so as to maintain topwise orthogonality wrt the string, but is somehow constrained to not swing leftward, despite the leftward movement of the pulley from which it is hanging), then when does it do that, and if then, why then? If it's not the case that it never moves leftward, then why wouldn't it move leftward immediately as ##M## and the pully do, given that the pulley moves leftward only as driven by the descent of falling ##m##?
 
  • #47
PeroK
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If falling ##m## doesn't move leftward exactly when and as much as the pulley does, but at some ##t>0## time later it does move leftward (unlike as in @A.T.'s simulation, in which it tilts so as to maintain topwise orthogonality wrt the string, but is somehow constrained to not swing leftward, despite the leftward movement of the pulley from which it is hanging), then when does it do that, and if then, why then? If it's not the case that it never moves leftward, then why wouldn't it move leftward immediately as ##M## and the pully do, given that the pulley moves leftward only as driven by the descent of falling ##m##?
Gravity acts on the hanging mass as soon as it is released. There is a non-zero acceleration at time ##t= 0##. If we drew a graph of the acceleration it would start at some value such as ##g/3## or something like that.

The force on the top mass is also instantaneous, so it moves with a non-zero acceleration at time ##t=0##.

And, the lateral force on the pulley is instantaneous, so the large block moves with a non-zero acceleration at time ##t = 0##.

The difference with the hanging mass is that there is zero acceleration at time ##t=0## and the acceleration must continuously increase from zero. It cannot, like the mass M, suddenly take off with non-zero acceleration.

If we look at the acceleration again after a very short time, then the side mass has non-zero acceleration leftwards: but it is still (much) less than the acceleration with which block ##M## began.

Technically, of course, the hanging mass is moving leftwards at any time ##t > 0##, but with less initial acceleration and hence less velocity than the mass ##M##.

And, in fact, the above analysis shows that it gradually catches up with the leftwards acceleration of ##M## until an equilibrium angle is reached.
 
  • #48
A.T.
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If it's not the case that it never moves leftward,
It moves leftward. Look at the animation and note where it ends up relative to the vertical grid line it started on.

then why wouldn't it move leftward immediately
For the hanging mass:

- initial horizontal velocity is zero.
- initial horizontal acceleration is zero (because the string is vertical)

But the above doesn't mean that all the further derivatives of horizontal position (jerk, snap, crackle, pop) are also zero. So it starts out moving horizontally very gently.
 
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  • #49
jbriggs444
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But the above doesn't mean that the all the further derivatives of horizontal position (jerk, snap, crackle, pop) are also zero.
Note that it is mathematically possible to have an displacement over time graph with in which all derivatives are zero but in which displacement is nonetheless only zero exactly at ##t=0##.

As I recall position ##s = e^{-1/t^2} - 1## for ##t \neq 0## and ##s=0## for ##t = 0## is a continuous and infinitely differentiable function satisfying those properties.
 
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  • #50
sysprog
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It moves leftward. Look at the animation and note where it ends up relative to the vertical grid line it started on.


For the hanging mass:

- initial horizontal velocity is zero.
- initial horizontal acceleration is zero (because the string is vertical)

But the above doesn't mean that all the further derivatives of horizontal position (jerk, snap, crackle, pop) are also zero. So it starts out moving horizontally very gently.
Thanks for the correction. I still have to finish mulling over over the consequences ##-## I think that I will probably soon wind up owing you, and @jbriggs444, @PeroK, @Dale, @berkeman, @PeterDonis, @Baluncore, @jedishrfu, and others, a big apology for my hard-headedness, and a big thank you for your patience and efforts.

I recognize and acknowledge that a higher derivative can be nonzero when a lower one is zero, e.g. jounce/snap nonzero when velocity, acceleration, and jerk are all zero:

Let ##f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1##.
Then ##f'(x) = 4x^3 - 12x^2 + 12x - 4##,
##f''(x) = 12x^2 - 24x + 12##,
##f'''(x) = 24x - 24##,
##f''''(x) = 24##.
So ##f(1) = f'(1) = f''(1) = f'''(1) = 0##, but ##f''''(1) = 24##.
 
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  • #51
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big apology for my hard-headedness, and a big thank you for your patience and efforts.
No need for the apology from my perspective, but a thank you is always well received!
 
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  • #52
berkeman
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I still have to finish mulling over over the consequences I think that I will probably soon wind up owing you, and @jbriggs444, @PeroK, @Dale, @berkeman, @PeterDonis,@ @Baluncore, @jedishrfu, and others
(for the record, I think all I contributed to this very interesting thread was fixing up the thread title early on...) :smile:
 
  • #53
A.T.
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I recognize and acknowledge that a higher derivative can be nonzero when a lower one is zero, e.g. jounce/snap nonzero when velocity, acceleration, and jerk are all zero:
Yes, but as @jbriggs444 notes in post#49, even all derivatives being zero at some point, doesn't imply the function is constant. The intuitive cause-effect interpretation of calculus (f' causes f to change) is not generally applicable.
 
  • #54
sysprog
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It has the right qualitative behavior. The falling block never moves rightward. It does (eventually) move leftward. The big block accelerates away leftward starting immediately.
Doesn't the falling block move leftward immediately, even if not as much as the big block with the pulley?
 
  • #55
Baluncore
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Doesn't the falling block move leftward immediately, even if not as much as the big block with the pulley?
Yes. In theory, everything starts to integrate at the instant; t = 0. The falling m block will not begin to accelerate horizontally until after the big M block has first moved ahead.
The falling m block accelerates at a zero rate initially, very much lower than the acceleration of the big M block, so it's velocity lags behind the big M block until the big M block has begun to move sufficiently far ahead for the falling m block to be pulled along by the string.
 
