3 balls in a moving mechanics problem

[Manolisjam]

Homework Statement

Consider 2 balls A,B on the same line . and they are connected to a third one G with a rope L. AG, AB. now the system monves in the effect of the mass of G and its projection to the line AB is in the middle. No friction. mass of A=mass of B=m and mass of G=2m
.FInd the time of impact and the velocity of G at that time A-B move only on the x axes.

Homework Equations

The Attempt at a Solution

Dont understand what it means the system moves on the effect of the mass G. Also I know the only forces are From the rope. 2 on the G ball and 1 in each of the balls.I understant i have to apply conservation law of energy but how? What is the potential energy(what if the force) ?

 

[kuruman]

Dont understand what it means the system moves on the effect of the mass G.

I think it means that mass Γ is pulling the the other two masses towards each other as it drops down. If masses A and B are moving horizontally does their potential energy change? What about mass Γ?

 

[Manolisjam]

I think it means that mass Γ is pulling the the other two masses towards each other as it drops down. If masses A and B are moving horizontally does their potential energy change? What about mass Γ?

so what is the force acting on G 2mg?

 

[kuruman]

so what is the force acting on G 2mg?

Yes, plus the tension from each string.

 

[Manolisjam]

Yes, plus the tension from each string.

So far so good?

 

[kuruman]

So far so good?

Nope. Why is T₁ = mg = T₂? Furthermore, the tension is not constant as mass G descends. If I were you, I would stick to the original idea of energy conservation and not try to derive the equation of motion. Note that the time of impact depends on how far apart masses A and B are when they are released (I assume) from rest. Does the problem give that distance?

 

[Manolisjam]

Nope. Why is T₁ = mg = T₂? Furthermore, the tension is not constant as mass G descends. If I were you, I would stick to the original idea of energy conservation and not try to derive the equation of motion. Note that the time of impact depends on how far apart masses A and B are when they are released (I assume) from rest. Does the problem give that distance?

From symmetry? ALso why isn't the tension constant? the rope isn't elastic .

 

[Manolisjam]

From symmetry? ALso why isn't the tension constant? the rope isn't elastic .

my Tx isn't constant

 

[Manolisjam]

Nope. Why is T₁ = mg = T₂? Furthermore, the tension is not constant as mass G descends. If I were you, I would stick to the original idea of energy conservation and not try to derive the equation of motion. Note that the time of impact depends on how far apart masses A and B are when they are released (I assume) from rest. Does the problem give that distance?

IM really confused...

 

[kuruman]

From symmetry? ALso why isn't the tension constant? the rope isn't elastic .

Symmetry demands that the tensions be equal, T₁ = T₂, but not that they are equal to mg which is what you had.

my Tx isn't constant

I agree, but neither T_y is constant. Is the acceleration constant or does it depend on θ?

 

[Manolisjam]

Symmetry demands that the tensions be equal, T₁ = T₂, but not that they are equal to mg which is what you had.

I agree, but neither T_y is constant. Is the acceleration constant or does it depend on θ?

neither Ty is constant what does that say? Also how do you break up a force equal=2mg to 2 equal forces? ANd the acceleration depends on θ

 

[kuruman]

neither Ty is constant what does that say? Also how do you break up a force equal=2mg to 2 equal forces? ANd the acceleration depends on θ

You have to consider two free body diagrams, one for the hanging mass G and one for A. You get two equations of motion. Note that the accelerations of A and G are different but related through angle θ. That relation you have to derive starting from y = (L² - x²)^1/2.

 

[Manolisjam]

You have to consider two free body diagrams, one for the hanging mass G and one for A. You get two equations of motion. Note that the accelerations of A and G are different but related through angle θ. That relation you have to derive starting from y = (L² - x²)^1/2.

dont get it. COuld you please post a pic with a complete answer I am trying 2 days now.

 

[Manolisjam]

dont get it. COuld you please post a pic with a complete answer I am trying 2 days now.

A, B only move in x don't fall

 

[Manolisjam]

A, B only move in x don't fall

why T1=T2=mg is wrong?

 

[kuruman]

dont get it. COuld you please post a pic with a complete answer I am trying 2 days now.

