How Is Velocity Calculated When a Rod Falls onto a Mass Connected by a Spring?

[Hamal_Arietis]

Homework Statement

The uniform rod mass m, length L falling into mass M. M is attacked with string k. Find the velocity u of system when the top of rod falling into M.

Homework Equations

I find this equation, but it seems wrong:
Chose origin at equilibrium position.
The rod has been divided into small segments, each of length has $dm=\frac{m}{L}dy$
At the time t, the mass M is effected of force F, with F is:
+ The total force of gravitation and the elastic force $F_1=ky$(magnitude)
+ The change of momentum of dm. It equals $F_2=\dfrac{dp'}{dt}=v\dfrac{dm}{dt}=v\dfrac{m}{L}\dfrac{dy}{dt}=v^2\dfrac{m}{L}$ with v is the velocity when dm touches M
So $F=-F_1+F_2=-ky+v^2\dfrac{m}{L}$
That force equals momentum changes with respect to time of system (M+dm)
$$ F=\frac{dp}{dt}$$
$$\Leftrightarrow -ky+v^2\frac{m}{L}=u\frac{dm}{dt}+(M+dm)\frac{du}{dt}=u\frac{dm}{dt}+M\frac{du}{dt}$$

The Attempt at a Solution

Where are wrong ? I don't have solution.
v is the velocity when dm touches M. Each of collision, String changes so I can't find v.
This way seems difficult

 

[ramzerimar]

Find the velocity u of system when the top of rod falling into M.

This last sentence makes no sense to me. I understand that your system is the mass-spring system. Do you want to find the velocity just after the rod hits the mass M?

 

[Hamal_Arietis]

This system is mass- rod. the velocity u at the time after the total length of the rod fall into mass M

 

[haruspex]

This system is mass- rod. the velocity u at the time after the total length of the rod fall into mass M

I do not understand the set-up.
What do you mean by the rod "falling into" M? Does it penetrate it? Does it fall over? If the rod remains vertical then what is the significance of its being a rod, not a point mass?
You write that M is attached (I assume you mean attached, not attacked) with 'string' k, but you show a spring under it, not a string.

 

[gneill]

Another question:

If the rod is falling (tipping if it's initially in contact with M), is there friction between the rod and mass M? Does the point of contact slide along the surface of mass M?

I can imagine a scenario where the mass rises as it rotates the rod and increases its angular rotation rate.

 

[TSny]

Looking at the attempt at a solution in the OP, maybe "rod" should be "chain".

 

[haruspex]

Looking at the attempt at a solution in the OP, maybe "rod" should be "chain".

Yes, that does seem like the interpretation, but that makes it quite a nasty problem. Conservation of energy cannot be used because of the way the chain coalesces into M; conservation of momentum cannot be used because of the way the normal reaction from the ground varies.
I get an equation of the form $At^2+Bs+C\ddot s+D(s\ddot s+{\dot s}^2)=0$.

 

[TSny]

Yes, that does seem like the interpretation, but that makes it quite a nasty problem. Conservation of energy cannot be used because of the way the chain coalesces into M; conservation of momentum cannot be used because of the way the normal reaction from the ground varies.

I agree.

I get an equation of the form $At^2+Bs+C\ddot s+D(s\ddot s+{\dot s}^2)=0$.

I get an equation of the same form as you except I get two additional terms involving time explicitly:

$At^2+Bs+C\ddot s+D(s\ddot s+{\dot s}^2) +Et \dot s + F t^2 \ddot s = 0$

I could be doing something wrong. Anyway, it's too complicated to deal with.

If m << M, then you could keep only terms in the differential equation that are of first order in certain small quantities. I think you then get an equation with just your first three terms: $At^2+Bs+C\ddot s = 0$, which we can solve. But of course there's no mention of any approximation in the statement of the problem.

 

[Hamal_Arietis]

Sorry about my words mistake, it is a rope :(( All collisions are inelastic collisions. I think we need use some approximations. If we seems m<<M
+ Firstly, pretermit the change coordinates of M so we can use $v=\frac{1}{2}gt^2$ of dm
+ Secondly, pretermit elastic force because the change coordinates very small.
That is the way of my thinking. Maybe it 's wrong
Thanks for all helping.