  • #56
sysprog
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Yes. In theory, everything starts to integrate at the instant; t = 0. The falling m block will not begin to accelerate horizontally until after the big M block has first moved ahead.
The falling m block accelerates at a zero rate initially, very much lower than the big M block, so it lags behind the big M block until the big M block has begun to move sufficiently far ahead for the falling m block to follow.
Isn't that 'after' no later than infinitesimally after? Don't both masses commence to move leftward exactly as soon as ##T>0##?
 
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  • #57
Baluncore
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Don't both masses commence to move leftward exactly as soon as T > 0 ?
Yes. All the plots of acceleration, velocity and position pass through the origin at t = 0.
On release at t = 0, the big M block gains a fixed acceleration.
At t = 0, the falling m block has zero horizontal acceleration.
 
  • #58
sysprog
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Yes. All the plots of acceleration, velocity and position pass through the origin at t = 0.
On release at t = 0, the big M block gains a fixed acceleration.
At t = 0, the falling m block has zero horizontal acceleration.
What about at ##t=0+\omega##? Isn't the leftward acceleration of ##m## at that point also greater than zero? I think that at any ##t>0## the leftward acceleration of falling ##m## is also greater than zero. Do you agree with that?
 
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  • #59
PeroK
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Isn't that 'after' no later than infinitesimally after? Don't both masses commence to move leftward exactly as soon as ##T>0##?
What you're doing now is simply wrestling with the fundamental concept of calculus and continuous change. There is no smallest number greater than zero. So, there is no finite initial time during which the block does not move. But, that is not mathematically incompatible with an initial velocity and initial acceleration of zero.

Think of the function ##y = t^3## as an example. We have ##y(0) = 0##, ##y'(0) = 0## and ##y''(0)= 0##. And, yet, for all ##t > 0##, we have ##t^3 > 0##.

The equations of motion in this case are too complicated to yield an analytic solution for the motion of the side block. We know from the physical constraints, however, that the initial velocity and acceleration are zero. And, we also know that for any time ##t > 0## both the velocity and the acceleration are non-zero. There is no contradiction there.
 
  • #60
A.T.
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The small angle approximation is, of course$$\theta = \frac{1}{2\mu + 1}$$I can't prove, however, that the solution tends asymptotically to that equilibrium angle.

I assumed that (in the rest frame of M) the sum of gravity and inertial force on the hanging m is parallel to the string. This lead me to the following relationship:
$$\mu= \frac{1}{2} \: cos(\theta) \: cot(\theta)$$
I compared it to your approximation, which would be:
$$\mu= \frac{1}{2} (\frac{1}{\theta} - 1)$$

block_angle.png


I think it makes more sense that when μ goes to 0, then θ approaches π/2 rather than 1 rad.
 
  • #61
PeroK
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I assumed that (in the rest frame of M) the sum of gravity and inertial force on the hanging m is parallel to the string. This lead me to the following relationship:
$$\mu= \frac{1}{2} \: cos(\theta) \: cot(\theta)$$
I get $$\mu= \frac{1}{2} (1- \sin(\theta)) \cot(\theta)$$I got that two different ways. Initially by an analysis of forces looking for a constant angle. Then, by using the E-L equations with constant angle.
I think it makes more sense that when μ goes to 0, then θ approaches π/2
The full formula above does this. But not, of course, the small angle approximation.
 
  • #62
A.T.
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I get $$\mu= \frac{1}{2} (1- \sin(\theta)) \cot(\theta)$$I got that two different ways. Initially by an analysis of forces looking for a constant angle. Then, by using the E-L equations with constant angle.
Yours also agrees better with the simulation. So I guess I made an error somewhere.
 
  • #63
sysprog
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PeroK said:
There is no smallest number greater than zero.
I didn't mean to suggest that I thought there to be such a thing as a least number among the positive reals; mea culpa for any notional abuse on my part; I am not at odds with the fundamental theorem of calculus, and I accept the epsilon-delta definition of limits, as described here: https://mathworld.wolfram.com/Epsilon-DeltaDefinition.html.

I was seeking to evoke an acknowledgment that at any finite time greater than zero, the leftward acceleration of falling ##m## is non-zero ##-## that despite its acceleration not being the same as that of the large block and pulley, it is just as true for it that it is non-zero at any arbitrarily small finite ##t>0##, as it is for the large block and pulley.

I think that there is no incompatibility between this statement of yours:
PeroK said:
So, there is no finite initial time during which the block does not move.
and this statement of mine:
sysprog said:
I think that at any ##t>0## the leftward acceleration of falling ##m## is also greater than zero.
That seeems to me to be obscured in some aspects of the following exchange:
I posted:
sysprog said:
Isn't the leftward acceleration of ##m## at that point also greater than zero? I think that at any ##t>0## the leftward acceleration of falling ##m## is also greater than zero. Do you agree with that?
and @Baluncore posted:
sysprog said:
Isn't the leftward acceleration of ##m## at that point also greater than zero?
Yes. All the plots of acceleration, velocity and position pass through the origin at t = 0.
On release at t = 0, the big M block gains a fixed acceleration.
At t = 0, the falling m block has zero horizontal acceleration.
I think that "on release at t = 0" might be more clearly expressed as "immediately at ##t>0##", and similarly, I think that the hanging ##m## block isn't rightly called "falling" until ##t>0##. However, I think that the first sentence of @Baluncore's response is abundantly clear: he said "Yes" to the question.
 

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