What is the smallest value of angular displacement of the raft?

[Lotto]

Homework Statement

There is a raft in the middle of the river. The mass of the raft is negligible, and it carries a crane on board. The crane moves boxes of building material of mass m from one river bank to another. In one cycle, the crane loads material at one side of the river, rotates to the other river bank, unloads the material there, and rotates back. Calculate the smallest value of angular displacement of the raft from its original position during one cycle. Approximate the crane by a homogenous cylinder of mass Mc and radius r, and a rotating jib in the shape of a slim rod of length kr. Assume that the velocity of the river and the „friction“ between the raft and the water are negligible.

Relevant Equations

torque1+torque2=0

What is meant by "the smallest value of angular displacement of the raft from its original position during one cycle"? I understand that I am supposed to solve this problem using torques of the crane and and of the boxes, but I am totally confused by that "smallest angular displacement". If it was the biggest angular displacement, I would understand it, but this is just strange. What does it mean?

I suppose I should use an equation that torque1+torque2=0, then the crane is stable. Torque1= Mga (a = distance of the gravity force Mg from the rotation axis ). Torque2=-mgb (b= distance of the gravity force mg from the rotation axis). But what am I to calculate anyway? I only need to know what that weird angle is.

 

[Lnewqban]

Welcome @Lotto !
Is there any diagram that is shown with this problem?
I can’t imagine what is holding the crane, so it can rotate 180°.
The massless raft could not do it, I believe.

 

[Lotto]

Welcome @Lotto !
Is there any diagram that is shown with this problem?
I can’t imagine what is holding the crane, so it can rotate 180°.
The massless raft could not do it, I believe.

I am sorry but this is the whole problem, I do not have any other information or pictures. I am disappointed as well.

 

[berkeman]

I am sorry but this is the whole problem, I do not have any other information or pictures.

What is the source of the problem? Is it from a university class that you are taking (if so, ask the professor or TA for clarification)? Or did you get it from some online problem set that you want to do on your own?

 

[bob012345]

I suppose I should use an equation that torque1+torque2=0, then the crane is stable. Torque1= Mga (a = distance of the gravity force Mg from the rotation axis ). Torque2=-mgb (b= distance of the gravity force mg from the rotation axis). But what am I to calculate anyway? I only need to know what that weird angle is.

Make some assumption as to what is actually moving and in what direction as a starting point. Draw a picture to get your bearings.

 

[berkeman]

as to what is actually moving

Yeah, I think that's got to be the key. It sounds like the crane "cylinder" is meant not to (counter-)rotate much, and only the thin "rod" is what rotates with the mass...

 

[Steve4Physics]

My interpretation is this...

I think ‘crane’ is a misnomer. It isn’t meant to do any vertical lifting or lowering. It can be thought of as a simple horizontal rod: one end is attached to the centre of the top of a cylinder (axis of cylinder vertical); the rod is only able to rotate in a horizontal plane.

A mass is attached to the free end of the rod.
The rod is rotated 180º (in a horizontal plane) carrying the mass to the other bank.
The mass is then detached and the rod is rotated -180º (in a horizontal plane) to return it to its starting position.

The cylinder is fixed to the raft and they rotate together. As the rod rotates clockwise the cylinder must rotate anticlockwise so that the system's angular momentum remains zero.

 

[TSny]

Here's an interpretation that makes for a nice problem.

Assume that the length of the jib is variable via the parameter$k$. If the jib is longer than half the width of the river, then the jib will not need to rotate through 180 degrees to move the load from one side of the river to the other side. I believe you can work out the amount of counter-rotation of the cylinder as a function of $k$ and the ratios $m/M_c$ and $r/W$, where $W$ is the width of the river. There will be a value of $k$ that minimizes the counter-rotation of the cylinder.

Of course, this interpretation might not be what was intended.

 

[kuruman]

The cylinder is fixed to the raft and they rotate together. As the rod rotates clockwise the cylinder must rotate anticlockwise so that the system's angular momentum remains zero.

Yes. When the mass is deposited on the opposite shore and the rod rotates to pick up the next mass, the raft does not return to its original orientation. Because the cylinder is fixed on the raft, the two rotate together during the "go" trip but not during the return trip because there is no load on the massless rod. I think the problem is asking for the incremental angle by which the raft + cylinder assembly rotates at the end of each "go" trip.

