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Ronno
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strings and velocity
« on: Feb 25th, 2009, 7:10pm » |
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v v
The two strings in the picture are going downward with constant velocity v. They go over two pulleys and are attached to a block, while making an angle theta (at some point of time, t) with the vertical. What will be the velocity of the block (also at time t)?
The pulleys and strings are considered ideal (massless, inextensible, frictionless).
I actually know two solutions to this problems, which are inconsistent (except at theta=Pi/4), but both appear to be right.
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towr
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Re: strings and velocity
« Reply #1 on: Feb 26th, 2009, 12:40am » |
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If the distance between the two pulleys is 2x, and the length of the rope from the pulley to the block at time t=0 is l
then your angle is
asin(x/(l-vt))
and position of the block (from top to block) is
sqrt((l-vt)2 - x2),
and the speed is the the derivative of this,
-v(l-tv)/sqrt((l-vt)2 - x2)
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| « Last Edit: Feb 26th, 2009, 12:41am by towr » |
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Ronno
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Re: strings and velocity
« Reply #2 on: Feb 26th, 2009, 12:57am » |
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The velocity is to expressed as a function of the angle theta. Sorry if I wasn't clear.
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towr
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Re: strings and velocity
« Reply #3 on: Feb 26th, 2009, 1:39am » |
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Oookay. So then we need to find time as a function of 
= asin(x/(l-vt))
=>
t = (l - x csc())/v
Fill it in, and simplify:
-v csc()/ (csc2() - 1)
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Eigenray
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Re: strings and velocity
« Reply #4 on: Feb 26th, 2009, 3:10am » |
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Which simplifies to -v/cos.
Maybe an easier way to see this: let the block be at a distance L from the pulley, with a vertical distance of y. Then L2 - y2 is constant, so LL' = yy', i.e., y' = L' * L/y = -v/cos.
I'm guessing the other answer is: y = L cos, so y' = L' cos. But this is wrong because is not constant.
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Ronno
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Re: strings and velocity
« Reply #5 on: Feb 26th, 2009, 4:19am » |
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Yes, v/cos(theta) is one answer.
This is the alternate, physical approach:
The block has two simultaneous velocities v, each making angle theta with the vertical. So its velocity will be their resultant, which is 2v*cos(theta).
I like the sec(theta) answer better but can't think of a way to dismiss this solution.
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rmsgrey
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Re: strings and velocity
« Reply #6 on: Feb 26th, 2009, 9:40am » |
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on Feb 26th, 2009, 4:19am, ronnodas wrote:This is the alternate, physical approach:
The block has two simultaneous velocities v, each making angle theta with the vertical. So its velocity will be their resultant, which is 2v*cos(theta).
I like the sec(theta) answer better but can't think of a way to dismiss this solution. |
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Uh, isn't it that the two v's are the (non-orthogonal) components of the block's velocity, so:
v=V*cos(theta) on both, so V=v*sec(theta)
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Ronno
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Re: strings and velocity
« Reply #7 on: Mar 1st, 2009, 11:43pm » |
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Since they are non-orthogonal, v=V*sin(theta)/sin(2*theta), which gives V=2*v*cos(theta)
You are actually applying the formula for orthogonal components.
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towr
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Re: strings and velocity
« Reply #8 on: Mar 2nd, 2009, 1:24am » |
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on Mar 1st, 2009, 11:43pm, ronnodas wrote:| Since they are non-orthogonal, v=V*sin(theta)/sin(2*theta), which gives V=2*v*cos(theta) |
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Let theta approach 0, V should then approach v. However with your formula it goes to twice that, so it can't be right.
I get the impression you're adding velocities as if they're forces.
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| « Last Edit: Mar 2nd, 2009, 1:25am by towr » |
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Ronno
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Re: strings and velocity
« Reply #9 on: Mar 2nd, 2009, 8:04am » |
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Why should V approach v?
And velocities are added just like forces, since they are both vectors.
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Grimbal
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Re: strings and velocity
« Reply #10 on: Mar 2nd, 2009, 8:26am » |
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You can not add velocities because when you pull one string a tiny bit, the bloc doesn't move in the direction where you pull. If the bloc is very low, if you pull one string it will swing to that side.
Also, when you are near the horizontal, the forces oppose and cancel each other as you reach the horizontal. But movement-wise, pulling a little bit on the string will result in a large movement.
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Ronno
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Re: strings and velocity
« Reply #11 on: Mar 2nd, 2009, 8:51am » |
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Oh yeah, I see the problem now. The inextensibility of the other string prevents the velocity to be transmitted to the block.
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towr
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Re: strings and velocity
« Reply #12 on: Mar 2nd, 2009, 9:02am » |
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on Mar 2nd, 2009, 8:51am, ronnodas wrote:| Oh yeah, I see the problem now. The inextensibility of the other string prevents the velocity to be transmitted to the block. |
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I'm not sure what the inextensibility has to do with, it's just that at theta=0 both ropes pull in the same direction, but the block won't go faster than the fastest rope; if it goes faster there is nothing pulling it.
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Immanuel_Bonfils
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Re: strings and velocity
« Reply #13 on: Mar 2nd, 2009, 1:16pm » |
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One can not add velocities just because they are vectors! There must be an physical reasoning: forces are added due superposition of effects. The corresponding situation for velocities would be adding relative velocity of P wrt to O1 and velocity of O1 wrt O2 => velocity of P wrt to O2.
Oi i=(1,2) (math he he he....) = Obsevers or References
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Ronno
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Re: strings and velocity
« Reply #14 on: Mar 2nd, 2009, 6:51pm » |
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Ok. So here the effects of the velocity do not completely add up.
Also, velocities are added due to superposition of effects too. For example, in projectile motion, you add the horizontal and vertical velocities vectorially.
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SMQ
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Re: strings and velocity
« Reply #15 on: Mar 3rd, 2009, 6:43am » |
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on Mar 2nd, 2009, 6:51pm, ronnodas wrote:| For example, in projectile motion, you add the horizontal and vertical velocities vectorially. |
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Ah, but that only works if the velocity vectors are all mutually orthogonal (90o apart, have a scalar product of 0). Otherwise you get some "overlap".
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