Bead on a rod which is being pulled by an ideal string with velocity v

[Null_Void]

Homework Statement

Pic attached below for reference

A bead of mass m, which can slide on a straight horizontal frictionless rod is being pulled by an ideal string of length $l$ such that the end P moved with velocity $v_{0}$ always directed along the thread.

Find an expression for the tensile force T in the thread at an instant when the thread makes an angle $θ$ with the rod.

Relevant Equations

$vcosθ$ = $v_{0}$

Along the vertical:

$mg$ + $Tsinθ$ = $N$

Along the horizontal:

$Tsinθ$ = $ma$

I know That the velocity of each particle of the thread along the thread must be $v_{0}$ since the thread is inextensible.

Now let's say the bead moves with some velocity $v$
For the particle attached to the bead to move with a velocity $v_{0}$ along the thread:
$v$cos$θ$= $v_{0}$
Is this inference right?

Though I'm not sure about the interplay of forces,
Along the vertical:
$mg$ + $Tsinθ$ = $N$
Along the horizontal:
$Tsinθ$ = $ma$
And this is where I'm also confused.

If the string is taut, then there must be some force exerted by the string on the bead. If there is a force, then it must accelerate. But if it accelerates, then velocity of the thread can't be $v_{0}$ throughout. I seem to be add odds with myself.
Beyond this I'm completely blank. I absolutely don't know how to proceed. There's not a single approach that comes to mind.

Any Help is deeply appreciated.

 

[haruspex]

For the particle attached to the bead to move with a velocity $v_{0}$ along the thread:
$v$#cos$θ$= $v_{0}$
Is this inference right?

Yes. (This is something many students get wrong, writing $v=v_0\cos(\theta)$.)
Btw, you can write the LaTeX more simply as "v_0\cos(\theta)" etc.

Along the vertical:
$mg$ + $Tsinθ$ = $N$

You do not care about N, so this equation won’t be useful.

Along the horizontal:
$Tsinθ$ = $ma$

If the string is taut, then there must be some force exerted by the string on the bead. If there is a force, then it must accelerate. But if it accelerates, then velocity of the thread can't be $v_{0}$ throughout.

You seem to be assuming $\theta$ is constant.

 

[Null_Void]

Yes. (This is something many students get wrong, writing $v=v_0\cos(\theta)$.)
Btw, you can write the LaTeX more simply as "v_0\cos(\theta)" etc.

You do not care about N, so this equation won’t be useful.

You seem to be assuming $\theta$ is constant.

So you mean that both $T$ and $θ$ will adjust themselves accordingly such that string velocity is $v_0$?
But even then how do I proceed after that?
This is one of the trickiest questions I've come across, with me having literally zero idea on how to proceed.

Also Thanks for the heads up with the LaTeX

 

[haruspex]

So you mean that both $T$ and $θ$ will adjust themselves accordingly such that string velocity is $v_0$?

Theta must change or P would move horizontally, not along the line of the string.

Just noticed you have $\sin(\theta)$ in both directions. That cannot be right.

 

[Null_Void]

Theta must change or P would move horizontally, not along the line of the string.

Just noticed you have $\sin(\theta)$ in both directions. That cannot be right.

The bead travels with a velocity $v$ horizontally.
If we resolve $v_0$ horizontally we get $v_0\cos(\theta)\ = v\cos^2(\theta)$
So essentially the bead is faster than the thread horizontally unless $θ$ is $0$ which is not the case here.

Perhaps is there something we can do with reference frames?
I remember a similar question being solved using a certain frame of reference though I don't know how.

This is the answer given in the book:

 

[haruspex]

This is the answer given in the book:
View attachment 352596

There is a strong clue in the presence of $l$ in the answer. What equation can you write involving that?

 

[Null_Void]

There is a strong clue in the presence of $l$ in the answer. What equation can you write involving that?

Circular motion maybe?
$m{v_0}^2/l$

 

[haruspex]

Circular motion maybe?
$m{v_0}^2/l$

Think about the displacement of the bead as a function of time and as a function of the angle.

 

[Null_Void]

Think about the displacement of the bead as a function of time and as a function of the angle.

I'm sorry but I'm not sure. Could you provide any hints? Won't the bead accelerate? Do I have to take that into account?

 

[PeroK]

I'm sorry but I'm not sure. Could you provide any hints? Won't the bead accelerate? Do I have to take that into account?

If the bead is not accelerating, then there is no.tension. This would be the case for $\theta =0$.

 

[Null_Void]

Think about the displacement of the bead as a function of time and as a function of the angle.

