Collision involving blocks and pulley

[Chestermiller]

I analyzed this problem independently, and ended up with the same analysis that TSny presented. To implement the analysis, I assumed that the radii of the two pulleys are very small compared to the lengths of the ropes, so that I didn't need to take into account the effect of the change in the wrap angle on the kinematics of the motion. I confirm TSny's result of g/18.

Chet

 

[insightful]

The speed of the particle before collision is v. v2= gL. Conservation of momentum says that 3/2 MU = M/2 v -->U = v/3.
Before the collision happened, the tension in the other string was Mg.
At the first instant, there is no displacement yet. A hangs vertically, and has a horizontal velocity - this is circular motion at that instant. To maintain that motion, the tension had to change The tension in the vertical string is T. The centripetal force is MU2/L=T-Mg.
T=Mg+Mv2/(9L)=Mg+MgL/(9L)=Mg+Mg/9.. The acceleration of B is determined by the upward tension and the downward force of gravity.

I find this to be correct, which means the answer is g/9.

 

[TSny]

Let $r$ be the length of the vertical string between the pulley and mass A. The point of disagreement seems to be with how one answers the following questions.

(1) If block B has an upward acceleration just after the collision, then does that mean that $\ddot{r} \neq 0$ at the same instant of time? If not, why not? If so, then proceed to question (2).

(2) Since $\ddot{r} \neq 0$, then doesn’t that mean that $\ddot{r}$ should contribute to the acceleration of mass A? If not, why not? If so, then proceed to question (3).

(3). [Edited to correct typo] If $\ddot{r}$ contributes to the acceleration of A, then doesn’t that mean that the acceleration of A just after the collision is $v^2/r - \ddot{r}$ rather than just $v^2/r$? If not, why not?

It seems odd to me to allow the effect of the velocity of A to be transmitted essentially instantaneously by the string to B (and causing B to accelerate) while not allowing the effect of the acceleration of B to be transmitted essentially instantaneously back to A.

If you allow any "time delays" due to elasticity of the string, then the question becomes ambiguous. I think this is ehild's point and why he doesn't like the question. You could then argue that the acceleration of B is zero immediately after the collision since there has not been enough time for the effect of the collision at A to propagate to B.

 

[insightful]

the acceleration of A just after the collision is mv2/r−r¨mv^2/r - \ddot{r} rather than just mv2/rmv^2/r? If not, why not?

Why is m in these expressions?

 

[TSny]

Why is m in these expressions?

The mass should not be there. Thanks!

 

[insightful]

So, the effect of the string accelerating is that the radius of movement of mass A is actually twice as long (2xl)? This would halve my tension to also give a=g/18 for mass B.

 

[TSny]

No, just after the collision the length of the string on the left side of the pulley is still $l$. So, the acceleration of mass A due to its circular motion is $v^2/l$. However, I believe there is an additional acceleration of A due to the acceleration of the string ($\ddot{r}$) associated with the acceleration of B.

 

[insightful]

But wouldn't that immediate additional acceleration immediately start mass A on a path different from a circle of radius l ?

 

[Chestermiller]

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.

Vertical Force Balance on mass B:
$$T = mg+m\frac{d^2r}{dt^2}$$

If we combine these equations, we get:

$$\frac{d^2r}{dt^2}=r\left(\frac{dθ}{dt}\right)^2-\frac{d^2r}{dt^2}$$

Solving for $d^2r/dt^2$, we obtain:

$$\frac{d^2r}{dt^2}=\frac{r}{2}\left(\frac{dθ}{dt}\right)^2=\frac{v^2}{2r}$$

In my judgement, this confirms TSny's g/18 result.

Chet

 

[Satvik Pandey]

Let $r$ be the length of the vertical string between the pulley and mass A. The point of disagreement seems to be with how one answers the following questions.

(1) If block B has an upward acceleration just after the collision, then does that mean that $\ddot{r} \neq 0$ at the same instant of time? If not, why not? If so, then proceed to question (2).

(2) Since $\ddot{r} \neq 0$, then doesn’t that mean that $\ddot{r}$ should contribute to the acceleration of mass A? If not, why not? If so, then proceed to question (3).

(3). [Edited to correct typo] If $\ddot{r}$ contributes to the acceleration of A, then doesn’t that mean that the acceleration of A just after the collision is $v^2/r - \ddot{r}$ rather than just $v^2/r$? If not, why not?

It seems odd to me to allow the effect of the velocity of A to be transmitted essentially instantaneously by the string to B (and causing B to accelerate) while not allowing the effect of the acceleration of B to be transmitted essentially instantaneously back to A.

If you allow any "time delays" due to elasticity of the string, then the question becomes ambiguous. I think this is ehild's point and why he doesn't like the question. You could then argue that the acceleration of B is zero immediately after the collision since there has not been enough time for the effect of the collision at A to propagate to B.

I too have same doubt. This question came in test papers of some coaching institution. And in the solution they simply wrote $a=v^(2) /l.$ I asked same question about the radial acceleration of $M_{a}$ here:(https://brilliant.org/discussions/thread/please-help-18/?ref_id=771297) but I didn't get any convincing answer.

