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Two masses on strings crossing each other

  • Thread starter Vibhor
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  • #1
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Homework Statement



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Homework Equations




The Attempt at a Solution



Suppose tension in the left string mAM be T1 and tension in the right string mBM be T2 .When the particles cross each other then the angle made by string mA and mB with vertical is α and that made by AM and BM be β . The speed of m is u and that of M is v .

The work done by tension would be zero , so

-(T1cosα)u + (T1cosβ)v - (T2cosα)u + (T2cosβ)v = 0

By symmetry I believe the tensions should be equal so if I put T1 = T2

I get , (cosα)u = (cosβ)v .

But now I do not think α is equal to β when the masses cross each other .

Am I thinking correctly ?

Thanks
 

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  • #2
TSny
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The work done by tension would be zero
Why?
so

-(T1cosα)u + (T1cosβ)v - (T2cosα)u + (T2cosβ)v = 0
Work involves force and distance, not force and speed. The magnitude of the tension in the string will change as the positions of the masses change and the direction of the tension forces on the masses will also change as the masses move. Calculating the work done by tension looks complicated!


See how far you can get with just kinematics and the fact that the total length of the string doesn't change. The pegs are given to be "small", so you can neglect the diameters of the pegs.
 
  • #3
haruspex
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Calculating the work done by tension looks complicated!
... but the method was valid and gave a correct answer. After cancelling out the equal tensions, it effectively reduces to the kinematic method. All that was needed further was to use the fact that the pegs are small, so the angles can be taken as equal.
 
  • #4
TSny
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Vibohr, I believe I was misinterpreting your solution.

When the particles cross each other then the angle made by string mA and mB with vertical is α and that made by AM and BM be β . The speed of m is u and that of M is v .

The work done by tension would be zero
I was interpreting this statement as saying that the work done by the tension forces on the masses was zero between the instant of release and the instant of coincidence of the masses.
, so

-(T1cosα)u + (T1cosβ)v - (T2cosα)u + (T2cosβ)v = 0
Rather, it appears that you are saying that the total rate at which work is being done by tension is zero at the instant the masses pass each other. How would you justify this statement?
 
  • #5
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Rather, it appears that you are saying that the total rate at which work is being done by tension is zero at the instant the masses pass each other.
Not only at the instant when the masses pass each other , but at all times.

How would you justify this statement?
Work done by both the tensions , individually , will be zero .

The string length is constant . If the displacement of 'm' is 'x' and the string makes an angle α with vertical , then component of displacement along the string would be xcosα . Work done on 'm' would be -T1xcosα . Similarly ,work done on 'M' would be T1ycosβ . The sum of total work done will be zero . If we diferentiate this expression , we can see that the rate of work done by tension will be zero at all instants . The tension in left and right string lengths would vary with time , but tension will be equal in the two string pieces and total work done will be zero .
 
  • #6
haruspex
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Not only at the instant when the masses pass each other , but at all times.
Yes, that's how I interpreted your approach. The net work done by the tensions must at all times be zero, therefore the rate of doing work is at all times zero.
 
  • #7
TSny
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Yes, that's how I interpreted your approach. The net work done by the tensions must at all times be zero, therefore the rate of doing work is at all times zero.
OK. But I think I'm being dense here. Help me out. With a little math, I see how kinematics and the fixed length of string implies that (cosα)u = (cosβ)v at each instant of time. Therefore u = v when the particles pass each other. Also, you can then conclude that the net rate of doing work by the string is zero at each instant of time.

But you and Vibhor seem to start from the premise that the rate of doing work is zero at each instant of time, as though this is obvious. Can you give me a little hint as to how you are thinking about it?
 
  • #8
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TSny ,

Didn't you find the justification in post#5 convincing ?
 
  • #9
TSny
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From post #5
The string length is constant . If the displacement of 'm' is 'x' and the string makes an angle α with vertical , then component of displacement along the string would be xcosα . Work done on 'm' would be -T1xcosα .
Is x a finite displacement? T1 and α would change during a finite displacement. If x is an infinitesimal displacement, then OK.

Similarly ,work done on 'M' would be T1ycosβ . The sum of total work done will be zero .
This is where I'm having trouble. What is your justification for stating that the sum of the total work equals zero?
 
  • #10
TSny
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I think I see how you are thinking about it. For infinitesimal displacements dx and dy, -dxcosα and dycosβ are displacements along the string. Since the string is inextensible, these must add to zero. (Therefore, the speeds must be equal at the instant of passing.) Then it follows that the total work for these displacements is zero. I got perplexed by the introduction of the concept of work. I thought I was missing some "dynamical" concept.
 
  • #11
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I think I see how you are thinking about it. For infinitesimal displacements dx and dy, -dxcosα and dycosβ are displacements along the string. Since the string is inextensible, these must add to zero. (Therefore, the speeds must be equal at the instant of passing.) Then it follows that the total work for these displacements is zero.
Exactly :smile:
 
  • #12
TSny
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Thanks, Vibhor :smile:
 

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