 

[haruspex]

the raft does not return to its original orientation

True, but if that becomes an issue it could counter that by occasionally going the long way around.
Unfortunately, neither will it return to its original location, and going the other way doesn't help. It will eventually reach the 'source' bank.

Anyway, I support @TSny's reading in post #8, and suggest the easiest approach (ignoring the cross-river creep) is to find the equation relating the two angles (sweep of arm relative to river, $\beta$, angular displacement of raft, $\alpha$) and set $\frac{d\alpha}{d\beta}=0$.
I get $\beta\sin(\beta)=1-\cos(\beta)$.

 

[Lotto]

I think I understand the task now. When the jib rotates, the radius of a circle it circumscribes is $kr$ and the angel it circumscribes is $\alpha_1=\pi$. So the angular momentum $L_1$ of the box dependent on time is $L_1=\frac {m\alpha_1 k^2 r^2}{t}$. The moment of inertia of the crane is $\frac 12 Mr^2$, so its angular momentum is $L_2=\frac{Mr^2\alpha_2}{2t}$ ($\alpha_2$ is the angel we want to calculate). Because of the conservation of angular momentum, $L_1=L_2$, so $\alpha_2=\frac{2m\pi k^2}{M}$. This is the smallest angel because the angular acceleration is zero. Is this solution right?

​

 

[Lnewqban]

I think I understand the task now. When the jib rotates, the radius of a circle it circumscribes is $kr$ and the angel it circumscribes is $\alpha_1=\pi$. So the angular momentum $L_1$ of the box dependent on time is $L_1=\frac {m\alpha_1 k^2 r^2}{t}$. The moment of inertia of the crane is $\frac 12 Mr^2$, so its angular momentum is $L_2=\frac{Mr^2\alpha_2}{2t}$ ($\alpha_2$ is the angel we want to calculate). Because of the conservation of angular momentum, $L_1=L_2$, so $\alpha_2=\frac{2m\pi k^2}{M}$. This is the smallest angel because the angular acceleration is zero. Is this solution right?

​

Then, the relevant equations shown in the OP do not apply, but those for angular momentum.
@haruspex , should not be a simultaneous translation of the raft as well?

 

[Lotto]

Then, the relevant equations shown in the OP do not apply, but those for angular momentum.
@haruspex , should not be a simultaneous translation of the raft as well?

Yes, this problem seems not to be about torques, that I thought for the first time. Is my solution right?

 

[haruspex]

When the jib rotates, the radius of a circle it circumscribes is $kr$ and the angel it circumscribes is $\alpha_1=\pi$.

No, it would only have rotate that far relative to the line of the river if the jib only just reaches each bank. (Of course, it will have to rotate more relative to the raft since the raft is rotating the other way.)
The point of the question is that increasing k reduces $\alpha_1$, but it also increases $\alpha_2$ relative to $\alpha_1$. So there can be an ideal $\alpha_1$ which minimises $\alpha_2$.

 

[haruspex]

Yes, this problem seems not to be about torques, that I thought for the first time. Is my solution right?

It can be done using torques because the torque equations lead to conservation of angular momentum. But one could argue it is not about that either because time is irrelevant.
In linear motion, conservation of momentum leads to the fact that the mass centre ofban isolated system does not move. Something analogous applies here to angular displacement. The equation is obtained by integrating the conservation of angular momentum equation wrt time.

should not be a simultaneous translation of the raft as well?

Yes, as I noted in post #10, but I doubt the question is meant to be that hard and we don’t know the width of the river.

 

[TSny]

(sweep of arm relative to river, $\beta$, angular displacement of raft, $\alpha$) and set $\frac{d\alpha}{d\beta}=0$.
I get $\beta\sin(\beta)=1-\cos(\beta)$.

I get the same condition, which can be written in other ways such as $\beta = \tan (\frac{\beta}{2})$.

Unfortunately, neither will it return to its original location, and going the other way doesn't help. It will eventually reach the 'source' bank.

@haruspex , should not be a simultaneous translation of the raft as well?

