Find the speed of a knot on rope

[VHAHAHA]

Homework Statement

http://s12.postimage.org/q2c8bqw97/image.png

http://i.imgur.com/6STfr.png

The part a is easy and i get the ans, the required speed should be v/cos30
but i really have no clue about how to do the part b
i know that the knot will move down and left but i just can't list any equation for the motion

Homework Equations

the length of string is constant

The Attempt at a Solution

i use √X^2 + Y^2 = constant is part a
and get the v/cos 30

but i really have no idea about b part

any clue please >??

 

[PeterO]

Homework Statement

The part a is easy and i get the ans, the required speed should be v/cos30
but i really have no clue about how to do the part b
i know that the knot will move down and left but i just can't list any equation for the motion

Homework Equations

the length of string is constant

The Attempt at a Solution

i use √X^2 + Y^2 = constant is part a
and get the v/cos 30

but i really have no idea about b part

any clue please >??

Not sure if you have noticed, but the image has not come through? Not here anyway.

 

[VHAHAHA]

Not sure if you have noticed, but the image has not come through? Not here anyway.

Sorry for my bad english

you can't see the image?

 

[PeterO]

Sorry for my bad english

you can't see the image?

Until I hit quote, I didn't even realize you had tried to include one. The imbedded image code showed up among your text.

EDIT: the image reference just appeared in the original post.

EDIT 2: But the image is too small to read, and has insufficient resolution to be enlarged ??

 

[VHAHAHA]

Until I hit quote, I didn't even realize you had tried to include one. The imbedded image code showed up among your text.

EDIT: the image reference just appeared in the original post.

i have now giving the website of image

can u see now?

maybe i can upload again

I am now updating the zip file

is it ok?

thx for your help

 

[VHAHAHA]

i am now uploading another image
hope it is all right

 

[PeterO]

i am now uploading another image
hope it is all right

Image works, but it is 2:00 am here and I am about to go to bed.

 

[VHAHAHA]

Image works, but it is 2:00 am here and I am about to go to bed.

thank you and have a good dream =D

 

[PeterO]

thank you and have a good dream =D

OK,

I see that the pulley is supposed to be infinitely small, so is to be represented as a point on the bottom rail matching the other geometry restraints.

Can you calculate how far K is from B [or from the pulley for that matter, since it is in the middle of the rope] at the "30^o" situation.
btw: you should be able to calculate that distance without reference to a pen, paper or calculator. [hint: 2,1,√3 ]

You can get a guide to the answer by establishing where K was when the angle was 29^o and 31^o and calculating the time taken for that change.

 

[PeterO]

i have now giving the website of image

can u see now?

maybe i can upload again

I am now updating the zip file

is it ok?

thx for your help

Just to let you know how I am visualising the situation.

Clearly as the string is drawn in, the knot, K, gets closer to the pulley, O, and at the same time the angle, θ, is going to get larger, until it reached 90^o.

Lets imagine those changes separately.
Considering only the angle change - I see K tracing out a circular arc, going up and left. [I can draw that in my mind or sketch it on paper]
I then imagine a series of radii drawn to that arc.
Considering that K is actually getting closer to O, I then see that K would actually be spiraling down towards O.

I am then faced with the calculus to describe that situation!

 

[VHAHAHA]

yes. i share the same idea with u. i think we can use polar coor. i am on my way back home and i will upload my calculation. thank you

 

[VHAHAHA]

Just to let you know how I am visualising the situation.

Clearly as the string is drawn in, the knot, K, gets closer to the pulley, O, and at the same time the angle, θ, is going to get larger, until it reached 90^o.

Lets imagine those changes separately.
Considering only the angle change - I see K tracing out a circular arc, going up and left. [I can draw that in my mind or sketch it on paper]
I then imagine a series of radii drawn to that arc.
Considering that K is actually getting closer to O, I then see that K would actually be spiraling down towards O.

I am then faced with the calculus to describe that situation!

I have figure out the solution ( maybe)

I use polar coor. to find the speed

the only problem is that i have to find the angular velocity

let the angle be θ and let the radius of the Sector be L

i found that if i consider the arc length and use
Ldθ = dx that would give a wrong ans

if i consider the area
so that (1/2) L^2 dθ = (1/2) dx H where H is the height

i will get the right ans

the attached file is my steps

please help me and thank you for your clue, It proved that my original idea is right , i just lack of confidence =D

 

[PeterO]

I have figure out the solution ( maybe)

I use polar coor. to find the speed

the only problem is that i have to find the angular velocity

let the angle be θ and let the radius of the Sector be L

i found that if i consider the arc length and use
Ldθ = dx that would give a wrong ans

if i consider the area
so that (1/2) L^2 dθ = (1/2) dx H where H is the height

i will get the right ans

the attached file is my steps

please help me and thank you for your clue, It proved that my original idea is right , i just lack of confidence =D

have you noticed that the geometry of the 30^o position gives a very special relationship between the distances OK or BK (they are equal of course since K is the midpoint) and H

 

[VHAHAHA]

have you noticed that the geometry of the 30^o position gives a very special relationship between the distances OK or BK (they are equal of course since K is the midpoint) and H

similar trian?

i am not good at geometry =(

do i get the right ans in my pdf? thank you

also, the horizontal distance between k and b is √3 /2 h

the vertial is h/2 ?

the angle 30 gives sin 30 =1/2 cos 30 = √3 /2 <-- this special relation ?

 

[PeterO]

similar trian?

i am not good at geometry =(

do i get the right ans in my pdf? thank you

also, the horizontal distance between k and b is √3 /2 h

the vertial is h/2 ?

the angle 30 gives sin 30 =1/2 cos 30 = √3 /2 <-- this special relation ?

