Why does the big block accelerate?

[haruspex]

It would be interesting to see if we do indeed get the oscillatory motion.

I think it must because as I posted there is a steady state angle the sine of which is given by s in $2Ms=m(1-s)^2$. If it starts at the vertical then it must swing past that angle, then back, etc.

 

[etotheipi]

I think it must because as I posted there is a steady state angle the sine of which is given by s in $2Ms=m(1-s)^2$. If it starts at the vertical then it must swing past that angle, then back, etc.

One thing I'm not sure about, is the acceleration of the large block uniform? It is in the case that the suspended mass is constrained to move down a rail attached to the side of the large block, but I'm not sure if it will still be uniform if the suspended mass can swing all over the place. A non-uniform acceleration of the larger block would seem to imply that the equilibrium angle is a function of time. Is that vaguely right?

 

[jbriggs444]

I think it must because as I posted there is a steady state angle the sine of which is given by s in $2Ms=m(1-s)^2$. If it starts at the vertical then it must swing past that angle, then back, etc.

I am less certain that the motion will be appear under-damped but have not thought the situation through in sufficient detail to argue with the conclusion. [The odds of catching @haruspex in error are not great on the best of days].

 

[jbriggs444]

One thing I'm not sure about, is the acceleration of the large block uniform?

No. It will not be. If you change the angle of the vertical[-ish] string segment, you change the vector sum of the tensions on the pulley and, hence, the lateral thrust on the large block.

There is no corresponding increase in string tension (which, if anything, would scale with the cosine of the angle from the vertical) to go with the change in lateral thrust due to angle (which would go with the sine of the angle).

 

[kuruman]

A lot has been said already, I agree with most but not all of it. When I first solved this problem I went back to basics to answer the original question, "Find the acceleration of the big block immediately after the system is released." To do that, one needs to draw a free body diagram (FBD) and then apply Newton's second law. Here is my FBD with the assumption that immediately after release, the hanging part of the string is vertical and the tension is $T=mg$. I did not include a normal force exerted by the block on the hanging mass because that force is zero at $t=0$ and remains zero as the block accelerates to the left at $t>0$.

This FBD and the $T=mg$ assumption do not result in the purported answer. Based on the above, one can find the horizontal acceleration is one's head. Has anyone obtained the answer $a=\dfrac{mg}{(2M+m)}$? Because of PF rules, I cannot post my derivation here, but I will be happy to send it to you by PM. However, if you find the error of my ways on the basis of what I have already said, I will welcome your reply here. As some of you already know, I have an occasional blind spot.

 

[haruspex]

A lot has been said already, I agree with most but not all of it. When I first solved this problem I went back to basics to answer the original question, "Find the acceleration of the big block immediately after the system is released." To do that, one needs to draw a free body diagram (FBD) and then apply Newton's second law. Here is my FBD with the assumption that immediately after release, the hanging part of the string is vertical and the tension is $T=mg$. I did not include a normal force exerted by the block on the hanging mass because that force is zero at $t=0$ and remains zero as the block accelerates to the left at $t>0$.
View attachment 267634
This FBD and the $T=mg$ assumption do not result in the purported answer. Based on the above, one can find the horizontal acceleration is one's head. Has anyone obtained the answer $a=\dfrac{mg}{(2M+m)}$? Because of PF rules, I cannot post my derivation here, but I will be happy to send it to you by PM. However, if you find the error of my ways on the basis of what I have already said, I will welcome your reply here. As some of you already know, I have an occasional blind spot.

I had done this for the more general case of three different masses and got $\frac{m_1m_2}{Mm_1+Mm_2+m_1m_2}g$. This reduces to that expression when $m_1=m_2$. Note the curious fact that my expression is symmetric in $m_1, m_2$.

 

[etotheipi]

I agree with your expression @kuruman. but I don't understand what you mean by saying that $T = mg$? I don't think that $T = mg$ at $t=0$.

Small masses are at positions $x_1$ and $x_2$, large box has CoM at $x_3$. Height of suspended mass w.r.t. pulley is $y$. Pulley is at $x_4 = x_3 + \delta$. At $t=0$,$$T - mg = m\ddot{y}$$ $$T = m\ddot{x}_1$$ $$-T = M\ddot{x}_3$$Conservation of string, $$(x_4 - x_1) + \sqrt{y^2 + (x_2 - x_4)^2} = \text{constant}$$since $|y| = -y$, at $t=0$,$$\dot{x}_4 - \dot{x}_1 - \dot{y} = 0 \implies \dot{x}_3 - \dot{x}_1 - \dot{y} = 0 \implies \ddot{y} = \ddot{x}_3 - \ddot{x}_1$$now back to the dynamical equations$$T - mg = m\ddot{x}_3 - m\ddot{x}_1$$ $$-M\ddot{x}_3 - mg = m\ddot{x}_3 + M\ddot{x}_3$$ $$\ddot{x}_3 = -\frac{mg}{2M+m}$$

 

[kuruman]

I agree with your expression @kuruman. but I don't understand what you mean by saying that $T = mg$? I don't think that $T = mg$ at $t=0$.

