Circular motion/momentum conservation

[ritwik06]

Homework Statement

A U shaped smooth wire has a semi circular bending between A and B (A and B are points on the U wire). A bead of mass 'm' moving with uniform speed v through the wire enters the semi circular bend at A and leaves at B. Find the average force exerted by the bead on the part AB of the wire. d is the perpendicular distance between the two arms of the U shape.

The Attempt at a Solution

Method 1:
As the part betwen A and B is circular
Force=mv^2/ (d/2)

Method 2:
The change in momentum is 2mv
time taken to go around AB=pi*(d/2)*(1/v)

Force =change in momentum/timeI get two different answers from these 2 methods. At least one of them is surely wrong. but which one? Both seem right to me.

 

[Carid]

Show your working then maybe someone can tell you where you are making an error.

 

[ritwik06]

Show your working then maybe someone can tell you where you are making an error.

Sir, If you take a careful look at Point 3: An attempt to the solution,
you will find all the work that I have done.

 

[Carid]

ritwik06

I get two different answers from these 2 methods.

You did not show your working.

I'm here to help not be subjected to impoliteness.

 

[ritwik06]

ritwik06



You did not show your working.

I'm here to help not be subjected to impoliteness.

Well sir, I apologise to you if you think I was impolite.

Actually, I have posted a problem.

I had 2 approaches for solving the problem. Both of them seem correct in my opinion- but they give two different answers. There lies the confusion. I cannot make out what went wrong with on of these. This is exactly where I need help.

For the sake of your words, I am going to repost this, adding some more details<which I thought might not be needed>:

Method 1:

d is the perpendicular distance between the two arms of the U shape wire
As the part between A and B is circular,
For circular motion the centripetal force is given by=(Mass*Velocity^2)/Radius
(Answer)Force=mv^2/ (d/2)

Method 2:
The change in momentum is 2mv
time taken to go around AB=pi*(d/2)*(1/v)

Force =change in momentum/time


Answer=(4*m*v^2)/(pi*d)

Please let me know if you need something else as well. Thanks in advance.

regards,
Ritwik

 

[Carid]

Well I'm going round in circles too

HELP

 

[ritwik06]

Well, I request the members of PF to help me out.My problem persists. I don't know why Carid troubled me...
Please rescue me.

 

[Carid]

ritwik

I don't know why Carid troubled me...

Manners maketh man. This may well be my last post to help you.

I've been turning your problem over in my mind for the last few hours and now have the solution.

Your first solution is the correct one.

The second fails because the force doing the reversal of momentum is only the vectorial part of the force which points in the direction of the arrival and departure of the bead.

 

[ritwik06]

Manners maketh man.

Well, sir. I never tried to be rude to you unless, you yourself went into circles.
See this situation yourself. I asked a question, I showed everything as I understood. And you without even caring to read it all, say "Where is your work?" And then when I mentioned that I had already done my work, you realising it, should have helped me if you at all desired. But to the contrary-you thought I had done something so callous that it resulted in one of the greatest dishonours faced by men.

Till then I didn't lose my patience. I reposted the whole thing (mind it-wasting my time as well as the valuable space of this server). Then you as the gentleman you are, told that you don't know the problem urself. How do you think I should have felt?

And now you continue to misguide me. Neither helping me, nor letting someone else help me. For your info, the first method is wrong. The question asks for average force. The force vector is rotating, but with constant magnitude. I need the average, so the second method is correct. I have learned that recently- and hopefully I am 99% correct.

I don't want to sound rude again- but please give statements you are confident about. If you at all have a doubt, mention you are not sure. Your reply is always welcome.

This may well be my last post to help you.

I don't know, whether you still think that its me who needs manners?
By the way, there are a lot of people around who can guide me correctly.

I've been turning your problem over in my mind for the last few hours and now have the solution.

Your first solution is the correct one.

The second fails because the force doing the reversal of momentum is only the vectorial part of the force which points in the direction of the arrival and departure of the bead.

Think it over, I recently learned that the second one was correct.
regards,
Ritwik

 

[Doc Al]

The 2nd solution is correct, as you already seem to realize. The problem with the first solution is that it just gives the magnitude of the force; but the direction of the force varies as the bead traverses the semi-circle.

 

[ritwik06]

The 2nd solution is correct, as you already seem to realize. The problem with the first solution is that it just gives the magnitude of the force; but the direction of the force varies as the bead traverses the semi-circle.

Thanks a lot. It went exactly as I thought it to be.
Thanks again for the effort. :D

 

[Carid]

My apologies for giving you an incorrect answer. I tripped over the word "average".

 

[ritwik06]

My apologies for giving you an incorrect answer. I tripped over the word "average".

There is no need. Dont you think, Einstein rightly said that we still know very little about the laws of nature.

I had no intention to ridicule you. Thanks for the effort.