Let's say I consider an infinitesimal time interval $dt$ where the angle between the thread and the bead increases by $dθ$

$dx/dt\ =\ v$

How do I represent $x$ in terms of $l$ and $cosθ$
Also is this the right approach? or am I still missing something?

 

[PeroK]

Let's say I consider an infinitesimal time interval $dt$ where the angle between the thread and the bead increases by $dθ$

$dx/dt\ =\ v$

How do I now try to represent $x$ in terms of $l$ and $cosθ$
Also is this the right approach? or am I still missing something?

I haven't looked at this problem yet, but isn't that just some trigonomery?

 

[Null_Void]

I haven't looked at this problem yet, but isn't that just some trigonomery?

Well, like I've already mentioned I don't really know how to proceed and this is the only thing that came to mind since @haruspex did ask me to represent the displacement in terms of $θ$ and $t$

 

[PeroK]

Well, like I've already mentioned I don't really know how to proceed and this is the only thing that came to mind since @haruspex did ask me to represent the displacement in terms of $θ$ and $t$

What's stopping you?

 

[haruspex]

How do I represent $x$ in terms of $l$ and $cosθ$

Simple trig. Drop a perpendicular from P to the rod. Choose an arbitrary starting position.

 

[Null_Void]

Simple trig. Drop a perpendicular from P to the rod. Choose an arbitrary starting position.

$dx$ = $-l\sin(\theta)dθ$ ?
$v$ = $-lsinθ dθ/dt$

 

[haruspex]

$dx$ = $-l\sin(\theta)dθ$ ?
$v$ = $-lsinθ dθ/dt$

That depends which way you are taking as positive for x. Presumably you want v to be positive, so change that minus to a plus.
In algebraic terms, $x=c-l\cos(\theta)$.

edit: whoops, no, that's rubbish. please ignore.
See @PeroK's post.

 

[PeroK]

Just to contribute something. I would let $x$ be the horizontal displacement of the mass $m$, then $X, Y$ be the horizontal and vertical displacements of the point $P$ relative to the mass $m$.

That means that:
$$\vec P = (x + X, Y) = (x + l\cos \theta, l\sin \theta)$$Then we can differentiate that to get the velocity of the point $P$, which we know from the constraint is given by:
$$\vec v_P = v_0(\cos \theta, \sin \theta)$$It should all come out with a bit of algebra from that.

 

[Null_Void]

Going by @PeroK 's answer,
I'm getting

$v_0\cos(\theta)\ = v - l\sin(\theta)dθ/dt$

Should I now try to equate this with

$a = dv/dt$ where $a = T\sin(\theta)/m$ ?

 

[PeroK]

Going by @PeroK 's answer,
I'm getting

$v_0\cos(\theta)\ = v - l\sin(\theta)dθ/dt$

Should I now try to equate this with

$a = dv/dt$ where $a = T\sin(\theta)/m$ ?

Try looking at the y-equation first.

 

[Null_Void]

Try looking at the y-equation first.

$v_0\sin(\theta) = l\cot(\theta)dθ/dt$

Should I try integrating now to obtain relation between $v_0, θ and t$.

Also, how am I supposed to approach such questions? What are we trying to get at here? How does representing the velocity in terms of $t\and\θ$ help us in obtaining the answer?

Thank You

 

[PeroK]

$v_0\sin(\theta) = l\cot(\theta)dθ/dt$

Should I try integrating now to obtain relation between $v_0, θ and t$.

Also, how am I supposed to approach such questions? What are we trying to get at here? How does representing the velocity in terms of $t\and\θ$ help us in obtaining the answer?

Thank You

You have to break the problem down. Some of it is technique - like setting the problem up. Some of it is doing things that you've done before - like trigonometry. Some of it is trying things to see what works in this particular case.

It seems to me you are struggling on all three.

In this case you got one equation:
$$v_0 = v\cos \theta$$But, I'm not sure where you got that from? Then you did part of the next step:

$v_0\cos(\theta)\ = v - l\sin(\theta)dθ/dt$

And the problem there is that you have the $\dot \theta$. I can't explain why you didn't differentiate the y-component as well, to give:
$$l(\cos \theta) \dot \theta = v_0\sin \theta$$That gives you an equation for $\dot \theta$:
$$\dot \theta = \frac {v_0}{l} \tan \theta$$Then, you just have to put that into your other equation.

This is technique. Why does it work? It works because that's how maths/physics problems are solved: by writing down equations, trying to find an expression for something you want to eliminate and then simplifying.

I'm not saying it's easy, but you have to persevere.