I completely overlooked this complication! I'm less certain than ever about the intended interpretation of the problem.

 

[kuruman]

Then, the relevant equations shown in the OP do not apply, but those for angular momentum.
@haruspex , should not be a simultaneous translation of the raft as well?

I have come to revise my thinking about this problem. When one talks about angular momentum, one must mention the reference point or axis about which this angular momentum is to be considered. Here, I think the appropriate point is the CoM of the cylinder - load system. The two masses orbit their CoM. I don't think that there is spin angular momentum of the cylinder because I cannot see what torque can change it. I am now considering that the angular displacement sought is that of the cylinder on the circle of its orbit around the CoM for a fixed length of the rod.

 

[Steve4Physics]

Because of the conservation of angular momentum, $L_1=L_2$

You mean $L_1 = - L_2$. If (for example) the rod rotates clockwise then the cylinder must rotate anticlockwise. Or you could write $|L_1|=|L_2|$.

Edit. Also, you may want to check the meanings of the two similar words 'angle' and 'angel'!

 

[TSny]

I have come to revise my thinking about this problem. When one talks about angular momentum, one must mention the reference point or axis about which this angular momentum is to be considered. Here, I think the appropriate point is the CoM of the cylinder - load system. The two masses orbit their CoM. I don't think that there is spin angular momentum of the cylinder because I cannot see what torque can change it. I am now considering that the angular displacement sought is that of the cylinder on the circle of its orbit around the CoM for a fixed length of the rod.

Interesting. For this interpretation, would it be necessary to state that the crane has the shape of a cylinder with a radius r?

Here's another option: assume that the raft is anchored in the middle of the river such that it can't translate through the water but it can freely rotate in the water.

Anyone's guess is at least as good as mine.

 

[Lotto]

Now I understand that I want an ideal angle $\alpha_1$ so that the product of $\alpha_1$ and $k^2$ is as small as it can be. But how to calculate it? Is it possible to solve it without calculus? Couldn't be the angle 90°?

 

[haruspex]

I don't think that there is spin angular momentum of the cylinder because I cannot see what torque can change it.

It's the rotation of the cylinder on its own axis that provides the torque forbtge rotation of the raft+boom+load system about its CoM. A motor on the raft will act to swing the boom around relative to the raft. This will necessarily rotate the raft in the opposite direction relative to the banks. I think it is clear that this last is the angle to be minimised. In principle, the raft could perform entire rotations for one sweep of the boom.

If the raft were mounted on an axle fixed to the bed of the river, that is all that would happen. As there is no such axle, yes, the raft+boom system will rotate about its CoM and this is what will be exactly balanced by the raft's rotation on its own axis.
It's the rotation about the CoM which leads to the displacement of the raft. On the return, the CoM is the centre of the raft. Indeed, the raft does not move at all in that phase.
This is what led me to ignore the fact that the boom rotation is about the CoM, but I see now that there is a compromise. We can take that into account, but just overlook that it leaves the raft displaced. I'll redo my analysis.

Edit: (I thought I added this yesterday but it has disappeared.)
Changing the pivot point to be the CoM produces the same result for $\beta$ as in posts #10 and #16. It does affect the answer for $\alpha$; see post #33.

 

[haruspex]

Now I understand that I want an ideal angle $\alpha_1$ so that the product of $\alpha_1$ and $k^2$ is as small as it can be. But how to calculate it? Is it possible to solve it without calculus? Couldn't be the angle 90°?

To avoid subscripts I will call the angles α (the angle to be minimised) and β.
If the raft counter-rotates α on its own axis, what is the rotation angle β of the raft+boom+load system?
What width of river will be traversed by the load?

 

[bob012345]

It's the rotation of the cylinder on its own axis that provides the torque forbtge rotation of the raft+boom+load system about its CoM. A motor on the raft will act to swing the boom around relative to the raft. This will necessarily rotate the raft in the opposite direction relative to the banks. I think it is clear that this last is the angle to be minimised. In principle, the raft could perform entire rotations for one sweep of the boom.

