Collision involving blocks and pulley

[theodoros.mihos]

Exercise ask for the acceleration of B exactly after collision, not ask for the equation of motion. The answer is zero because θ=0 exactly after collision.

 

[Chestermiller]

Exercise ask for the acceleration of B exactly after collision, not ask for the equation of motion. The answer is zero because θ=0 exactly after collision.

Sir,

Suppose I told you that the spatial location of mass A as a function of time after the collision is described in Cartesian coordinates by the equations:
$$x=\frac{\sqrt{gl}}{3}t$$
$$y=\frac{g}{36}t^2$$
Would you say that (a) mass A is not accelerating upward at time t = 0 because its initial direction of travel is horizontal or (b) mass A is accelerating upward at time t = 0, even though its initial direction of travel is horizontal?

If mass A is accelerating upward at time t = 0, would you say that this (a) is caused by a change in the tension in the rope above mass A or (b) is not caused by a change in the tension in the rope above mass A?

If the tension in the rope above mass A causes the upward acceleration of mass A, would you say that this (a) causes an acceleration of mass B or (b) does not cause an acceleration of mass B?

Chet

 

[theodoros.mihos]

I don't care about the functions of move. The above analysis is very good but solves a much larger problem. At t=0 exact after collision, A have only horizontal velocity, so vertical acceleration of m_A is zero.

 

[Chestermiller]

I don't care about the functions of move. The above analysis is very good but solves a much larger problem. At t=0 exact after collision, A have only horizontal velocity, so vertical acceleration of m_A is zero.

Are you aware that, if the direction in which an object is moving is changing with time, that constitutes acceleration? To get it to change direction with time, you need to exert a force. Have you ever heard the term centripetal acceleration?

Chet

 

[theodoros.mihos]

I hear but I can't see it before A change its position horizontally. Sorry, I 'm a farmer and my mind works simply.

 

[Chestermiller]

I hear but I can't see it before A change its position horizontally. Sorry, I 'm a farmer and my mind works simply.

OK. From your experience as a farmer, if you throw a tomato off a cliff horizontally, does the tomato start to fall with increasing speed downward as soon as you release it, or is there some magical time that you need to wait before it starts to speed up in the downward direction? If, instead of throwing the tomato horizontally, you simply let it drop off the cliff, would it take less time to hit the ground below than when you throw it horizontally? Or would the amount of time be about the same? What are your thoughts on this?

Chet

 

[theodoros.mihos]

You are right. Sorry for my English. A body can have acceleration while its velocity is steel zero.
But this problem is more complicate because the "radius" of A movement is not constant. I can communicate better by equations so system Lagrangian with origin on top is:
$$ L = m \left( \frac{1}{2}\dot{x}_A^2+\frac{1}{2}\dot{y}_A^2+\frac{1}{2}\dot{y}_B^2+gy_A^2(x_A^2+y_A^2)^{-1/2}+gy_B \right) \Rightarrow \frac{d}{dt}\frac{\partial{L}}{\partial{\dot{y}_B}} = mg \Rightarrow \ddot{y}_B = g $$

This agree with the (a) answer but exist a condition:
$$ -y_B + \sqrt{x_A^2+y_A^2} = 2l $$
where $2l$ the total rope length. Adding this condition the Lagrangian takes the form:
$$ L = m \left( \frac{1}{2}\dot{x}_A^2+\frac{1}{2}\dot{y}_A^2+\frac{1}{2}\dot{y}_B^2+g\frac{y_A^2}{2l+y_B}+gy_B \right) $$
(we want to keep the information for $y_B$ on Lagrangian) so the derivation gives:
$$ \ddot{y}_B = g\left(1-\frac{y_A^2}{(2l+y_B)^2}\right) $$
That mean the acceleration of B when $y_A=y_B=-l$ is zero, so the correct answer is (d). As I say on previous post, needed a small amount of $y_A$ so of $x_A$ to make acceleration on B. On computational physics there is no $\Delta{t}\to0$ and this fact is absolutely clear.
This is very interesting problem, thanks to all.

 

[theodoros.mihos]

https://drive.google.com/file/d/0Bxn0albRsY_RVEFleTlQVmNnb2c/view?usp=sharing
The red line is the trajectory of A and green line is just the $y_B$ value for the same time for $x_A$.
There is no upward acceleration to both A and B because rope cannot be less than 2l. All works with assumption about "circular" movement with radius l probably are basically wrong. System goes out of stability even few initial energy. It is very easy to try.