 

[Hamal_Arietis]

I get an equation of the form $At^2+Bs+C\ddot s+D(s\ddot s+{\dot s}^2)=0$.

How you get this equation?

 

[Hamal_Arietis]

Use this approximate I make solution:
Firstly, we find the force when $dm_i$ impact M, this force include:
+ The gravity of falled part.
+ The chage of momentum of dm
So we have:
$$F_i=F_1+F_2=\frac{m}{L}yg+v\frac{dm}{dt}=\frac{3m}{L}yg=\frac{3m}{2L}g^2t^2$$
The acceleration at time t:
$$a=\frac{F_i}{M+\dfrac{m}Ly}=\frac{\dfrac{3m}{2L}g^2t^2}{M+\dfrac{m}{2L}gt^2}=\frac{du}{dt}$$
$$\Leftrightarrow \frac{u}{g}= \int^{\sqrt{\dfrac{2L}{g}}}_0 \frac{3t^2}{\dfrac{2ML}{mg}+t^2}dt$$
Let $a=\sqrt{\dfrac{2ML}{mg}}$
$$\Leftrightarrow \frac{u}{g} =3\int^{\sqrt{\dfrac{2L}{g}}}_0 dt-3a^2\int^{\sqrt{\dfrac{2L}{g}}}_0 \frac{dt}{a^2+t^2}$$
$$ \Leftrightarrow \frac{u}{g} =3\sqrt{\dfrac{2L}{g}}-3a\arctan{\frac{t}{a}}$$
$$\Leftrightarrow \frac{u}{g}=3\sqrt{\dfrac{2L}{g}}-3\sqrt{\dfrac{2ML}{mg}}\arctan{\sqrt{\frac{m}{M}}}$$

Is right?

 

[haruspex]

How you get this equation?

Similarly to the way you got yours, but you made a few mistakes.
First, you mentioned gravity but it does not appear in your equation.
Secondly, there is a bit of a trap when working with changing masses. dp/dt=mdv/dt+vdm/dt is not really valid. That would apply if a mass could magically change without any inflow or outflow of material. In the present case, the mass change comes from inflow, so it comes with its own momentum. Let's break this down a bit. M is falling at speed u, the chain falling with relative speed v. In time dt, extra mass vρdt arrives at speed u+v, adding vρdt(u+v) to the total momentum. Ignoring gravity and the spring, that would take the accumulated mass to speed u+du:
vρdt(u+v)+mu=(m+ρdt)(u+du).
Discarding second order small terms, v²ρdt=mdu. There is no udm term.

Once you have the right equation, think about the speed of the top of the chain at time t. That should allow you to write the relationship between u and v. Similarly, if you let s be the length of the chain that has already settled on the mass (this gave me the simplest formulation) you can relate that to the distance the spring has been compressed.

 

[Hamal_Arietis]

First, you mentioned gravity but it does not appear in your equation.

No, At equilibrium position, $k\Delta l=Mg$
So if M move, the force total of gravity and elastic force is F=ky because I chose origin at equilibrium position.

 

[Hamal_Arietis]

Once you have the right equation, think about the speed of the top of the chain at time t. That should allow you to write the relationship between u and v. Similarly, if you let s be the length of the chain that has already settled on the mass (this gave me the simplest formulation) you can relate that to the distance the spring has been compressed.

It means:
$ky=\frac{m}{L}s$?
If this equation, I think that process will occur very slow. So finally the velocity of system is zero ?

 

[haruspex]

No, At equilibrium position, kΔl=MgkΔl=Mgk\Delta l=Mg
So if M move, the force total of gravity and elastic force is F=ky because

Ok.

It means:
$ky=\frac{m}{L}s$ ?

No. You first have to figure out where the top of the chain is at time t.

 

[Hamal_Arietis]

At time t, the coordinates of the top of the chain is $y=L-\frac{1}{2}gt^2$ so $v=-gt$ ? but how find distance the spring?

 

[haruspex]

v=-gt ?

That depends whether you are defining v as the speed of the chain or its speed relative to M.

how find distance the spring

You know the position of the top of the chain at time t, and the original length, and how much chain (s) is collapsed onto M. That tells you the height M has fallen.

