Pulleys moving down with constant speed


by amal
Tags: constant, moving, pulleys, speed
amal
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#1
Nov19-11, 07:49 PM
P: 34
1. The problem statement, all variables and given/known data
Please refer diagram attached.
The pulleys are moving down with constant speed u each. Pulleys are light , string ideal. What is the speed with the mass moves up?






2. Relevant equations



3. The attempt at a solution
Now, I got the answer as ucosΘ, obviously. But the answer is u/cosΘ.
the solution too is given. But I feel that the latter is wrong as the velocity goes to to infinity as the mass goes upwards. Please tell me which is correct.
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File Type: bmp pulleys.bmp (263.7 KB, 17 views)
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gneill
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#2
Nov20-11, 12:40 AM
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Why don't you explain your reasoning for the 'obvious' result, u*cos(θ)?
ehild
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#3
Nov20-11, 12:48 AM
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If the half-length of the rope between the pulleys is L, the mass is at depth y=L*cos(Θ), and dL/dt = -u. The rate of change of y is dy/dt=d(L*cos(Θ))/dt. Both L and Θ are changing as the rope is pulled while x = LsinΘ stays constant. If you take these into account you will get the correct result, that the mass raises wit u/cosΘ.

ehild

amal
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#4
Nov20-11, 01:13 AM
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Pulleys moving down with constant speed


Quote Quote by gneill View Post
Why don't you explain your reasoning for the 'obvious' result, u*cos(θ)?
Well, the string section 'r' should move with speed u only as it is a light pulley. Then the vertical component is ucosθ.
amal
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#5
Nov20-11, 01:17 AM
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Quote Quote by ehild View Post
If the half-length of the rope between the pulleys is L, the mass is at depth y=L*cos(Θ), and dL/dt = -u. The rate of change of y is dy/dt=d(L*cos(Θ))/dt. Both L and Θ are changing as the rope is pulled while x = LsinΘ stays constant. If you take these into account you will get the correct result, that the mass raises wit u/cosΘ.
But then is my interpretation of decreasing vertical velocity wrong?
ehild
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#6
Nov20-11, 01:25 AM
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Quote Quote by amal View Post
But then is my interpretation of decreasing vertical velocity wrong?
It is wrong. And you did not prove it.

ehild
gneill
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#7
Nov20-11, 07:26 AM
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Quote Quote by amal View Post
Well, the string section 'r' should move with speed u only as it is a light pulley. Then the vertical component is ucosθ.
The problem with this analysis (as tempting as it is) is that the rope end at D will NOT move along the line DA (or DB), as it is constrained by its mirror-image partner. While your length r may be getting shorter, point D has no velocity along r. So trying to treat dr/dt = u as a velocity component for point D is not right.
gneill
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#8
Nov20-11, 07:30 AM
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Quote Quote by amal View Post
But then is my interpretation of decreasing vertical velocity wrong?
The vertical velocity will decrease, but not for the reason you think! As the mass rises and the angle θ approaches 90, the tension in the rope required to maintain the rope speed u will head to infinity. IF the rope speed could be maintained at u (or any positive real value greater than zero) then the velocity of the mass would go infinite, but rope speed cannot be so maintained by any real apparatus.
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#9
Nov20-11, 07:36 AM
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Quote Quote by amal View Post
Well, the string section 'r' should move with speed u only as it is a light pulley. Then the vertical component is ucosθ.
The rate at which the length of segment DA changes will equal u. You need to relate the rate of change of segment DC to the rate of change of DA. How are those lengths related?
amal
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#10
Nov20-11, 09:52 PM
P: 34
OK, I see now. Using Pythagoras theorem we can relate the legths as
[itex]DA^{2}[/itex]=[itex]DC^{2}[/itex]+[itex]AC^{2}[/itex]
then differentiating,
[itex]2DA\frac{dDA}{dt}[/itex]=[itex]0+2AC\frac{dDC}{dt}[/itex]
And finally the answer will come u/cosθ. Right?
ehild
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#11
Nov20-11, 11:08 PM
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Quote Quote by amal View Post
OK, I see now. Using Pythagoras theorem we can relate the legths as
[itex]DA^{2}[/itex]=[itex]DC^{2}[/itex]+[itex]AC^{2}[/itex]
then differentiating,
[itex]2DA\frac{dDA}{dt}[/itex]=[itex]0+2AC\frac{dDC}{dt}[/itex]
And finally the answer will come u/cosθ. Right?
If you replace AC with DC in the second equation it will be right.

ehild
amal
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#12
Nov21-11, 04:35 AM
P: 34
Yes, thank you.


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