Constrained motion of a ring along a horizontal rod

[arham_jain_hsr]

Homework Statement

If the velocity of the ring is v as shown in the figure then find the velocity u with which the rope is being pulled?

Relevant Equations

N/A

The correct answer is u=vcos\theta. I have understood so far to be able to conclude that \text{displacement of string} = PA - PC \approx AB
Also, \overline{AB}=\overline{AC}cos\theta
or, more generally, \vec{S}_{along\ the\ string}=(\vec{S}_{along\ the\ horizontal})cos\theta

Now, I had hoped that simply differentiating both sides of the last equation with respect to time would give me the expected answer. But, since cos\theta is not constant, doing so does not yield u=vcos\theta. How do I proceed from here?

 

[erobz]

Homework Statement: If the velocity of the ring is v as shown in the figure then find the velocity u with which the rope is being pulled?
Relevant Equations: N/A

View attachment 338963

The correct answer is u=vcos\theta. I have understood so far to be able to conclude that \text{displacement of string} = PA - PC \approx AB
Also, \overline{AB}=\overline{AC}cos\theta
or, more generally, \vec{S}_{along\ the\ string}=(\vec{S}_{along\ the\ horizontal})cos\theta

View attachment 338964

Now, I had hoped that simply differentiating both sides of the last equation with respect to time would give me the expected answer. But, since cos\theta is not constant, doing so does not yield u=vcos\theta. How do I proceed from here?

Drop a line perpendicular to the rod at the pulley, it has a fixed length $d$. From the ring to the intersection (along the rod ) is $x$. From the ring to the pulley is $\ell$.

Examine that triangle with implicit differentiation w.r.t. time.

 

[haruspex]

Now, I had hoped that simply differentiating both sides of the last equation with respect to time would give me the expected answer

Why differentiate? That would be appropriate if your equation related position of the ring to position of whatever is pulling the string, but instead you have found the incremental change to each of those. That is, you have already differentiated.

An easier approach is to say that the rate at which the ring is approaching the pulley is simply the component of v in that direction.

 

[arham_jain_hsr]

Why differentiate? That would be appropriate if your equation related position of the ring to position of whatever is pulling the string, but instead you have found the incremental change to each of those. That is, you have already differentiated.

An easier approach is to say that the rate at which the ring is approaching the pulley is simply the component of v in that direction.

Don't I have to at least divide both sides by \Delta t and find the limit \lim_{\Delta t\to 0}?

 

[erobz]

Out of curiosity, have you had any success with the method I described?

 

[haruspex]

Don't I have to at least divide both sides by \Delta t and find the limit \lim_{\Delta t\to 0}?

You did not write any $\Delta$s in post #1. You could write AB and AC in terms of u, v and $\Delta t$, then divide by that last.
And you did not define $\vec S$. Is that supposed to be speed or displacement?

 

[arham_jain_hsr]

You did not write any $\Delta$s in post #1. You could write AB and AC in terms of u, v and $\Delta t$, then divide by that last.
And you did not define $\vec S$. Is that supposed to be speed or displacement?

Displacement

 

[haruspex]

Displacement

Then you can write those in terms of u, v and $\Delta t$, then divide by that.

 

[arham_jain_hsr]

Then you can write those in terms of u, v and $\Delta t$, then divide by that.

You mean like \vec{S}_{along\ the\ string}=(\vec{S}_{along\ the\ horizontal})cos\theta \implies u\Delta t = (v\Delta t) cos\theta \implies u=vcos\theta?

 

[Delta2]

An easier approach is to say that the rate at which the ring is approaching the pulley is simply the component of v in that direction

This is what I call dense and (possibly vague !!?!), intuitive-qualitative explanation.

 

[arham_jain_hsr]

Out of curiosity, have you had any success with the method I described?

View attachment 338989

Sorry. Well, yes, I did, and was indeed able to find the correct answer. But, I don't understand where I'm going wrong in the current approach.
Like, as @haruspex pointed out:

Why differentiate? That would be appropriate if your equation related position of the ring to position of whatever is pulling the string, but instead you have found the incremental change to each of those. That is, you have already differentiated.

An easier approach is to say that the rate at which the ring is approaching the pulley is simply the component of v in that direction.

That makes sense. But, why cannot we do this instead:

Don't I have to at least divide both sides by \Delta t and find the limit \lim_{\Delta t\to 0}?

Like, where is the fallacy?

 

[haruspex]

You mean like \vec{S}_{along\ the\ string}=(\vec{S}_{along\ the\ horizontal})cos\theta \implies u\Delta t = (v\Delta t) cos\theta \implies u=vcos\theta?

Yes.

why cannot we do this instead:

In post #7, you stated that the S variables were "displacements”, so vectors. I guess you meant distances, since they are in different directions. If so, they would be OA and AP, i.e. $constant-x, l$. But that is not how you used them.

The general equation relating the ring's position and its distance from the pulley is $x^2+d^2=l^2$. Differentiating, $x\dot x=l\dot l$, so $xv=lu$.
Since $x=l\cos(\theta)$, $v\cos(\theta)=u$.

Instead, the way you used them, they were small changes in those distances. In considering such, you had already, in effect, differentiated.

 

[arham_jain_hsr]

If so, they would be OA and AP, i.e. $constant-x, l$. But that is not how you used them.

What is OA?

 

[Delta2]

You mean like \vec{S}_{along\ the\ string}=(\vec{S}_{along\ the\ horizontal})cos\theta \implies u\Delta t = (v\Delta t) cos\theta \implies u=vcos\theta?

That is correct but it is within infinitesimal logic which is kind of rigorous but not as rigorous (you have to say that the angle theta remains almost constant during the time interval dt) as what @erobz describes at post #2 and haruspex explains it in full detail with the equations in post #12.

 

[haruspex]

What is OA?

I was calling the left end of the rod O. I hoped that would be evident.

 

[arham_jain_hsr]

Instead, the way you used them, they were small changes in those distances. In considering such, you had already, in effect, differentiated.

I understand the part where you say that the way I did it, they are not actually position vectors, but changes in distances. But, what do you mean by "already differentiated"? How is this "already differentiated" when clearly we use differentiation to compute the "rates" of change?

 

[erobz]

I understand the part where you say that the way I did it, they are not actually position vectors, but changes in distances. But, what do you mean by "already differentiated"? How is this "already differentiated" when clearly we use differentiation to compute the "rates" of change?

You pretended it was a differential change in position when you said $\bar{PA}-\bar{PC} \approx \bar{AB}$. So $\theta + \Delta \theta \approx \theta$. So all that left to do is divide by $\Delta t$ and take the limit.

 

[arham_jain_hsr]

Got it. Thank you @erobz @haruspex @Delta2