Mass A is not pivoted about a fixed point so here mass A will not have centripetal acceleration only. So I think it is not evident to consider centripetal acceleration of block only. I think the answer should be $g/18$.

 

[ehild]

If you allow any "time delays" due to elasticity of the string, then the question becomes ambiguous. I think this is ehild's point and why he doesn't like the question. You could then argue that the acceleration of B is zero immediately after the collision since there has not been enough time for the effect of the collision at A to propagate to B.

Yes, it is my point. Collision between two bodies can be solved by applying conservation of momentum and the condition for energy but the problem is undetermined in case of three or more bodies without knowing how the mechanical deformation propagates in them.
In this problem, the pan gets an impulse. Its mass is negligible, so the impulse is transmitted to the string around the left pulley to block A.Block A hangs on the other string, and the tension in it just compensates its weight, so A starts to move horizontally. That is the outcome of the first stage of collision.
I do not know how you would solve the problem if all bodies take part in the collision instantaneously.
A nice example for collision among more bodies is Newton's cradle. What we see is that after the first ball hits the next, the last ball emerges. http://en.wikipedia.org/wiki/Newton's_cradle. That result is obtained if we assume collisions happening one after the other, between a pair of balls as if the balls do not touch each other. But they do. So the true solution would consider elastic wave traveling in the balls.

Block A hangs on the vertical string around the right pulley. If the pulley had mass the block was able to start its motion along a circle. So its motion would not change the length L₀ of the string, but changes the tension in it. Well, the pulley has negligible mass, but the situation could be the same.

Assuming A moves horizontally with speed u. With TSny notation, r he length of the left piece of the rope changes as $r=\sqrt{L^2+(ut)^2}$.
If t=0, the rate of change is zero, and the acceleration is $\ddot r = \frac{u^2}{L }$. It implies that the second derivative of the right piece is the negative of it, $\ddot s = - \frac{u^2}{L } = a_B$.

But my opinion is that such problems can not be solved and should not be given to high-school students.
For collision among more bodies, Newton's cradle is a nice example. The simple solution assumes that collisions happen between two balls, one after other, as if the balls do not touch each other. But they do, and the exact solution should include the propagation of the elastic wave inside the balls. http://en.wikipedia.org/wiki/Newton's_cradle

 

[TSny]

Assuming A moves horizontally with speed u. With TSny notation, r he length of the left piece of the rope changes as $r=\sqrt{L^2+(ut)^2}$.
If t=0, the rate of change is zero, and the acceleration is $\ddot r = \frac{u^2}{L }$. It implies that the second derivative of the right piece is the negative of it, $\ddot s = - \frac{u^2}{L } = a_B$.

It seems to me that your expression for $r$ as a function of time is equivalent to treating A as moving along a horizontal line with constant speed just after the collision. Thus, A would have zero vertical component of acceleration. Another way to see this is to use the standard expression for radial acceleration in polar coordinates to write $a_{A,y} = r \dot{\theta}^2 - \ddot{r}$ where y is taken as positive upward. According to your result, $\ddot{r} = \frac{u^2}{L }$. Therefore, $a_{A,y} = r \dot{\theta}^2 - \ddot{r} = \frac{u^2}{L } - \frac{u^2}{L } = 0$.

But if $a_{A,y} = 0$, then $T-Mg = Ma_{A,y} = 0$. So, $T-Mg = 0$ for A. But then $T-Mg = 0$ also for B if we assume the tension is the same throughout the string. Therefore the net force on B is zero and B would have zero acceleration, which is not consistent with $a_B = \ddot{r}$

 

[ehild]

It seems to me that your expression for $r$ as a function of time is equivalent to treating A as moving along a horizontal line with constant speed just after the collision. Thus, A would have zero vertical component of acceleration. Another way to see this is to use the standard expression for radial acceleration in polar coordinates to write $a_{A,y} = r \dot{\theta}^2 - \ddot{r}$ where y is taken as positive upward. According to your result, $\ddot{r} = \frac{u^2}{L }$. Therefore, $a_{A,y} = r \dot{\theta}^2 - \ddot{r} = \frac{u^2}{L } - \frac{u^2}{L } = 0$.

But if $a_{A,y} = 0$, then $T-Mg = Ma_{A,y} = 0$. So, $T-Mg = 0$ for A. But then $T-Mg = 0$ also for B if we assume the tension is the same throughout the string. Therefore the net force on B is zero and B would have zero acceleration.

That is what I assumed. B does not accelerate during the collision between the falling particle, pan and A, but starts to accelerate after as consequence of the motion of A.
Of course, it is an approximation. In reality, everything is involved but then you have to take elasticity of the strings and mass of the pulleys into account.

 

[TSny]

That is what I assumed. B does not accelerate during the collision between the falling particle, pan and A, but starts to accelerate after as consequence of the motion of A.

How do you make $a_B = 0$ consistent with $a_B = \ddot{r} \neq 0$?

 

[ehild]

a_B=0 during the collision, a_B≠0 after the collision.
There is some time delay before B starts to move. The limit of it is zero when the string is ideal. See the video when a ball collides with two others, connected by a spring. The collision happens between the touching balls, and the other ball starts to move a bit later, when it feels the tension built up in the string.

 

[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.