A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 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
 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

This Question is Closed

exploringphysics
 2 years ago
Best ResponseYou've already chosen the best response.0dw: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/uv ;

Saikam
 one year ago
Best ResponseYou've already chosen the best response.1Very 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.

Saikam
 one year ago
Best ResponseYou've already chosen the best response.1As 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 = \[(xV)/(L+Vt)\]

Saikam
 one year ago
Best ResponseYou've already chosen the best response.1Therefore, 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.

Saikam
 one year ago
Best ResponseYou've already chosen the best response.1Therefore, 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 xt 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!

Saikam
 one year ago
Best ResponseYou've already chosen the best response.1The 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.

adityalakhe
 one year ago
Best ResponseYou've already chosen the best response.0thanks 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...
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.