 

[PeroK]

Actually, that gives you the equation that you got from somewhere else:
$$v = \frac{v_0}{\cos \theta}$$But, now you also have an equation for $\dot \theta$, so you can differentiate that and use the substitution for $\dot \theta$ again. Then, there is still a bit more work to do.

This is what I meant about breaking the problem down. You are never going to get to $T$ in one step.

 

[Null_Void]

You have to break the problem down. Some of it is technique - like setting the problem up. Some of it is doing things that you've done before - like trigonometry.

I guess the first part was where I was lagging the most. Introducing a coordinate system and and then deriving a relation between the velocites was definitely out of the box for me. So anyway, Thank You for that
I finally solved the problem, so huge thanks to everyone in this thread. I just have one last query.
I came across a different approach to this question and I do understand that it should have been brought up earlier but I would deeply appreciate it if someone can clarify one point.

So the question is solved in the frame of P where the bead undergoes circular motion which is very confusing to me. The bead is constrained to move along the rod, so how can you assume it will undergo circular motion?

 

[PeroK]

I guess the first part was where I was lagging the most. Introducing a coordinate system and and then deriving a relation between the velocites was definitely out of the box for me. So anyway, Thank You for that
I finally solved the problem, so huge thanks to everyone in this thread. I just have one last query.
I came across a different approach to this question and I do understand that it should have been brought up earlier but I would deeply appreciate it if someone can clarify one point.

So the question is solved in the frame of P where the bead undergoes circular motion which is very confusing to me. The bead is constrained to move along the rod, so how can you assume it will undergo circular motion?

View attachment 352644

That's a completely different solution. That's the shortcut for getting $v\cos \theta = v_0$.

P is accelerating, so I wouldn't be confident of that solution. I must admit I don't see how you can transform forces from an inertial frame to an accelerating frame in that way.

My solution is more straightforward - and, dare I say, more reliable. It's a real issue that you can't do it "my" way after all the hints and tips, because there was nothing conceptually challenging or tricky that way.

I'd be concerned that you are so far out of your depth, that even when look at the full solution, you don't understand it. It seems you were looking at both the clever book solution and my approach, but you can't ultimately do problems of this difficulty yourself?

 

[Null_Void]

That's a completely different solution. That's the shortcut for getting $v\cos \theta = v_0$.

P is accelerating, so I wouldn't be confident of that solution. I must admit I don't see how you can transform forces from an inertial frame to an accelerating frame in that way.

My solution is more straightforward - and, dare I say, more reliable. It's a real issue that you can't do it "my" way after all the hints and tips, because there was nothing conceptually challenging or tricky that way.

I'd be concerned that you are so far out of your depth, that even when look at the full solution, you don't understand it. It seems you were looking at both the clever book solution and my approach, but you can't ultimately do problems of this difficulty yourself?

No, You're mistaken. I vaguely remembered a similar problem (not the same problem) and just had in mind that the solution involved some complex frame manipulation and put it off. It's literally been months since that. Since I came across this question, i thought I would look up it and fortunately there was a solution for this problem though I wasn't aware of that at the time I made this post.
I deeply appreciate your answer since it is actually more straightforward and overall a better one in my opinion.
But since I made this post I thought I might as well clear up the confusion in the complex method just for the sake of knowing what it was.

 

[PeroK]

But since I made this post I thought I might as well clear up the confusion in the complex method just for the sake of knowing what it was.

When you change to an accelerating reference frame, you have to add a fictictious force to everything. Where is that done here? I don't understand why $T\cos^2 \theta$ works. Not without further analysis.

Here is the analysis. If we use my method, we see that the acceleration of $P$ is perpendicular to the string. I think you need this fact for the clever solution to work. When we add the fictitious force, that is perpendicular to the line between $P$ and $m$ and does not affect the centripetal acceleration.

You'd need to ask whoever came up with the clever solution whether they saw that - or whether they just got lucky.

The real net force on $m$ is $T\cos \theta$ along the line of the wire, hence $T \cos^2 \theta$ in the direction of P.; plus a component of $T\cos \theta \sin \theta$ perpendicular to the line between $m$ and $P$ - which again doesn't affect the centripetal acceleration.

From that, we know that $T \cos^2 \theta$ provides the centripetal acceleration in the frame of $P$ and the result follows.

 

[PeroK]

I deeply appreciate your answer since it is actually more straightforward and overall a better one in my opinion.

Thanks for that. Also, if different criteria were specified for $P$, then that approach would work. You would just equate the two expressions for the velocity of $P$. No need for tricky accelerating reference frames.