I shouldn't have thought so either. At $t=0$, the tension changes (almost) discontinuously from $mg$ to something less than $mg$. My blind spot was that I approached $t=0$ from the negative side, i.e. when the tension is still $mg$. That is clearly incorrect because "immediately after" (as opposed to "immediately before") means $t=0+\delta t$ in the limit $\delta t\rightarrow 0$. Thanks for finding my blind spot.

 

[Orodruin]

I think that "you may neglect" means that the reader has the option of neglecting it or not; not that it is specified as definitely of a negligible magnitude

No, it does not generally have this implication. The implication is that the frictional forces are to be considered small enough to be negligible.

Regardless, this is irrelevant in relation to your erroneous claim that there will not be an initial horizontal separation between the large block and the hanging block. Verifying that this will be the case is a simple matter of considering the free-body diagrams of the two blocks. There will be a horizontal force on the large block due to the forces on the pulley and the large block will therefore have a non-zero horizontal acceleration $a$. Its horizontal displacement a short time $\tau$ after release is therefore $a\tau^2/2+\mathcal O(\tau^3)$. The forces on the hanging block are all vertical, meaning that it has zero instantaneous acceleration and therefore its horizontal displacement after the short time $\tau$ is $\mathcal O(\tau^3)$. The difference between the horizontal displacements are therefore $a\tau^2/2 +\mathcal O(\tau^3)$ which is non-zero for small enough values of $\tau$. Your assertion is simply unfounded.

 

[etotheipi]

On an unrelated note, I'm glad you're back @Orodruin!

 

[Orodruin]

From your post:
The large block cannot accelerate laterally away from the descending block; it is attached to the pulley from the tangent of which the descending block depends, and it gets all of its lateral acceleration from the conversion of the potential energy that the pulley mediates from the descent of the descending block.

This is, again, an unfounded and erroneous claim. Yes, it gets its acceleration from the action of the string on the pulley, but you have failed to analyze the implications properly. The hanging block does not accelerate horizontally at the initial moment while the large block does. The only way to avoid this is to introduce additional horizontal forces keeping the hanging block in contact with the large block.

Yes, the large block is attached to the pulley, but the hanging block is not fixed to the pulley. The rope connecting the hanging block to the pulley is free to move both by lengthening the part of the rope below the pulley and by changing the angle at which the rope exits the pulley. In order to impart horizontal acceleration on the hanging weight, the angle of the rope must first change, meaning that this becomes a higher order effect not relevant to the initial acceleration.

On an unrelated note, I'm glad you're back @Orodruin!

At least someone noticed I was gone. Cannot promise I will hang around, I historically do PF on my daily commute to work and there is no such commute when working from home ...

 

[etotheipi]

The lateral movement of the descending block is not from belatedly reacting to the lateral movement of the pulley; mediated by the tension of the string, the downward movement of the pulley is driving the lateral movement of the pulley and large block, and of the descending block itself. The lateral movement of the descending block can't lag behind the lateral movement of the pulley and large block assemblage the lateral movement of which is driven by the descent of the descending block.

The onset of lateral movement of the large block and pulley assemblage cannot precede the onset of the descent of the descending block. The onset of lateral movement of the descending block cannot delay from the moment of the onset of its descent. All of the movement in the system is contemporaneous.

@sysprog, with respect,

 

[jbriggs444]

How does the large block get laterally ahead of the descending block when the only thing moving the large block is mediation of the descent of the descending block, which can't descend without simultaneously moving laterally, which it has to do to be able to drive the pulley and large block?

What makes you assert the highlighted claim? It can descend and swing free at the same time. Nothing prevents it.

I have run the experiment. It works.

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

Moderator's note: A misinformation warning has been issued and a whole gaggle of posts in the subthread based on that misinformation have been deleted.

I am going to leave the thread closed since it seems like the OP's question has been well answered. If anyone has further contributions they would like to make, please PM me and I can reopen the thread.