Mechanics problem -- 2 masses and a string in motion

[Kumar Singh]

[Mod Note: moved to homework forum, template missing but homework-type problem]

There are two masses (m) tied at two ends of a string of length 'l'. Whole system is placed on a friction-less surface and force 'F' is applied at the centre of string in direction perpendicular to the string. Find the time taken by masses to come together and nature of path followed.

 

[davenn]

HI there
welcome to PF

There are two masses (m) tied at two ends of a string of length 'l'. Whole system is placed on a friction-less surface and force 'F' is applied at the centre of string in direction perpendicular to the string. Find the time taken by masses to come together and nature of path followed.

is this homework ?
what have you done so far yourself to work towards an answer ?

also you shouldn't have used the advanced tag

Dave

 

[Kumar Singh]

Basically I worked over it by taking some angle at some moment and resolving the force then writing down velocity and acceleration equation for x and y-axis but I am getting entangled in calculus part.

 

[Kumar Singh]

HI there
welcome to PF

is this homework ?
what have you done so fart yourself to work towards an answer ?

also you shouldn't have used the advanced tag

Dave

Ok.

I am new here so bit unfamiliar with the prevailing norms.

 

[davenn]

Ok.

I am new here so bit unfamiliar with the prevailing norms.

that's OK
so is it homework ? if so, start a new thread in the homework section and use the template method supplied there to lay
out your question and what work you have done so far

as a note, you will notice at the top of most of the forum sections including the one you posted in ... a "Please don't post homework in here" thread Dave

 

[Kumar Singh]

that's OK
so is it homework ? if so, start a new thread in the homework section and use the template method supplied there to lay
out your question and what work you have done so far Dave

Certainly not homework. I am not somebody's teacher to give someone some homework. I thought this is a forum for Physics lovers to exchange the ideas and shared problem solving.

 

[davenn]

Certainly not homework, I am not somebody's teacher to give someone some homework.

That's OK ... it's just the way you worded you question it looked like the normal homework Q that the forum gets 1000's of
whether you are a teacher or not isn't the issue

I thought this is a forum for Physics lovers to exchange the ideas and shared problem solving.

it is, but a major part of PF is that we like people to think for themselves and rather than just putting out answers, we encourage the poster ... you ...
to think about your problem, do some research, maybe come up with a formula or 2 that may be appropriate in helping to solve you query, then members here can guide you towards an answer, with good effort from you

Dave

 

[Kumar Singh]

That's OK ... it's just the way you worded you question it looked like the normal homework Q that the forum gets 1000's of
whether you are a teacher or not isn't the issue
it is, but a major part of PF is that we like people to think for themselves and rather than just putting out answers, we encourage the poster ... you ...
to think about your problem, do some research, maybe come up with a formula or 2 that may be appropriate in helping to solve you query, then members here can guide you towards an answer, with good effort from you

Dave

Ok. Got it

 

[davenn]

sooooooooooooooooo

with all that behind us, what things do you think are relevant in working towards a solution ?

 

[Kumar Singh]

Ok. This is what I did.

I have assumed whole case at some angle θ with vertical and then resolved the force

http://filesystem:https://web.telegram.org/temporary/851516233_42736_5351388760177376947.jpg

so,
F = 2Tcosθ
T = F/(2cosθ)
Now, for an individual mass
F(x) = T sinθ = (F tanθ)/2
a(x) = (F tanθ)/2m ↔ ∫dv(x) = (F/2m) ∫tan ωt .dt [ I don't know what to do next at this point]
F(y) = T cosθ
a(y) = F/2m ↔ v(y) = Ft/2m

Perhaps I can use one more equation

x/y = tan θ
x = y tanθ
dx = y sec^2 θ. dθ
dx/dt = d/dt ( y sec^2 θ. dθ) [Again I am stuck here..what next]

Now I have equations for v(x) and v(y) but v(x) is function of θ/ω, and I don't know if angular velocity is really constant.

again if i approach it from different method time for meeting can be calculating by assuming situation for motion along x axis

l = (1/2)[{a(x) + a(y)} (t^2)]

but here answer will come in terms of θ.

 

[ehild]

Ok. This is what I did.

