Beads on Semi-Circular wire

Pranav-Arora

1. The problem statement, all variables and given/known data
Two beads A and B move along a semicircular wire frame as shown in the figure as shown in figure. The beads are connected by an inelastic string which always remains tight. At an instant the speed of A is u, angle BAC is 45 degrees and angle BOC is 75 degrees, where O is the centre of semicircular arc. The speed of bead B at that instant is.



2. Relevant equations



3. The attempt at a solution
I really have no idea on how to begin with this. How should i go on forming equations?
Please point me in the right direction.

Thanks!

Infinitum

Hi again

Since the string is not extensible, the vertical velocity of B will always have to be equal to the velocity of A. Can you frame an equation for this?

Pranav-Arora

I did that once but i thought i am trying my own foolish things.
I assumed that at any instant the distance of A from O is x and that of B is y.
l=(x+y)cos45
Differentiating with respect to time, i get:
(dy/dt)=-(dx/dt)

How can i proceed further?

Infinitum

Umm noo..how did you get that?

You can simply equate the vertical velocity component of B to u. The velocity of B will be have its direction tangential to the ring...what does this tell you about its vertical component?

TSny

Hello, Infinitum. I don't see how your statement can be true. If the vertical component of velocity of B equals the velocity of A, then during a time Δt, both particles would move the same distance vertically. But B will also move horizontally during this time. So, it seems to me that this would change the distance between A and B.

TSny

You might give us some context for the problem since there are different methods for solving it. If it's a "related rates problem" from a calculus course then you would be expected to use calculus. Or if it's a problem where you can use any method you want, then you can solve it with just geometry, trig, and some insight.

Pranav-Arora

Sorry, i think i did not explain it correctly. My assumption can be better shown by this pic:

So what i did was incorrect? and why?

Well, it isn't specified that i need to do the problem by a specified method, any method will do.

Would you be so kind to tell me how i can solve by this?

ehild

Hi Pranav,

The angles change with time so just call them α and β or anything else. Find relation between them and x. You know that u=dx/dt. The speed of the other bead on the circle is v=R|dβ/dt|. You need to differentiate, using that the length of the cord is constant in time, and so is the radius of the circle.

ehild

Attached Thumbnails

 

Pranav-Arora

Hello ehild!

As you said i tried finding some relations,
Ist relation, Lsin(α)=Rsin(β)=d
IInd Relation, Lcos(α)-Rcos(β)=x
Differentiating the second relation with respect to time,

Rsin(β)(dβ/dt)-Lsin(α)(dα/dt)=dx/dt
I can use the first relation here but i don't seem to reach the answer.

azizlwl

Instead of using law of sines, why not try using law of cosines.

Infinitum

Hi TSny. You are right, I meant to write that the velocity component of A along the length of the rod has to be equal to the velocity component of B along the length of the rod, so that the rod length remains the same.

Infinitum

Oh, I see. You had mentioned that the distance of B from O is y earlier, that led me into confusion

I also made a little mistake while saying their vertical components are equal. Instead, their velocities along the rod have to be the same, for no extension.

Pranav-Arora

Umm..so what i did is incorrect?

How should i proceed now?

Infinitum

Yes, you got "l=(x+y)cos45"

But it should be [itex]lcos45 = x+y[/itex]

And you cannot simply differentiate this because the angle is changing continuously, as ehild indicated.

An easier approach is equating the velocities of A and B along the rod. Let the velocity of B be v, and then try to solve the problem.

Pranav-Arora

Oh, that's a nice hint, thanks for the help, i got the answer.

Infinitum

Way to go!

TSny

That's a very nice way to solve it.

I had solved it by setting up a velocity triangle based on

V_B = V_A + V_B/A . (vector addition!)

where V_B/A is the velcity of B relative to A. V_B/A must be perpendicular to line AB in order for AB to keep a fixed distance apart. This is similar to your statement that the velocity components of A and B along line AB must be equal. From the velocity triangle you can then use the law of sines to get the result.

But I like your approach much better. It gets the answer right away.

Thanks.