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adityalakhe

  • 2 years ago

one end of a rubber band is attatched to a wall and a bead is inserted from other end a force is applied on the free end of rubber band which gives it a velocity v and a velocity u is given to bead in opposite direction. find the time taken by the bead to reach the wall

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  1. exploringphysics
    • 2 years ago
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    |dw:1355851540442:dw| dl/dt=v; dx/dt=u; if the speed of increasing the lengh of thread is greater than that of bead ,it will never reach othe end. otherwise udt=l+vdt. t=l/u-v ;

  2. Saikam
    • one year ago
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    Very good question . and the answer given above is not correct. Here is what I understand from the data provided: The end of the rubber band i constantly pulled with a speed v, and the bead has a constant speed u on the rubberband. OK. Let the initial length of the rubber band be L You first need to analyze how any point on the rubber band moves when the end is pulled with velocity V.

  3. Saikam
    • one year ago
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    As it is being pulled, The free end moves with speed V and the fixed end moves with speed 0. Therefore, any point on the rubberband that is at a distance x from the fixed end moves with a speed = \[(xV)/(l)\] where l is the length of the rubberband at that instant., which can be also written as: l = L+Vt , which makes the speed = \[(xV)/(L+Vt)\]

  4. Saikam
    • one year ago
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    Therefore, the speed of the bead w.r.t ground = (bead speed w.r.t rubberband) - (speed of the point on which the bead is sliding) Which can be written as: U - (xV)/(L+Vt) This is the net speed at which the bead is moving towards the wall, when it is at a distance x from it. Therefore, the rate of change of x, -(dx/dt) = U - (xV)/(L+Vt) and we have used the minus sign before dx/dt since x is decreasing.

  5. Saikam
    • one year ago
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    Therefore, we have, dx/dt = (xV)/(L+Vt) - U That is, x'(L+Vt) = xV - UL - UVt This can be solved using the differential equation solver by Wolfram available at www.wolframalpha.com The solution of this equation would be: x = C(L+Vt) -(U/V)*(L+Vt)*ln(L+Vt) The constant C can be calculated by substituting x = L at t = 0; We get C = 1 + (U/V)*ln(L) Therefore, x = [ 1 + (U/V)*ln(L)] * [L+Vt] - (U/V)*(L+Vt)*ln(L+Vt) This is the required x-t relation. Now when the bead reaches the wall, x = 0. We need the value of t that satisfoes x =0. On substituting x =0 and a little bit of rearranging, you can get the value of t as: \[(L/V)*(e ^{V/U} - 1) \] which is also dimensionally correct. Good problem. Took quite some effort to analyze!

  6. Saikam
    • one year ago
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    The mistake in the solution given by @exploringphysics is that the rubberband was assumed to be something like a rod moving with a velocity V, which is wrong. The rubber band stretches and the movement of points on the rubber band must be analyzed in a different way. All the points on the rubber band would not move at the same speed.

  7. adityalakhe
    • one year ago
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    thanks a lot @saikam.... great analysis.... even i derived that differential equation... and wanted to confirm whether i ws ryt or wrong n i got my answer...... thanks again...

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