Author |
Topic: strings and velocity (Read 704 times) |
|
Ronno
Junior Member
 

Gender: 
Posts: 140
|
 |
strings and velocity
« on: Feb 25th, 2009, 7:10pm » |
Quote Modify
|
. . | \ | / | | \ | / | | \ | / | | \_| / | | \ |/ | | [_] | 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.
|
|
IP Logged |
Sarchasm: The gulf between the author of sarcastic wit and the person who doesn't get it..
|
|
|
towr
wu::riddles Moderator Uberpuzzler
    
 Some people are average, some are just mean.
Gender: 
Posts: 13730
|
 |
Re: strings and velocity
« Reply #1 on: Feb 26th, 2009, 12:40am » |
Quote Modify
|
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)
|
« Last Edit: Feb 26th, 2009, 12:41am by towr » |
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Ronno
Junior Member
 

Gender: 
Posts: 140
|
 |
Re: strings and velocity
« Reply #2 on: Feb 26th, 2009, 12:57am » |
Quote Modify
|
The velocity is to expressed as a function of the angle theta. Sorry if I wasn't clear.
|
|
IP Logged |
Sarchasm: The gulf between the author of sarcastic wit and the person who doesn't get it..
|
|
|
towr
wu::riddles Moderator Uberpuzzler
    
 Some people are average, some are just mean.
Gender: 
Posts: 13730
|
 |
Re: strings and velocity
« Reply #3 on: Feb 26th, 2009, 1:39am » |
Quote Modify
|
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)
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Eigenray
wu::riddles Moderator Uberpuzzler
    

Gender: 
Posts: 1948
|
 |
Re: strings and velocity
« Reply #4 on: Feb 26th, 2009, 3:10am » |
Quote Modify
|
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.
|
|
IP Logged |
|
|
|
Ronno
Junior Member
 

Gender: 
Posts: 140
|
 |
Re: strings and velocity
« Reply #5 on: Feb 26th, 2009, 4:19am » |
Quote Modify
|
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.
|
|
IP Logged |
Sarchasm: The gulf between the author of sarcastic wit and the person who doesn't get it..
|
|
|
rmsgrey
Uberpuzzler
    


Gender: 
Posts: 2874
|
 |
Re: strings and velocity
« Reply #6 on: Feb 26th, 2009, 9:40am » |
Quote Modify
|
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. |
| 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)
|
|
IP Logged |
|
|
|
Ronno
Junior Member
 

Gender: 
Posts: 140
|
 |
Re: strings and velocity
« Reply #7 on: Mar 1st, 2009, 11:43pm » |
Quote Modify
|
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.
|
|
IP Logged |
Sarchasm: The gulf between the author of sarcastic wit and the person who doesn't get it..
|
|
|
towr
wu::riddles Moderator Uberpuzzler
    
 Some people are average, some are just mean.
Gender: 
Posts: 13730
|
 |
Re: strings and velocity
« Reply #8 on: Mar 2nd, 2009, 1:24am » |
Quote Modify
|
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) |
| 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.
|
« Last Edit: Mar 2nd, 2009, 1:25am by towr » |
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Ronno
Junior Member
 

Gender: 
Posts: 140
|
 |
Re: strings and velocity
« Reply #9 on: Mar 2nd, 2009, 8:04am » |
Quote Modify
|
Why should V approach v? And velocities are added just like forces, since they are both vectors.
|
|
IP Logged |
Sarchasm: The gulf between the author of sarcastic wit and the person who doesn't get it..
|
|
|
Grimbal
wu::riddles Moderator Uberpuzzler
    

Gender: 
Posts: 7527
|
 |
Re: strings and velocity
« Reply #10 on: Mar 2nd, 2009, 8:26am » |
Quote Modify
|
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.
|
|
IP Logged |
|
|
|
Ronno
Junior Member
 

Gender: 
Posts: 140
|
 |
Re: strings and velocity
« Reply #11 on: Mar 2nd, 2009, 8:51am » |
Quote Modify
|
Oh yeah, I see the problem now. The inextensibility of the other string prevents the velocity to be transmitted to the block.
|
|
IP Logged |
Sarchasm: The gulf between the author of sarcastic wit and the person who doesn't get it..
|
|
|
towr
wu::riddles Moderator Uberpuzzler
    
 Some people are average, some are just mean.
Gender: 
Posts: 13730
|
 |
Re: strings and velocity
« Reply #12 on: Mar 2nd, 2009, 9:02am » |
Quote Modify
|
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. |
| 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.
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Immanuel_Bonfils
Junior Member
 

Posts: 114
|
 |
Re: strings and velocity
« Reply #13 on: Mar 2nd, 2009, 1:16pm » |
Quote Modify
|
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
|
|
IP Logged |
|
|
|
Ronno
Junior Member
 

Gender: 
Posts: 140
|
 |
Re: strings and velocity
« Reply #14 on: Mar 2nd, 2009, 6:51pm » |
Quote Modify
|
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.
|
|
IP Logged |
Sarchasm: The gulf between the author of sarcastic wit and the person who doesn't get it..
|
|
|
SMQ
wu::riddles Moderator Uberpuzzler
    

Gender: 
Posts: 2084
|
 |
Re: strings and velocity
« Reply #15 on: Mar 3rd, 2009, 6:43am » |
Quote Modify
|
on Mar 2nd, 2009, 6:51pm, ronnodas wrote:For example, in projectile motion, you add the horizontal and vertical velocities vectorially. |
| 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". --SMQ
|
|
IP Logged |
--SMQ
|
|
|
|