Sorry, it's against forum rules to post complete answers.

why T1=T2=mg is wrong?

Because in the vertical direction, the forces acting on mass G are 2T_y up and 2mg down. What is the net force on G? What is the net force equal to according to Newton's 3rd law?

 

[Manolisjam]

Sorry, it's against forum rules to post complete answers.

Because in the vertical direction, the forces acting on mass G are 2T_y up and 2mg down. What is the net force on G? What is the net force equal to according to Newton's 3rd law?

 

[kuruman]

You don't really need to find T. You want to write two equations involving T the masses, and the accelerations. These would be F_net = Ma in the vertical and horizontal directions. So in the vertical direction, you get 2T cosθ - 2mg = - 2ma_G. Can you write an analogous equation for the horizontal direction?

 

[Manolisjam]

You don't really need to find T. You want to write two equations involving T the masses, and the accelerations. These would be F_net = Ma in the vertical and horizontal directions. So in the vertical direction, you get 2T cosθ - 2mg = - 2ma_G. Can you write an analogous equation for the horizontal direction?

mind that A-B don't move on the y axes

 

[Manolisjam]

mind that A-B don't move on the y axes

ALso is it possible to find T without knowning a_g ? just asking cause i want to undersand what is happening

 

[Manolisjam]

ALso is it possible to find T without knowning a_g ? just asking cause i want to undersand what is happening

Also i found Ty on an angle θ' not on θ

 

[kuruman]

mind that A-B don't move on the y axes

I understand that. However, there is horizontal tension T_x that accelerates A and B in the horizontal direction.
No

ALso is it possible to find T without knowning a_g ?

No, because they are related through Newton's 2nd Law equation. What's happening is that as mass G descends, it pulls the other masses together until they collide.

 

[kuruman]

Also i found Ty on an angle θ' not on θ

It doesn't matter because they are related, θ + θ' = 90°. You can choose either one.

 

[Manolisjam]

It doesn't matter because they are related, θ + θ' = 90°. You can choose either one.

 

[Manolisjam]

It doesn't matter because they are related, θ + θ' = 90°. You can choose either one.

i osted my progress so far.

 

[kuruman]

Let's say the gravitational potential energy U is zero when all three masses are at the same height and mass G is at the origin O. What is the potential energy when the distance between A and B is 2x and mass G is distance y below the origin? How are x and y related?

 

[Manolisjam]

Let's say the gravitational potential energy U is zero when all three masses are at the same height and mass G is at the origin O. What is the potential energy when the distance between A and B is 2x and mass G is distance y below the origin? How are x and y related?

so all i did is a waste?

 

[Manolisjam]

Let's say the gravitational potential energy U is zero when all three masses are at the same height and mass G is at the origin O. What is the potential energy when the distance between A and B is 2x and mass G is distance y below the origin? How are x and y related?

ive been trying 2 days now straight .l . I've asked other people no1 will give me a full answer so i can study it and understant it and learn from it . because that the policy .. i give up...

 

[kuruman]

ive been trying 2 days now straight .l . I've asked other people no1 will give me a full answer so i can study it and understant it and learn from it . because that the policy .. i give up...

I am sorry you feel that you should give up. In this forum we do not give answers but guide people to the answers so they could learn how to do things on their own. I tried pointing you in the right direction when I posted #26, but instead of following my lead and answering my questions, you became concerned about your "wasted" time. So be it, I will not waste any more of your time.

 

[Manolisjam]

I am sorry you feel that you should give up. In this forum we do not give answers but guide people to the answers so they could learn how to do things on their own. I tried pointing you in the right direction when I posted #26, but instead of following my lead and answering my questions, you became concerned about your "wasted" time. So be it, I will not waste any more of your time.

sorry for being rude I am just frustrated.. really appreciate you wasting your time with me.

 

[kuruman]

I do not consider the time I gave freely to you wasted. If you are willing to pick up at post #26 and reconsider your expressions for the initial and final potential energy, please do so and I will continue guiding you. If not, not.