I believe at this position, the distance you are calling L is equal to H, so there is one less variable than you have.
I can't follow your calculation: looks to me to be more of an advanced maths problem than Introductory Physics one.

 

[VHAHAHA]

I believe at this position, the distance you are calling L is equal to H, so there is one less variable than you have.
I can't follow your calculation: looks to me to be more of an advanced maths problem than Introductory Physics one.

so i need to post to the advanced mechanics ?
i am moving now please del this post

and thank you for your help! =D

 

[VHAHAHA]

1. Homework Statement [/b]

http://s12.postimage.org/q2c8bqw97/image.png

http://i.imgur.com/6STfr.png

this post is moved from the introductory physics as this actually involves advanced mechanics

The part a is easy and i get the ans, the required speed should be v/cos30

but the part b is the thing that i can't understand

i actually use polor coor. to get b ans and it seems right as when i assume the speed v be a certain number, and the find the speed after 0.000001 , the ans is equal that ans calculated from from my equation

Homework Equations

the length of string is constant
the area of the triangle

The Attempt at a Solution

i use √X^2 + Y^2 = constant is part a
and get the v/cos 30

in part b

it's difficult to explain and i just upload my steps

the problem is that if i use s=rθ the equation will be wrong

but if i use the area of sector = 1/2 l^2 θ = 1/2 b x h where b is the base and h is the height and l is the radius

i will get correct ans

can anyone help ?

the attached pdf is my steps
http://pdfcast.org/pdf/dsada-dsad-as-das <-- this is the steps that plugging the valve v and assure my equation is correct

 

[VHAHAHA]

the admin said that it doesn't involve advanced.math
can anyone provide a ans to me? my answer is right or wrong?

 

[haruspex]

You have the speed and direction of the string at B. Think about the speed and direction of a point on the string OB near to O. Point K is halfway between the two. What do you think the relation would be between the three velocities?

 

[VHAHAHA]

the speed of a pt near O is pointing downward? with speed v ?

 

[haruspex]

the speed of a pt near O is pointing downward? with speed v ?

Speed v, yes, but not downwards exactly. Think again.

 

[VHAHAHA]

Speed v, yes, but not downwards exactly. Think again.

along the string ?

 

[haruspex]

along the string ?

Yes. So you know the speed and direction of points on the string near B and near O. K is halfway between B and O, and OKB remains a straight line. Can you work out the speed and direction of movement at K?

 

[VHAHAHA]

Yes. So you know the speed and direction of points on the string near B and near O. K is halfway between B and O, and OKB remains a straight line. Can you work out the speed and direction of movement at K?

Does it involves vectors? or sin cos tan?
or just using a^2 + b^2 = c^2 ?
i am really bad at this

 

[haruspex]

Does it involves vectors?

Yes, it certainly involves vectors. Are you uncomfortable with that?
For ease of discussion, let me give a name to a point on the string which is almost at O. I'll call it S. As discussed, it's moving towards O.
I felt it was obvious that the knot, which is fixed for short time at halfway between S and B, would be moving at their vectorial average. If that that's not clear to you, I'll see if I can explain it another way.
Work out the x and y components of the speeds at B and S. (Please put this in your next post.)
Now imagine you are on the string at S. You are moving with the string, so you are staying a constant distance from both K and B (yes?). This means that they must appear to be going in a circle around you. Since K is half as far away as B, it would look like it is traveling at half the speed.
How fast, in the x and y directions, would B appear to move relative to where you are at S?

 

[VHAHAHA]

let. leftwards be positive and downward be positive

b: v/cos 30 i + 0j
s: vcos30 i + vsin30 j

the speed of b relative to s: v/cos30 - vcos30 i - vsin30j ?


I felt it was obvious that the knot, which is fixed for short time at halfway between S and B, would be moving at their vectorial averag. <== y move at vectorial ave? thanks for u help

 

[haruspex]

let. leftwards be positive and downward be positive

b: v/cos 30 i + 0j
s: vcos30 i + vsin30 j

the speed of b relative to s: v/cos30 - vcos30 i - vsin30j ?

Right.
[As a check, we can work out the apparent velocity along the line of the string:
(v/cos30 - vcos30)*cos30 + - vsin30*sin30 = 0
as required by the fact that SB is a constant length - until S reaches the pulley.]

K is, and remains for a short time, halfway between S and B. Therefore its velocity relative to S is exactly half of B's:
v/(2cos30)i - vcos30 i/2 - vsin30 j/2
To get the velocity of K relative to the static frame, add back in the velocity of S.

 

[VHAHAHA]

y the velocity is half? y the distance is half tells me that velocity is half? i don't know if i am correct: i think that if b and k is moving in a circular path within a short time t relative to s , as v=rw they line on same string so they hv same w, as the r is halfed , v is halfed?

how can u know its related to relative velocity? i am not bad at vectors but i really can't think that its is related until u give me clue:( i really like physics and i want to know how to think logically in physics, so knowing the way of getting ans is important, thx

 

[haruspex]

If I know the velocity vector for each end of a rigid rod, it seems obvious to me that the velocity vector of its midpoint will be the average. But seeming obvious isn't good enough, so I tried to make it clearer by working in terms of relative velocities. Since in those terms B is going in a circle around S at radius SB, and K must be going around S at the same angular rate at half the radius, its relative speed must be half. Looks like you understand that.
Shifting reference frame is a common trick, though you sometimes have to be careful not to choose one that's accelerating.

 

[VHAHAHA]

thank you for ur help. i will study hard in physics:)