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*LouLou*
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Classical Mechanics Homework question
Question - A light elastic string AB of natural length L and spring constant K, lies slack on a horisontal plane. A particle of mass m also at rest, is attached to end A of the string. The other end B is pulled along the plane with constant velocity V away from A in a straight line passing through A.
Find the time that elapses after the string first becomes taut until it becomes slack again.
Show that the particle will have traveled a total distance [itex]2L + V\pi({\frac{m}{k}})^{\frac{1}{2}}[/itex] by the time it catches up with end B.
Thoughts - I think I have to start with Newton's 2nd law, I'm just not sure what the force is a function of. I think it could be F(x) where x is the distance traveled but it can also depend on the speed that the string is being pulled so it would be F(V)?
If the force depends on the velocity of the end of te string then N's 2nd law would be
F(V) = ma = -Kx = m(dv/dt)
where v is the velocity of the particle at time t and V is the velocity of the end of the string. If I equate the two terms on the right I get...
-Kx = m(dv/dt) and if I rearrange this and integrate it I get...
t*(K/m)^0.5 = ln(x) + c
But I don't know where to go from here. I need to somehow eliminate x or t so that I only have 1 unknown. Or maybe I have to split the problem up into sections from the start and deal with separate parts?
Question - A light elastic string AB of natural length L and spring constant K, lies slack on a horisontal plane. A particle of mass m also at rest, is attached to end A of the string. The other end B is pulled along the plane with constant velocity V away from A in a straight line passing through A.
Find the time that elapses after the string first becomes taut until it becomes slack again.
Show that the particle will have traveled a total distance [itex]2L + V\pi({\frac{m}{k}})^{\frac{1}{2}}[/itex] by the time it catches up with end B.
Thoughts - I think I have to start with Newton's 2nd law, I'm just not sure what the force is a function of. I think it could be F(x) where x is the distance traveled but it can also depend on the speed that the string is being pulled so it would be F(V)?
If the force depends on the velocity of the end of te string then N's 2nd law would be
F(V) = ma = -Kx = m(dv/dt)
where v is the velocity of the particle at time t and V is the velocity of the end of the string. If I equate the two terms on the right I get...
-Kx = m(dv/dt) and if I rearrange this and integrate it I get...
t*(K/m)^0.5 = ln(x) + c
But I don't know where to go from here. I need to somehow eliminate x or t so that I only have 1 unknown. Or maybe I have to split the problem up into sections from the start and deal with separate parts?
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