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

[Dale]

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!

 

[berkeman]

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...)

 

[A.T.]

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.

 

[sysprog]

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?

 

[Baluncore]

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.

 

[sysprog]

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 $T>0$?

 

[Baluncore]

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.

 

[sysprog]

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?

 

[PeroK]

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.

 

[A.T.]

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:
$$\mu= \frac{1}{2} (\frac{1}{\theta} - 1)$$

I think it makes more sense that when μ goes to 0, then θ approaches π/2 rather than 1 rad.

 

[PeroK]

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.

 

[A.T.]

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.

 

[sysprog]

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:

So, there is no finite initial time during which the block does not move.

and this statement of mine:

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:

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:

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.