Body sliding down an accelerating hemisphere

[chester20080]

So in the occasion where we don't rotate the system,the initial potential energy would be mg_effd, d would be what?

 

[TSny]

I probably should use a prime on θ to denote the angle from the y' axis in the rotated figure. So, in the rotated figure, the body starts at θ' = β, which corresponds to θ = 0 in the top figure. In the rotated figure, the body falls off for some θ' > β, which means it falls off for some θ > 0 in the upper figure.

 

[TSny]

So in the occasion where we don't rotate the system,the initial potential energy would be mg_effd, d would be what?

Draw a figure showing the initial position and final position of the block (where it leaves the surface). Draw a vector between these two positions. This is the displacement vector Δr of the body. d is the component of Δr in the direction of g_eff.

 

[chester20080]

I think I found an equation for the relative velocity,considering the rotated system as you advised me to.I have:
u_relative=\sqrt{}2R(g-gcosθ+Wsinθ). It yields Urelative=0 for θ=0.

 

[TSny]

I think I found an equation for the relative velocity,considering the rotated system as you advised me to.I have:
u_relative=\sqrt{}2(g-gcosθ+Wsinθ). It yields Urelative=0 for θ=0.

Except for a missing R that looks good!

 

[chester20080]

Yes,I forgot it,but I edited the post.

 

[chester20080]

And the equation for the angle I got:
sinθ(2-\sqrt{}2)-2cosθ+2=0 and for θ=17 degrees it yields 0.2587=0,which considering that 17 degrees is an approximation,is very good!

 

[chester20080]

I hope that is the result for this exercise.Thank you very very much for your time and your invaluable help,I appreciate it!

 

[TSny]

And the equation for the angle I got:
sinθ(2-\sqrt{}2)-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 g_eff rather than g. You should find g_eff cancels out. (But the "angle of tilt" β of g_eff does enter the solution.)

 

[chester20080]

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.

 

[TSny]

I find

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

$\beta$ is determined from g and W.

 

[chester20080]

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?

 

[TSny]

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.

 

[chester20080]

Ok,thank you for everything!