If the raft were mounted on an axle fixed to the bed of the river, that is all that would happen. As there is no such axle, yes, the raft+boom system will rotate about its CoM and this is what will be exactly balanced by the raft's rotation on its own axis.
It's the rotation about the CoM which leads to the displacement of the raft. On the return, the CoM is the centre of the raft. Indeed, the raft does not move at all in that phase.
This is what led me to ignore the fact that the boom rotation is about the CoM, but I see now that there is a compromise. We can take that into account, but just overlook that it leaves the raft displaced. I'll redo my analysis.

Suppose the crane cylinder was just as wide as the river? In other words W=2r in the nice rendition by @TSny in post #8?

 

[haruspex]

Suppose the crane cylinder was just as wide as the river? In other words W=2r in the nice rendition by @TSny in post #8?

View attachment 316897

What are you suggesting? A way of preventing the linear displacement? But in that case, why not just anchor it to the bank so that it can't rotate either?

 

[Lnewqban]

I have come to revise my thinking about this problem. When one talks about angular momentum, one must mention the reference point or axis about which this angular momentum is to be considered. Here, I think the appropriate point is the CoM of the cylinder - load system. The two masses orbit their CoM. I don't think that there is spin angular momentum of the cylinder because I cannot see what torque can change it. I am now considering that the angular displacement sought is that of the cylinder on the circle of its orbit around the CoM for a fixed length of the rod.

There is also the problem of changing angular momentum with each load-unload cycle.

First loaded boom: Crane’s body-boom-load rotation happens about a common (loaded) COM; therefore, the crane’s body rotates mirroring the load, but describing a smaller circle.

First unloaded boom: Crane’s body-boom rotation happens about a new common (unloaded) COM’, this time located closer to the body, which describes a smaller circle in reverse.

Second loaded boom: As the second load is fixed respect to the ground, and next to where the previous load was located, but the crane’s body has relocated itself, the lenght of the boom must be adjusted for the second cycle, thus modifying the angular momentum and subsequent circles.

Thank you, @kuruman , @TSny and @haruspex !

 

[Steve4Physics]

Following this thread with interest and can’t stop myself sharing a few thoughts…

FWIW, I suspect that the question is simply a (very) poorly posed introductory-level problem with two key unstated assumptions:

1. Since the width of the river (or more correctly the distance from pick-up to drop-off points) is unspecified, we are meant to assume it is 2kR. The rod’s angles of rotation (relative to the ground) can then be taken as 180º (transferring) and -180º (returning).

2. The linear displacement of the crane’s centre of mass during transfers can be ignored.

With these assumptions @Lotto’s answer in Post #11 would be the required one. (Though I fully appreciate the more rigorous analyses in the various posts.)

The minimum angle of rotation of the cylinder is achieved providing each mass (m) is released when the rod has completely stopped rotating – else the mass (m) will still have some angular momentum when released – this would increase the necessary angle of rotation of the cylinder. That may answer the question originally asked by @Lotto about the use of the word 'minimum'.

Intuitively I don’t see how @haruspex's $\beta\sin(\beta)=1-\cos(\beta)$ can be correct because its solutions ($\beta = 0, 2.331rad, ...)$ are discrete. Surely $\beta$ must be a (continuous) function of [some or all of] the masses R and k? Or have I misunderstood something?

But, in the words of @TSny, anyone's guess is at least as good as mine.

To @Lotto: If and when you get the official answer, I’m sure we would be interested to hear it. (And may the angles be with you!)

Edits: Minor changes.

 

[bob012345]

What are you suggesting? A way of preventing the linear displacement? But in that case, why not just anchor it to the bank so that it can't rotate either?

Since the original wording asks to calculate the minimum angular displacement I think it assumes no linear displacement. If we assume $m<<M_c$ we can ignore the CM translation and assume everything just rotates around the axis of the raft/crane. One way of minimizing the angle of the jib is to shrink the river to $2r$ for a given k.

 

[bob012345]

Now I understand that I want an ideal angle $\alpha_1$ so that the product of $\alpha_1$ and $k^2$ is as small as it can be. But how to calculate it? Is it possible to solve it without calculus? Couldn't be the angle 90°?

Does your course use Calculus? The ideal angle would not be 90° unfortunately. The question is whether the teacher/problem expects you to use $m,k,M_c$ as fixed constants or whether they (or at least $k$) can be optimized. If they are fixed constants and if only rotation is involved then I think you were on the right track earlier.

