Collision involving blocks and pulley

[Chestermiller]

One way to get a handle on all this is to see what a simple model of the inelastic collision between the mass and the pan tells you. I have done such a model, by inserting a small viscous damper between the mass and the pan, and allowing the mass to hit the damper at the initial contact velocity $v_0=\sqrt{gl}$. The force exerted by the viscous damper is given by $F=C(v-v_p)=C(v-v_A)$, where C is the damper constant, v is the velocity of the falling mass during the collision, v_p is the downward velocity of the pan at time t during the collision, and v_A=v_p is the horizontal velocity of mass A during the collision. From force balances on the falling mass m/2 and on mass A (m), I obtain:
$$\frac{m}{2}\left(\frac{dv}{dt}\right)=\frac{m}{2}g-C(v-v_A)$$
$$m\frac{dv_A}{dt}=+C(v-v_A)$$
Initial conditions are v(0)=v₀ and v_A=0.

If we neglect the term representing the weight of the falling mass during the very short time interval of the collision, the solution for the horizontal velocity of mass A as a function of time is found to be:
$$v=\frac{v_0}{3}\left(1-e^{-\frac{3C}{m}t}\right)$$
So, based on what I presented in post #44, during the collision, we have:
$$\frac{d^2r}{dt^2}=\frac{g}{18}\left(1-e^{-\frac{3C}{m}t}\right)^2$$
This result indicates that, at the beginning of the collision, the acceleration of mass B is zero, but at times $t>>\frac{m}{3C}$ (i.e., by the end of the collision), the acceleration of mass B is g/18.

Chet

 

[ehild]

The equations I obtained for time zero were:

Radial Force Balance from upper pulley to mass A:
$$T-mg=m\left[r\left(\frac{dθ}{dt}\right)^2-\frac{d^2r}{dt^2}\right]$$
where r is the distance between the upper pulley and mass A, θ is the angle that the rope segment from the upper pulley to mass A makes with the vertical, and T is the tension in the upper rope.

The left rope also exerts force on A. (And you have to take the radial component of mg.)

 

[Chestermiller]

The left rope also exerts force on A. (And you have to take the radial component of mg.)

And you think that this is significant on the time scale of the collision, correct? If so, do you think you can provide us with an estimate of what that radial force would be at say time t =4m/3C (when the collision is effectively complete, and the lower mass and pan are now traveling at essentially the same speed) so that we can compare it with the value mg/18? If you don't think you can do it, I think that I can.

Chet

 

[ehild]

I think it significant. I tried but I failed.

 

[Chestermiller]

I think it significant. I tried but I failed.

OK. I know how to do this, but I won't be able to get back here until later today.

Chet

 

[Chestermiller]

I think it significant. I tried but I failed.

OK. The objective is to get an estimate of the effect of the mg/2 portion of the tension in the lower rope on the radial component of tension in the upper rope. I'm going to start out by writing down the force balance equations on mass A, including the angulation effect of the two ropes (which was neglected in the previous development):
$$T'cosθ'-Tsinθ=ma_x $$
$$-T'sinθ'+Tcosθ-mg=ma_y$$
The first equation is the force balance in the horizontal direction, the second equation is the force balance in the vertical direction, the x-direction is to the left, the y-direction is upward, T' is the tension in the lower rope, and θ' is the angle that the lower rope makes with the horizontal. If we multiply the second equation by cosθ and the first equation by sinθ, and then subtract the first equation from the second equation, we obtain the component of the force balance in the radial direction:
$$-T'sin(θ+θ')+T-mg\cosθ=-ma_r=m\left[r\left(\frac{dθ}{dt}\right)^2-\frac{d^2r}{dt^2}\right]$$
The focus of the present analysis is the first term on the left hand side of this equation. Our first objective is to obtain an upper bound to the angle θ+θ' at the end of the collision, effectively at t =4τ, where t = (m/3C). If we integrate our equation for the x-velocity over this time interval, we obtain $$δx=v_0τ$$where δx is the x displacement during the time period. From this, it follows that the angle θ is given by:
$$θ=\frac{v_0τ}{l}$$
We can obtain an upper bound to the angle θ' by assuming that the mass A follows a circular arc, rather than also including the outward radial component of velocity along the upper rope. This will result in an upper bound to the predicted vertical displacement δ_y of mass A. So,
$$δ_y=l-lcosθ=l(1-cosθ)=l\frac{θ^2}{2}=\frac{(v_0τ)^2}{2l}$$
From this it follows that the angle θ' is given by:
$$θ'=\frac{(v_0τ)^2}{2ll'}$$
where l' is the horizontal distance between the lower pulley and mass A.
If we now sum the angles θ and θ', we obtain:
$$θ+θ'=\left(\frac{v_0τ}{l}\right)\left[1+\frac{v_0τ}{2l'}\right]$$
Note that, if the length l' is on the same order as l, the contribution of θ' to the sum of the two angles will be considerably smaller than the contribution of θ. If we substitute the equation $v_0=\sqrt{gl}$ into the the equation for θ+θ', we obtain:
$$θ+θ'=\sqrt{\frac{g}{l}}τ\left[1+\frac{\sqrt{gl}}{2l'}τ\right]≈\sqrt{\frac{g}{l}}τ$$
From this it follows that the contribution of the lower weight mg/2 to the radial component of tension on the upper rope is given by:
$$δT=\frac{mg}{2}\sqrt{\frac{g}{l}}τ$$
If the length l is 1 m, and the characteristic collision contact time is equal to ~0.001 sec, this equation predicts that δT≈0.0016 mg. This compares with the total tension of T=mg/18 = 0.055 mg calculated without accounting for the angulation of the ropes. With these values for the data, the contribution of the weight of the falling mass to the tension on mass B is on the order of a factor of about 35 lower than mg/18.

Chet

 

[ehild]

Thank you Chet. I will think it over.

 

[theodoros.mihos]

Why an horizontal acceleration on A makes vertical acceleration on B at t=0 ?
This is a multiple choice task and must be answered by few or non calculations.

 

[insightful]

Why an horizontal acceleration on A makes vertical acceleration on B at t=0 ?

Because it is not a horizontal acceleration on A, but a horizontal velocity given to A at t=0, which gives A a centripetal acceleration.

 

[ehild]

Why an horizontal acceleration on A makes vertical acceleration on B at t=0 ?
This is a multiple choice task and must be answered by few or non calculations.

See figure. If A travels horizontally, the left piece of the string gets longer, so the vertical piece has to shorten by the same amount. You can get the vertical acceleration of B $\ddot y_2 = \ddot r$ in terms of the horizontal velocity of A.

 

[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