Body sliding down an accelerating hemisphere

In summary, the conversation discusses the scenario of a frictionless hemisphere with a body on top of it. The hemisphere is given a horizontal constant acceleration, causing the body to slide down the accelerating hemisphere. The relative velocity of the body with respect to the hemisphere when it leaves the hemisphere is found to be ##\sqrt{R*(2g-W)\sqrt{W/g}}##, where ##R## is the radius of the hemisphere and ##W## is the acceleration given to the hemisphere. The angle ##\theta## made by the body when it leaves the hemisphere is also calculated to be ##\cos^{-1}(\sqrt{W/g})##. The conversation then discusses the principles
  • #36
Yes,I forgot it,but I edited the post.
 
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  • #37
And the equation for the angle I got:
sinθ(2-[itex]\sqrt{}2[/itex])-2cosθ+2=0 and for θ=17 degrees it yields 0.2587=0,which considering that 17 degrees is an approximation,is very good!
 
  • #38
I hope that is the result for this exercise.Thank you very very much for your time and your invaluable help,I appreciate it!
 
  • #39
chester20080 said:
And the equation for the angle I got:
sinθ(2-[itex]\sqrt{}2[/itex])-2cosθ+2=0

I don't think that's correct.

I find the algebra to be much simpler if I work with the rotated figure and derive an expression for the angle θ' (measured from the y' axis) at which the block falls off. It is then easy to convert this to the angle θ in the non-rotated figure. Also, I think it is easier to keep the expressions in terms of geff rather than g. You should find geff cancels out. (But the "angle of tilt" β of geff does enter the solution.)
 
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  • #40
Can you please give me the angle equation you found,in order to work on my equations and to compare what I will find with yours and tell you tomorrow what I will have found?Unfortunately here (in Greece) it is almost midnight and I am too tired to have a clear mind right now.
 
  • #41
I find

##\theta = \cos^{-1}(\frac{2}{3}\cos\beta) - \beta##

##\beta## is determined from g and W.
 
  • #42
Hey,I found 3cosθ-3sinθ-2=0 which is equivalent as I checked with your equation (just use the formula cos(θ+β)=cosθ*cosβ-sinθ*sinβ and cosβ=g/sqrt{g2+W2} and sinβ=W/sqrt{g2+W2}).And for approximately 17 degrees it yields -0.0027=0 which is excellent!My only doubt is that this angle equation for θ=0 gives 3=2,which is not valid.Is it something to worry about?
 
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  • #43
chester20080 said:
Hey,I found 3cosθ-3sinθ-2=0 which is equivalent
Good.

My only doubt is that this angle equation for θ=0 gives 3=2,which is not valid.Is it something to worry about?

This just shows that θ = 0 is not a solution of the equation 3cosθ-3sinθ-2=0.
So, the body does not leave the surface at θ = 0.
 
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  • #44
Ok,thank you for everything!
 
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