# Tipping a Tumbler

https://isaacphysics.org/questions/tipping_tumbler?board=848c14ef-ea98-465d-a6e9-d1fafb62a526&stage=further_a

For part c of this question, I have no idea how to get it into the form f= (...)^(2/3)
I've tried different methods but none require using f^(3/2 )so im quite confused.

Weirdly, inside the brackets my final answer is kinda close to the actual answer so I think im roughly on the right track?
I just cant see how you end up with an expression of f with some power, much less with f to the power of a fraction of 3/2.

Any pointers would be greatly appreciated.
Working (in advance sorry for the messy working);

The actual answer is at the very bottom and the answer I got is above it.
Pretty sure you have to integrate this because the fluid is a continuous object, https://isaacphysics.org/concepts/cp_centre_mass?stage=all#centre_mass_tipping has a drop down section on this "Continuous Objects and Non-Uniform density".
sorry, but tbh im still not really following.

I think the fluid has a constant density and hint 3 in part c tells you the COM of a right angled triangle is (a/3,b/3), so I dont really get how you would need to integrate - im probably missing something crucial or just misunderstanding completely

Besides even if you do integrate, I cant really see a way for it to be f^(3/2)=.... because you would need to start with f^(1/2)=...and then integrate to get f^(3/2)=..., and I cant really see how you would start of with f^(1/2)=...
Original post by mosaurlodon
sorry, but tbh im still not really following.
I think the fluid has a constant density and hint 3 in part c tells you the COM of a right angled triangle is (a/3,b/3), so I dont really get how you would need to integrate - im probably missing something crucial or just misunderstanding completely
Besides even if you do integrate, I cant really see a way for it to be f^(3/2)=.... because you would need to start with f^(1/2)=...and then integrate to get f^(3/2)=..., and I cant really see how you would start of with f^(1/2)=...

Not really following what youve done for the fluid part but Id agree that you shouldnt need to integrate, just use a bit of geometry.

The com for the fluld only must correspond to theta=45, so the angle you want lies between 45 and 66. For the fluid only part you have the com at (h/3, h/3tan) which seems to assume you have one side of length h? I would have thought youd need to introduce another varaible z for one side of the fluid prism and do something similar to what you have done?
(edited 3 weeks ago)
Yup, that was what I was missing.

I thought that would make the algebra a lot more complicated (it did) but chugging through, and getting the final equation felt amazing.

Thank you
Original post by mosaurlodon
Yup, that was what I was missing.
I thought that would make the algebra a lot more complicated (it did) but chugging through, and getting the final equation felt amazing.
Thank you

Sometimes its hard to spot the problem when its how you set it up. Couple of things that may help

Do some simple thought experiments for the key variables, so the com angle for the beaker only is worked out in part a ~66, and it should be obvious that the com for the water only would correspond to 45. When theta is 45, the water fills up the "v" on both sides precisely and theta < 45 would mean that the water would no longer be a triangular prism so thats ruled out in the question and means you can draw the water triangular prism "easily".

So the com for the composite lies between the two, and it will depend on how full it is (f). If f~0, then the water will correspond to a tiny prism at the point of rotation and theta ~66 (though you dont know its value). To work it out more precisely you need to get a relationship between the depth or the length of the prism on one side, then use tan (as you did) to get the other one and ....

Key things are a clear sketch (or two) "always" helps and think about simplifying the problem (or splitting it up or considering extreme values) to make sure youre approaching it in the right way.