I have assumed whole case at some angle θ with vertical and then resolved the force

http://filesystem:https://web.telegram.org/temporary/851516233_42736_5351388760177376947.jpg

so,
F = 2Tcosθ
T = F/(2cosθ)
Now, for an individual mass
F(x) = T sinθ = (F tanθ)/2
a(x) = (F tanθ)/2m ↔ ∫dv(x) = (F/2m) ∫tan ωt .dt [ I don't know what to do next at this point]
F(y) = T cosθ
a(y) = F/2m ↔ v(y) = Ft/2m

Perhaps I can use one more equation

x/y = tan θ

What do you call x and y? What is theta? Make a sketch and show, please. The picture in the post does not show.

 

[Kumar Singh]

Ok. This is the diagram I used to make equations..

On the other hand I used some other method to solve it...

Let the point of application moves distance d before masses meet, worked done on system would be F.d

By work energy theorem
F.d = 2 (1/2 m v^2)
v^2 = Fd/m
again for motion along y-axis
a(y) = F/2m, u = 0, s = d - l/2

So using equation v^2 = u^2 + 2as

Fd/m = 2 (F/2m) (d-l/2)

well this equation is meaningless..reaches nowhere

 

[ehild]

Ok. This is the diagram I used to make equations..

On the other hand I used some other method to solve it...

Let the point of application moves distance d before masses meet, worked done on system would be F.d

By work energy theorem
F.d = 2 (1/2 m v^2)2v^2 = Fd/m

v is the total speed, so v²=v_x²+v_y² here.

again for motion along y axis
a(y) = F/2m, u = 0, s = d - l/2

So using equation v^2 = u^2 + 2as

Along the y axis, you have to use the y component of velocity.

 

[Kumar Singh]

v is the total speed, so v²=v_x²+v_y² here.

Along the y axis, you have to use the y component of velocity.

Well...you see at the moment they meet, speed along x-axis will be zero and as I am taking energy concept in use I have to care only about initial and final positions. initially speed along x as well as y-axis was zero and finally speed is only along y axis.

 

[Kumar Singh]

v is the total speed, so v²=v_x²+v_y² here.

Along the y axis, you have to use the y component of velocity.

Finally whole speed is in y-axis only, i hope.

 

[ehild]

Well...you see at the moment they meet, speed along x-axis will be zero and as I am taking energy concept in use I have to care only about initial and final positions. initially speed along x as well as y-axis was zero and finally speed is only along y axis.

The horizontal components of velocity becomes zero when the balls collide. During their motion the x component of acceleration is towards the middle which makes the magnitude of the horizontal velocity increase, How can it become zero, if the acceleration never changes sign?

 

[Kumar Singh]

The horizontal components of velocity becomes zero when the balls collide. During their motion the x component of acceleration is towards the middle which makes the magnitude of the horizontal velocity increase, How can it become zero, if the acceleration never changes sign?

Yes , you are right. But acceleration perpetually changes magnitude along x axis...and x component of acceleration / force will vanish ultimately. But what's the nature of acceleration and velocity along x axis..i really don't know. I feel calculus is involved here. Can you help?

 

[ehild]

Sorry, I do not think this problem is properly set. You said, correctly, that F=2T cos(θ). The problem stated that initially a finite force F acted perpendicularly at the centre of the string. At the beginning, θ=pi/2, and cos (θ)=0. That would mean infinite tension in the string. I think, you should assume some initial angle between the strings and the vertical. Have you found this problem somewhere, or was it your idea?

 

[Kumar Singh]

Sorry, I do not think this problem is properly set. You said, correctly, that F=2T cos(θ). The problem stated that initially a finite force F acted perpendicularly at the centre of the string. At the beginning, θ=pi/2, and cos (θ)=0. That would mean infinite tension in the string. I think, you should assume some initial angle between the strings and the vertical. Have you found this problem somewhere, or was it your idea?

Sorry! question set up is right. A simpler version (only a(x)) has already been asked in IITJEE 2007. See here question number 3.

http://www.kshitij-iitjee.com/IITJEE-Past-Year-Papers/IITJEE-2007-Paper-1

This question has been taken from a test paper of a reputed coaching institute for IITJEE.

 

[ehild]

This is quite different problem from that in the OP. No need to consider the motion of the balls. It simply asks a_x, x component of the acceleration, at the moment when the separation between the balls is x, and the balls are connected to the ends of a string of length a and a force F acts vertically at the middle of the string. A fairly easy problem. The one in the OP needs calculus - a differential equation to solve. Use the equation you have already : a_x = - (F tanθ)/2m, but write tan(θ) in terms of x and the length of the string.