 

[TSny]

Following this thread with interest and can’t stop myself sharing a few thoughts…

FWIW, I suspect that the question is simply a (very) poorly posed introductory-level problem with two key unstated assumptions:

1. Since the width of the river (or more correctly the distance from pick-up to drop-off points) is unspecified, we are meant to assume it is 2kR. The rod’s angles of rotation (relative to the ground) can then be taken as 180º (transferring) and -180º (returning).

2. The linear displacement of the crane’s centre of mass during transfers can be ignored.

With these assumptions @Lotto’s answer in Post #11 would be the required one. (Though I fully appreciate the more rigorous analyses in the various posts.)

Yes, I suspect this is the intended interpretation.

 

[haruspex]

Since the width of the river (or more correctly the distance from pick-up to drop-off points) is unspecified, we are meant to assume it is 2kR.

I cannot agree.
We are asked to find the minimum angle, implying there is some variable which needs to be tuned to achieve that. Of the variables christened in post #1, only k seems like something the operator could tune.
It turns out that with this interpretation the width of the river, and r, M and m, are all irrelevant (Edit: for finding $\beta$ - thanks TSny), so the fact that we are not told the width of the river does not cast doubt on it. Moreover, it creates quite an interesting problem to solve.
Edit: Question is, did the author forget to specify a variable for the river width, or by mistake asked for the value of the angle minimised instead of the angle the boom sweeps for that minimum value?

(And, of course, if the width of the river is 2kr then the task becomes impossible. When the boom has swung 180° the raft will have shifted a bit the other way, leaving the load short of the target shore!)

 

[TSny]

We are asked to find the minimum angle, implying there is some variable which needs to be tuned to achieve that. Of the variables christened in post #1, only k seems like something the operator could tune.
It turns out that with this interpretation the width of the river, and r, M and m, are all irrelevant

The value of $\beta$ that minimizes $\alpha$ doesn't depend on the width of the river, nor does it depend on r, M, m. But doesn't the minimum value of $\alpha$ itself depend on the width of the river as well as r, M, and m? This is one thing that bothered me about the problem statement, which doesn't mention the width of the river as being relevant.

 

[Tom.G]

Relevant Equations:: torque1+torque2=0

What is meant by "the smallest value of angular displacement of the raft from its original position during one cycle"?

Since that is the Only question asked, the planar movement around the COM of crane+load is not needed.

In the limit of a miniscle (0) load mass, the answer would be _____.

In the limit of an infinite load mass, the answer would be ______.

If the assumption of a massless boom is valid, I think that reduces to a function of the mass ratios, m and Mc, and the variable k.

 

[haruspex]

The value of $\beta$ that minimizes $\alpha$ doesn't depend on the width of the river, nor does it depend on r, M, m. But doesn't the minimum value of $\alpha$ itself depend on the width of the river as well as r, M, and m? This is one thing that bothered me about the problem statement, which doesn't mention the width of the river as being relevant.

Ah, good point.
$\alpha=\frac{W^2m}{r^2(M+m)\sin(\beta)}$.
But I still do not see any more reasonable interpretation.

 

[haruspex]

Since that is the Only question asked, the planar movement around the COM of crane+load is not needed.

Except that the linear movement means the answer changes with each cycle, so it should ask for the first cycle, not "one" cycle,

In the limit of a miniscle (0) load mass, the answer would be

Zero

In the limit of an infinite load mass, the answer would be

Infinity
How does that help?

 

[Tom.G]

How does that help?

By giving a hint of the form of the equation needed to answer the problem. (I hoped)

 

[Steve4Physics]

It may not be the original intention of the question, but assume that the question
- specifies the river’s width (or more correctly the distance between pick-up and drop-off points) as $W$;
- specifies lateral drift of the raft is negligible;
- specifies that $k$ is variable;
- requires us to find the minimum value of the cylinder’s angular displacement ($\alpha$) resulting from varying $k$. I.e. we need to minimise $\alpha (k)$.

$\alpha$= angular displacement of cylinder (mass M) relative to the ground.
$\beta$ = angle swept by rod relative to ground.