 

[ehild]

But this problem is more complicate because the "radius" of A movement is not constant. I can communicate better by equations so system Lagrangian with origin on top is:
$$ L = m \left( \frac{1}{2}\dot{x}_A^2+\frac{1}{2}\dot{y}_A^2+\frac{1}{2}\dot{y}_B^2+gy_A^2(x_A^2+y_A^2)^{-1/2}+gy_B \right) \Rightarrow \frac{d}{dt}\frac{\partial{L}}{\partial{\dot{y}_B}} = mg \Rightarrow \ddot{y}_B = g $$

Do you try to write the Langranigian of theoriginal setup? There are three bodies, A,B of mass m and C of mass m/2, sticked to the pan and connected by an other rope to A. You ignored it.

This agree with the (a) answer but exist a condition:
$$ -y_B + \sqrt{x_A^2+y_A^2} = 2l $$
where $2l$ the total rope length. Adding this condition the Lagrangian takes the form:
$$ L = m \left( \frac{1}{2}\dot{x}_A^2+\frac{1}{2}\dot{y}_A^2+\frac{1}{2}\dot{y}_B^2+g\frac{y_A^2}{2l+y_B}+gy_B \right) $$
(we want to keep the information for $y_B$ on Lagrangian) so the derivation gives:
$$ \ddot{y}_B = g\left(1-\frac{y_A^2}{(2l+y_B)^2}\right) $$

Considering m = unity, the potential energy of A is -gy_A instead of $-gy_A^2(x_A^2+y_A^2)^{-1/2}$.
In the constraint equation, you should use +y_B.

You can not treat the constraints as you did. Apply the method of Lagrange Multipliers.
And there is an other constraint, the constant length of the rope connecting C and A.

It would be very nice if you could show a numerical solution of the real problem.

 

[theodoros.mihos]

Before and after colission energy is not conserved so the Lagrangian is different. I start with the initial velocity of A computed, as others doing before me. I said the origin is on the top, sign is ok. I use Lagrangian method only for $\ddot{y}_B$ so for accelerations take what ever for $x_0$ or $v_0$.
I ignore the plate because this problem not ask for the movement equation but only what is the acceleration on B just after collission. If steel exist someone which believes that the A body start with vertical acceleration to up, I have no anything to say. We have do more much work that the initial problem ask.

 

[Chestermiller]

You are right. Sorry for my English. A body can have acceleration while its velocity is steel zero.
But this problem is more complicate because the "radius" of A movement is not constant.

Thank you for finally acknowledging that mass A can be accelerating upward even though its initial upward velocity is zero.

All works with assumption about "circular" movement with radius l probably are basically wrong.

Excuse me, but my analysis does not assume that the movement of mass A is circular with radius l, and it does take into account the fact that the radial distance of mass A from the pulley is not constant. The equations that I presented in post #62 give the actual trajectory of mass A at short times, where x is the distance mass A moves to the left, and y is the distance mass A moves upward. If we eliminate time t from these equations, we obtain:
$$y=\frac{x^2}{4l}$$
This is consistent with a radius of curvature of the circular arc of mass A of 2l, rather than l. So mass A does execute a circular arc at short times, but, as a result of the increasing radial distance of mass A from the pulley, it is initially traveling along an arc of radius 2l rather than l.

Chet

 

[ehild]

Theodoros inspired me to do the Lagrangian method. I have to admit that TSny and Chet were right, the upward acceleration of B is v₀²/(2L) at t=0, just after the collision .

@theodoros.mihos : I see, you considered upward positive for y, so the sign is OK in the constraint.
But why did you write that the potential energy of A is $\frac{gy_A^2}{\sqrt{x_A^2+y_A^2}}$ instead of gy_A?

 

[theodoros.mihos]

@ehild you are right. This mistake comes for a previous look about the direct force.
@chet. I see match more works except yours. Nothing personal.
Now all differs.

 

[Chestermiller]

@ehild you are right. This mistake comes for a previous look about the direct force.
@chet. I see match more works except yours. Nothing personal.
Now all differs.

I have no idea what you are saying. But I can tell you that I am totally thrilled that ehild, TSny, and I are now in total agreement.

Now that full consensus has been reached, I am closing this thread.

Chet