Bead sliding down a smooth cord.

[peripatein]

Hello,

Homework Statement

I am asked to show that the time it would take a bead to slide down a smooth cord, positioned at an angle beta wrt the vertical axis, is independent on that angle (between the cord and the
axis). The bead starts its slide from rest.

Homework Equations

The Attempt at a Solution

I was initially not sure whether I should write down force equations or use energy conservation, so I have tried both (should I have used impulse, insead?).
In any case, for the forces acting on the bead sliding down I got:
mgsin(beta) = N ; mgcos(beta) - T = md²x/dt²
And from conservation of energy I got:
mgy = 1/2*mv²

My problem is that I am really not sure any of these equations are flawless as they stand, AND they all seem to be dependent on the angle beta.
Could someone please advise?

 

[lewando]

In any case, for the forces acting on the bead sliding down I got:
mgsin(beta) = N ;

Looks okay.

mgcos(beta) - T = md²x/dt²

Where did this come from? Is "T" tension? If so, not relevant.

And from conservation of energy I got:
mgy = 1/2*mv²

True. May not need it, though.

Since time is important, you'll need an expression containing this. Use one of the standard kinematic equations. Try to come up with a general expression for time as a function of β. If you can do that then time is dependent on β. If your general expression for time is only dependent on m,g, and L, then you have shown otherwise.

 

[peripatein]

Why would T, tension, be irrelevant here? Doesn't "smooth" merely imply without friction?
Moreover, supposing my choise of coordinate system is such where my x-axis is parallel to the slope, then will the bead not be accelerating only along the x-axis (hence, a_y = 0)?
In any case, x = 1/2*a_x*t^2, where x should be L*cos(beta) with L denoting the distance the bead has passed along the cord. Hence, t = sqrt(2Lcos(beta)/a_x) where a_x = gcos(beta) (from the equation in my previous post). Thus, t = sqrt(2gL).
Is it correct?

 

[lewando]

Comments in blue

Why would T, tension, be irrelevant here? [Because there is no force related to tension acting on the bead] Doesn't "smooth" merely imply without friction?[Yes]
Moreover, supposing my choise of coordinate system is such where my x-axis is parallel to the slope, then will the bead not be accelerating only along the x-axis (hence, a_y = 0)?[Okay]
In any case, x = 1/2*a_x*t^2, where x should be L*cos(beta) with L denoting the distance the bead has passed along the cord. [In your new coordinate system, why wouldn't x = L, the length of the cord? We are looking for the time it takes to traverse the cord.] Hence, t = sqrt(2Lcos(beta)/a_x) where a_x = gcos(beta) [a_x looks right. t does not look right based on definition of x] (from the equation in my previous postC). Thus, t = sqrt(2gL).
Is it correct?[So what would t be if β = 90 degrees?]

 

[peripatein]

Are you suggesting that t should have been 2L/[gcos(beta)], yielding t infinite for cos pi/2? Please remember that I am to show that the time of the slide from the top of the cord of slope tan(beta) is independent on beta.

 

[lewando]

if you consider two cases, β = 90 and β = 0, you will get two different times, yes?

 

[peripatein]

It should, yes. But again, the time has to be independent on beta as that is how the question is formulated: "show that the time of the slide from A to B is not dependent on the angle beta." Just to make sure, you are stating that it should depend on the angle, right?

 

[lewando]

Yes, that is what I am asserting. And you should be comfortable asserting that too, based on the 0/90 cases.

So I am reviewing the original question and the assumption I am making is that the cord is in tension such that it is essentially rigid and stays that way (no letting go of the bottom end upon release of the bead). I can't think of any more gotchas.

 

[peripatein]

Would the expression for the time remain the same?

 

[lewando]

I think you were trying to ask if t = sqrt(2L/(gcos(β))) was the result. I think it is.

 

[peripatein]

I have attached the question's diagram. Hopefully it would assist in somewhat clarifying the set up. The bead is sliding, from rest, from A to B, down the cord, as shown.
I doubt I'd be asked to prove the independence of the time on the angle beta, were it not the case.
I'd be happy to share your thoughts on this, as I do deem it rather perplexing.

 

[peripatein]

I mean, having looked at the diagram, can you now come up with an explanation why t = sqrt(2L/(gcos(β))) might not be the right answer?

 

[lewando]

I mean, having looked at the diagram, can you now come up with an explanation why t = sqrt(2L/(gcos(β))) might not be the right answer?

It appears that L is not constant. It also appears that we are talking about a chord not a cord. Thanks for sharing the diagram, it will be a big help moving forward.

 

[peripatein]

I never wrote L was constant. I actually explicitly stated that it was the distance traversed by the bead along the chord.
Should it then indeed be sqrt(2L/g)?

 

[lewando]

I never wrote L was constant. I actually explicitly stated that it was the distance traversed by the bead along the chord.
Should it then indeed be sqrt(2L/g)?

What is L in this expression then? It appears to be a constant. What it really is is the length of chord AB as a function of β.

 

[peripatein]

I am a bit confused myself. Are you claiming that t is dependent on the angle beta still?

 

[lewando]

I am going to hold off on making any new claims for now . That was my claim based on incomplete information and making the assumption that L was constant. Now that all the cards are on the table, I am observing (rather than claiming) that the lenghth of the chord is a function of β, as you can observe as well. To move forward, can you come up with this relationship?

 

[peripatein]

I believe it should be L*cos(beta). I am simply wondering how you would account for the fact that the question specifically asks to demonstrate that t is NOT dependent on beta (assuming you are still convinced that it is). To put it more clearly - I am not suggesting you are indubitably wrong, simply that there seems to be a contradiction, wouldn't you reckon?

 

[lewando]

I am not convinced it is. in my original understanding of the question, it was. Questions can be badly worded every now and then, so I was not alarmed at that possibility. Let's just see where this goes. Another thing to consider... I originally discounted the value of the energy approach, I now think that will come back into play.

 

[lewando]

So after some scribbling, I see that t is not dependent on β. Presently, I cannot communicate it elegantly (though I'm pretty sure that's not my job ). I can tell you the general approach I used was:
1. Equate the PE lost in descending a vertical distance, h (from point A), to the KE at B.
2. Find V_y at point B.
3. Use that result in the kinematic equation h = 1/2(V_y)t, solve for t. Trouble is, you are left with a cosβ term and an h term (both dependent on β). For specific angles, 0, 30, and 45, you get specific h's (at fractions of the constant diameter) and these reduce to an expression for time with no β and no h. Unfortunately, I am unable to state this in a more general way at this time.

 

[lewando]

Since the PE to KE method is not giving up the general formula as directly as I'd like, let's revisit the original approach--trying to find the length of the chord as a function of β, etc.

I believe it should be L*cos(beta)...

The length of the chord, L, is a function of β and the diameter. Once you get this, using a little bit of geometry and trig, everything should fall into place quickly.

 

[rcgldr]

Try to solve for time t as a function of g (acceleration of gravity), r (radius of the circle), and β. If β doesn't matter, than t is some function of g and r. lewando gave you some hints in the other thread, including an image that should help you determine the length L in terms of r and β. You've already determined acceleration in terms of g and β.