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

[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.