 

[Manolisjam]

I do not consider the time I gave freely to you wasted. If you are willing to pick up at post #26 and reconsider your expressions for the initial and final potential energy, please do so and I will continue guiding you. If not, not.

are you sure conservation will lead me to finding the time of impact?

 

[Manolisjam]

are you sure conservation will lead me to finding the time of impact?

some1 else told me to llok at the relaton between the accaleration o A in x axes withs its component to the direction AG and then the relation between Gs accelaration and that component

 

[Manolisjam]

I do not consider the time I gave freely to you wasted. If you are willing to pick up at post #26 and reconsider your expressions for the initial and final potential energy, please do so and I will continue guiding you. If not, not.

Ok I am trying to do what you said i get the same with i did already .what is different? MY potential energy was zero when G was at L distance so if its zero at the same height i sa the final potential energy is the initial i found.

 

[Manolisjam]

I do not consider the time I gave freely to you wasted. If you are willing to pick up at post #26 and reconsider your expressions for the initial and final potential energy, please do so and I will continue guiding you. If not, not.

Laso could you classify the diffculty of the problem 1-10. for a math undergrad .

 

[Manolisjam]

I do not consider the time I gave freely to you wasted. If you are willing to pick up at post #26 and reconsider your expressions for the initial and final potential energy, please do so and I will continue guiding you. If not, not.

Solved it!

 

[haruspex]

Solved it!

This is a duplicate of
https://www.physicsforums.com/threads/classical-mechanics-problem-with-balls.940763/
@Manolisjam , please do not duplicate threads to garner a wider audience. If you wish to bring others in you can either use the "@" link to bring in specific people, or ask your current respondent (me, in this case) to do it. Or even "report" the thread to the mentors.

Anyway, you claim to have solved it, and I think that may be true for the collision velocity if you figured out how to write the energy equation correctly, but I do not see how you will have found the time that way.
Have you found time to collision?

 

[Manolisjam]

This is a duplicate of
https://www.physicsforums.com/threads/classical-mechanics-problem-with-balls.940763/
@Manolisjam , please do not duplicate threads to garner a wider audience. If you wish to bring others in you can either use the "@" link to bring in specific people, or ask your current respondent (me, in this case) to do it. Or even "report" the thread to the mentors.

Anyway, you claim to have solved it, and I think that may be true for the collision velocity if you figured out how to write the energy equation correctly, but I do not see how you will have found the time that way.
Have you found time to collision?

At the 0 point using conservation energy i get something like u_a^2+u_g^2=2gsinθl now i know dx/dt=u_a=lsinθdθ/dt . find same way u_G now squaring those. andi plugging them in conservation i get Ldθ/dt=sqrt(2sinθ) this is a differential eq separable. find θ(t) and solve for θ=π/2 but i can't solve the integral

 

[Manolisjam]

At the 0 point using conservation energy i get something like u_a^2+u_g^2=2gsinθl now i know dx/dt=u_a=lsinθdθ/dt . find same way u_G now squaring those. andi plugging them in conservation i get Ldθ/dt=sqrt(2sinθ) this is a differential eq separable. find θ(t) and solve for θ=π/2 but i can't solve the integral

if you can show me another way for the time. i still haven't understant what you are trying to help me do.

 

[haruspex]

if you can show me another way for the time. i still haven't understant what you are trying to help me do.

Ok, you have found another route to the same equation. (Your final equation is not quite right: check the powers of L and g in it.)
In case it helps in future, if two objects remain a constant distance apart then they must have the same velocities and accelerations along the line joining them. Hence a_Acos(θ)=a_Gsin(θ).
I'll get back to you on solving the integral.

 

[haruspex]

I'll get back to you on solving the integral

It seems to involve elliptic integrals. Nasty.
http://www.wolframalpha.com/input/?i=integrate+(sin(x))^(-1/2)dx+from+x=0+to+pi/2

 

[kuruman]

It seems to involve elliptic integrals. Nasty.

Yes, it is nasty. It's a bit simpler if all masses start from rest and along the horizontal line through the origin, but still an elliptic integral. Considering that OP is a math undergrad, this is perhaps a math exercise that assumes understanding of physics to get to the math. I would be curious to see what the solution is according to the person who assigned the problem.