Two masses attached by a string.

  • #1
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Two masses m_A and m_B are connected by a string of length l and lie on a frcitionless table. the system is twirled and released with m_A instantatneously at rest, and m_B moving with instantatneously velocity v0at right angles to the line of centres.
find the subsequent motion of the system and the tension in the string.
in order to find the motion it's easy to use energies:
m_Bv0^2/2=m_A*v_A^2/2+m_B*v_B^2/2+I_Aw_A^2/2+I_B*w_B^2/2
where by conservation of angular momentum: we have around the centre of mass of two masses: m_B*v0R_B=m_A*R_A*v_A+m_B*R_B*v_B
where v_B=w_B*R_B v_A=w_A*R_A, where the R's are the displacements of the masses from the centre of mass.
now to compute the tension i think that T=(m_Av_A^2/R_A)-(m_Bv_B^2/R_B)
the problem is that it's obviously not correct, can someone help me in this, thanks.
 

Answers and Replies

  • #2
Doc Al
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I'm having a bit of trouble following what you are doing. I don't see the utility of using energy--and what do you mean by I_A and I_B? Nor do see the need for conservation of angular momentum--nothing's changing, so of course it's conserved.

Instead, try this: What's the motion of the center of mass? (That should be easy.) What's the angular speed of the system about the center of mass?
 
  • #3
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well it depends on the masses, for example if both of them have the same mass then the motion is only circular without translation (i.e the centre of mass isnt moving), and then the angular speed would be 2v0/l, but when we have different masses, we would have both m_A and m_B circling the centre of mass with diifferent angular speeds, but the same velocity v0 (or so i think).
if this is correct, i still need to figure out the tension force.
 
  • #4
Doc Al
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Use the given data to find the speed of the center of mass. Hint: What's the linear momentum of the system?
 
  • #5
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v_cm=m_Bv0/(m_A+m_B)
now the speed of A is: v0-v_cm.
is this correct?
still the tension must be equal to the centrifugal force here, but i still dont get the quantative answer right.
 
  • #6
Doc Al
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v_cm=m_Bv0/(m_A+m_B)
now the speed of A is: v0-v_cm.
is this correct?
Yes. You've found the speed of A with respect to the center of mass, which is good. How far is A from the center of mass?

still the tension must be equal to the centrifugal force here, but i still dont get the quantative answer right.
Show what you've done. Consider things from the center of mass frame.
 
  • #7
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well we have the centrifigual force by m_A which is m_Av_A^2/R_A
and by m_B is m_Bv_0^2/R_B
where R_A=m_B*l/(m_A+m_B) R_B=m_A*l/(m_A+m_B) iv'e assumed that m_A>=m_B, now i think the tension should be the sum of these, but i tried it even before i posted here, but when i insert the figures in the answer clue i should have:
if m_A=M_B=2 v0=3 l=0.5 then T=18, but i dont get this answer.
when i think of this the answer clue points to the cetrifugal force on m_A, shouldn't we include the cent force on m_B?
 
  • #8
Doc Al
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There is a single tension in the string; a single centripetal force. The system rotates about its center of mass. Find the centripetal force by considering the rotation of A (or B) about the center of mass.
 
  • #9
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i just have one probelm, when calculating it with regard to B, the velocity now v_B is unchanged, i.e v0, correct?
this doesnt seem to fit the answer clue.
 
  • #10
Doc Al
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i just have one probelm, when calculating it with regard to B, the velocity now v_B is unchanged, i.e v0, correct?
this doesnt seem to fit the answer clue.
This is not correct. v_B = v0 with respect to the table; you need v_B with respect to the center of mass.

v_cm=m_Bv0/(m_A+m_B)
now the speed of A is: v0-v_cm.
is this correct?
Earlier I said this was correct, but I was wrong. (I mixed up A and B.) Find the speed of A with respect to the center of mass: A is the one initially at rest with respect to the table. (The center of mass speed is correct.)
 
  • #11
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so it should be:
v_B=v0-v_cm
v_A=0-v_cm
is this correct, or wrong again?
 
  • #12
Doc Al
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That looks correct. Now you should be able to figure out the angular speed of the system about its center of mass and calculate the tension in the string.
 

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