Conserving angular momentum gives:

$\frac 12 Mr^2 α + m(kr)^2 β = 0$

$α = -\frac {2mk^2} M β$

The geometry gives:

$\sin ({\frac β2}) = \frac {W/2}{kr} = \frac W {2kr}$

$β = 2\sin^{-1} (\frac W {2kr})$

We can adopt a sign-convention which makes $β$ negative (in order to make $\alpha$ positive, for neatness); so use:

$β = -2\sin^{-1} (\frac W {2kr})$

Combining the equations for $α$ and $β$:

$α = \frac {4mk^2} M \sin^{-1} (\frac W {2kr})$

Anyone so inlined can then determine the minimum value of α(k) by starting with $\frac {dα}{dk} = 0$.

 

[bob012345]

$α = \frac {4mk^2} M \sin^{-1} (\frac W {2kr})$

Anyone so inlined can then determine the minimum value of α(k) by starting with $\frac {dα}{dk} = 0$.

This gives

$$\frac {dα}{dk}=\frac{8mk}{M_c} sin^{-1}\left(\frac {W} {2kr}\right) +\frac {4mk^2}{ M_c} \left(\frac{-W}{rk^2\sqrt{4- \frac{W^2}{r^2k^2}}}\right)=0$$ or

$$2k sin^{-1}\left(\frac {W} {2kr}\right) + \left(\frac{-W}{r \sqrt{4- \frac{W^2}{r^2k^2}}}\right)=0$$ or

$$2 \sqrt{1- \left(\frac{W }{2rk }\right)^2} sin^{-1}\left(\frac {W} {2kr}\right) = \left(\frac{W}{2kr }\right) $$
letting $z=\large\frac{W }{2rk }$ we have;

$$ z=2 sin^{-1}(z)\sqrt{1- z^2}$$ then the root is $z=\frac{W }{2rk }=0.9190$ and $1/z =1.088$

thus $k=1.088 \large\frac{W }{2r }$ so $k$ will be minimum when $W=2r$ (since $W≥2r$) then

$$α_{min} = \frac {4m(1.088)^2} M \sin^{-1} (0.9190) =5.52 \frac {m}{M} rads$$

This is only a little smaller than @Lotto 's original answer $\frac{2mk^2 \pi}{M_c}$ when $k=1$ giving $6.28 \frac {m}{M} rads$

https://www.wolframalpha.com/input?i=solve+sin^-1+(x)+=x/(2sqrt(1-x^2))

 

[haruspex]

Anyone so inlined can then determine the minimum value of α(k)

as was done in posts #10, #16 and #33.

 

[Steve4Physics]

as was done in posts #10, #16 and #33.

Calculating the minimum value of $\alpha (k)$ wasn't actually done in those posts.

There has been some confusion about what the precise question is – as evidenced in the 30+ posts.

In Post #36 I thought it would be helpful to present a (hopefully) unambiguous version of the question and suggest an appropriate method of solving it. The method directly gives $k$ and then $\alpha_{min}$ (the key objective of the question).

I’m not clear if the equations in Posts #10, #16 and #33 apply to the specific question formulated in Post #36 - but they may.

Edit - minor changes.

 

[haruspex]

Calculating the minimum value of α(k) wasn't actually done in those posts.

The result for $\alpha(k)$ is given in post ##33, but see below.

not clear if the equations in Posts #10, #16 and #33 apply to the specific question formulated in Post #36

Those posts are all predicated on the interpretation you lay out in post #36, except that post #33 does allow that the pivot for the first half of the cycle will be about the CoM of the system, not about the cylinder's centre. (Note that this does not affect $\beta$.)
The difference this makes to the answer is M+m instead of M.

Here's my working for the post #36 view:
$2kr\sin(\beta/2)=W$
$mk^2r^2\beta=\frac 12Mr^2\alpha$
Whence
$2M\alpha r^2\sin^2(\beta/2)=mW^2\beta$
Since $\beta$ depends directly on k, we can treat it as the independent variable and differentiate wrt it. $\frac{d\alpha}{d\beta}=0$ gives
$2M\alpha r^2\sin(\beta/2)\cos(\beta/2)=mW^2$
Combining the differentiated and undifferentiated forms produces
$\beta=\tan(\frac 12\beta)$.