 

[Hamal_Arietis]

I use in room reference frame. Let $y_0$ is the coordinates of the top of the chain so $\ddot{y_0}=-g$
We have (because y<0) : $$-y=L-s-y_0\Rightarrow \dot{s}+\dot{y_0}=\dot{y}\Rightarrow -g+\ddot{s}=\ddot{y}$$
And $$F=\frac{dp}{dt}$$
$$\Leftrightarrow k(\frac{1}{2}gt^2-s)-\frac{m}{L} 2g(L-y_0)=M(\ddot{s}+g)$$
$$\Leftrightarrow (\frac{k}{2}g-g^2\frac{m}{L})t^2-ks-M\ddot{s}+(2m+M)g=0$$
This is different with your equation but this is the same form with:

I think you then get an equation with just your first three terms: $At^2+Bs+C\ddot s = 0$, which we can solve. But of course there's no mention of any approximation in the statement of the problem.

 

[Hamal_Arietis]

Use this approximate I make solution:
Firstly, we find the force when $dm_i$ impact M, this force include:
+ The gravity of falled part.
+ The chage of momentum of dm
So we have:
$$F_i=F_1+F_2=\frac{m}{L}yg+v\frac{dm}{dt}=\frac{3m}{L}yg=\frac{3m}{2L}g^2t^2$$
The acceleration at time t:
$$a=\frac{F_i}{M+\dfrac{m}Ly}=\frac{\dfrac{3m}{2L}g^2t^2}{M+\dfrac{m}{2L}gt^2}=\frac{du}{dt}$$
$$\Leftrightarrow \frac{u}{g}= \int^{\sqrt{\dfrac{2L}{g}}}_0 \frac{3t^2}{\dfrac{2ML}{mg}+t^2}dt$$
Let $a=\sqrt{\dfrac{2ML}{mg}}$
$$\Leftrightarrow \frac{u}{g} =3\int^{\sqrt{\dfrac{2L}{g}}}_0 dt-3a^2\int^{\sqrt{\dfrac{2L}{g}}}_0 \frac{dt}{a^2+t^2}$$
$$ \Leftrightarrow \frac{u}{g} =3\sqrt{\dfrac{2L}{g}}-3a\arctan{\frac{t}{a}}$$
$$\Leftrightarrow \frac{u}{g}=3\sqrt{\dfrac{2L}{g}}-3\sqrt{\dfrac{2ML}{mg}}\arctan{\sqrt{\frac{m}{M}}}$$

Is right?

How about that ?

 

[TSny]

if you let s be the length of the chain that has already settled on the mass (this gave me the simplest formulation) ...

OK, I see how you are defining $s$. I was thinking $s$ was the displacement of M. When I change variables to your $s$, my extra terms with $t$ cancel out. That's nice!

But I do get an additional constant term. Shouldn't $\ddot s$ be nonzero at $t = 0$?
[EDIT: But then I could just redefine $s$ by a constant shift and get rid of the additional constant term. So, I finally agree with your equation!]

 

[Hamal_Arietis]

How to slove this problem?

 

[haruspex]

How about that ?

I'm not ignoring you, just struggling to follow your algebra. Some of your variable definitions are unclear. What is y₀ exactly? Which way is it measured and from where? The 0 suffix would normally mean initial position, but you seem to be using it as a variable.
How are you using the m<<M approximation? What term are you omitting on that basis?
Where does the myg/L term come from? Shouldn't it be msg/L, s being the collapsed length of the chain? I thought y was the displacement of M.
You are defining v as the absolute velocity of the chain, right? If so, the change in velocity of dm is not v.
I have no idea how you get 3myg/L.

 

[Hamal_Arietis]

Sorry because some my variable definitions are unclear.
Use some approximations: m<<M and the falling-time's small, so I seem M doesn't move on process, just moves after the process finishes. The 3myg/L is the result I finded in book " A guild to physics problems". This is the solution:

 

[haruspex]

Sorry because some my variable definitions are unclear.
Use some approximations: m<<M and the falling-time's small, so I seem M doesn't move on process, just moves after the process finishes. The 3myg/L is the result I finded in book " A guild to physics problems". This is the solution:

Ok, that's different from what you posted originally.
The scale spring is assumed to be very stiff, so the vertical movement of the scale pan can be ignored.
The mass of the scale pan becomes irrelevant. M here is the mass of the chain.