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

[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.

 

[bob012345]

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

Not really. That value of ##\beta## is specific and was derived by two different methods (which I think are equivalent). The ratio to minimize ##\beta## is ##\large(\frac{ W} {2kr}\large)=0.9190## as shown in posts #37 and #41. This means'

$$\large k=1.088 \frac{W }{2r }$$

The way I read this is that for a given ##W,r## that gives the ##k## value for minimum angle ##\beta##. Then you get the minimum ##\alpha## given those constraints but the absolute minimum of angle ##\alpha## is found by playing with the ratio ##\frac{W }{2r }## the minimum being 1. Physically, to me that means if the crane is as wide as it can get, as wide as the river, its moment of inertia is largest possible thus the angle turned is the smallest possible.

 

[haruspex]

The ratio to minimize β

But we're not asked to minimise β; we are asked to minimise α.

 

[bob012345]

But we're not asked to minimise β; we are asked to minimise α.

True. Let me clarify, I meant this is the ##\beta## that makes ##\alpha## minimum. As you showed in post #40, it has a specific value ##\beta= tan(\frac{\beta}{2}) ≈133.6°##

 

[haruspex]

True. Let me clarify, I meant this is the ##\beta## that makes ##\alpha## minimum. As you showed in post #40, it has a specific value ##\beta= tan(\frac{\beta}{2}) ≈133.6°##

Right, which is why I queried what you wrote in post #49; it was not what you meant.

 

[bob012345]

Right, which is why I queried what you wrote in post #49; it was not what you meant.

Correct. I was trying to make a point about the importance of that ratio that defines ##\beta##. In my mind I summarize all the results above as follows;

By conservation and geometry we have

$$α = \frac {4mk^2} {M_c} \sin^{-1} \left(\frac {W} {2kr}\right)$$ letting ##\large z=\frac {W} {2kr}## we have;

$$α = \frac {4m} {M_c} \left(\frac{W}{2r}\right)^2 \frac{\sin^{-1}(z)}{z^2}$$ therefore ##\alpha## is minimized for given constraints with ##z=0.9190## giving the fixed ##\beta## above.