 

[Steve4Physics]

$\beta=\tan(\frac 12\beta)$.

Thanks. I understand what you are saying.

If we consider say $0 \lt \beta \lt 2\pi$ the only solution to $\beta=\tan(\frac 12\beta)$ is $\beta ≈ 2.331$rad $≈ 134^o$.

I know it comes out of the maths but I haven’t got an intuition for why, to minimise $\alpha$, there is only one value of $\beta$ and it is completely independent of $m, M, r$ and $W$. It bothers me. I’ll think about it!

 

[bob012345]

Since $\beta$ depends directly on k, we can treat it as the independent variable and differentiate wrt it. $\frac{d\alpha}{d\beta}=0$ gives

It seems to me doing that violates the original assumption that angular momentum is conserved does it not? $$\frac{d\alpha}{d\beta}= \frac{2mk^2}{M_c}≠0$$

 

[haruspex]

It seems to me doing that violates the original assumption that angular momentum is conserved does it not? $$\frac{d\alpha}{d\beta}= \frac{2mk^2}{M_c}≠0$$

no, k is a variable. By making $\beta$ the independent variable k becomes $k=k(\beta)$.

 

[bob012345]

Thanks. I understand what you are saying.

If we consider say $0 \lt \beta \lt 2\pi$ the only solution to $\beta=\tan(\frac 12\beta)$ is $\beta ≈ 2.331$rad $≈ 134^o$.

I know it comes out of the maths but I haven’t got an intuition for why, to minimise $\alpha$, there is only one value of $\beta$ and it is completely independent of $m, M, r$ and $W$. It bothers me. I’ll think about it!

Physically, $\beta$ depends on the ratio of $\frac{W}{2kr}$ which has a minimum. I get the same value for $\beta$ in post #37 where $z=\frac{W }{2rk }=0.9190$ thus $\beta = 2sin^{-1} (0.9190)= 2.331$ rad or ≈133.6°

But that is only part of minimizing $\alpha$. You still have $\alpha= \frac{2mk^2}{M_c}\beta$ and $k=1.088 \large\frac{W }{2r }$ thus $k_{min}= 1.088$

For a wider river, you get a larger $\alpha$ for the same minimum $\beta$ because the moment of inertia of the mass $m$ is larger.

 

[bob012345]

no, k is a variable. By making $\beta$ the independent variable k becomes $k=k(\beta)$.

Ok, I think I was doing the same thing implicitly in post #37...

 

[Lnewqban]

Edit: (I thought I added this yesterday but it has disappeared.)
Changing the pivot point to be the CoM produces the same result for $\beta$ as in posts #10 and #16. It does affect the answer for $\alpha$; see post #33.

Another reason to disregard the small translation of the raft:
From a lateral point of view, the common CoM of crane and load should be located well on the footprint of the raft, so the whole thing does not capsize.
That makes the mass of the crane bigger than the mass of the load, as much as the kr distance of the boom is bigger than the radius r of the raft.

 

[Lotto]

Another reason to disregard the small translation of the raft:
From a lateral point of view, the common CoM of crane and load should be located well on the footprint of the raft, so the whole thing does not capsize.
That makes the mass of the crane bigger than the mass of the load, as much as the kr distance of the boom is bigger than the radius r of the raft.

So you all claim that the angle beta is dependent on the width of the river? Isn't the width insignificant?

 

[Lnewqban]

So you all claim that the angle beta is dependent on the width of the river? Isn't the width insignificant?

The concept of that angle was introduced in post #10, based on proposal in post #8.
The width of the river represents the linear distance between the points of picking up and releasing the load, we all have guessed.

 

[bob012345]

So you all claim that the angle beta is dependent on the width of the river? Isn't the width insignificant?

As has been shown in several posts, $$\large β = 2\sin^{-1} (\frac {W} {2kr})$$

It is the ratio $\large(\frac{ W} {2kr}\large)$ that has a minimum.

 

[haruspex]

It is the ratio $\large(\frac{ W} {2kr}\large)$ that has a minimum.

Did you mean that? If W and r are fixed, that would be minimised by letting k tend to infinity.