Conservation of Energy in circular motion

[haruspex]

As Θ increases, V increases since it is sinus function (in 0-90 degree quadrant).

Hmm.. I think I almost figured it out, right ? The only thing I miss is just the launch angle (Let's call this α )
http://www.sumoware.com/images/temp/xzihalbxxfsximxp.png
Could you please tell me what is the relation between α and Θ ?

As theta increases, the block is rising, so must be losing KE and slowing.
Further, the fraction of the speed which is vertical is also decreasing, so it can't be as sin theta.

 

[terryds]

As theta increases, the block is rising, so must be losing KE and slowing.
Further, the fraction of the speed which is vertical is also decreasing, so it can't be as sin theta.

So, it is cos theta ? Right ?

 

[haruspex]

So, it is cos theta ? Right ?

Yes v_y = v cos(theta). Now, what about the value of v as a function of theta?

 

[terryds]

Yes v_y = v cos(theta). Now, what about the value of v as a function of theta?

Do you mean the velocity as a function of theta when it's still on the track ?
E = E'
mg (2R) = mgh' + 1/2 mv^2
g (2R) = g(R+R sinΘ) + 1/2 v^2
2gR - gR - gR sinΘ = 1/2 v^2
gR (1-sin Θ) = 1/2 v^2
v = √(2 gR(1-sinΘ))

Is it right ?

 

[haruspex]

_y

Do you mean the velocity as a function of theta when it's still on the track ?
E = E'
mg (2R) = mgh' + 1/2 mv^2
g (2R) = g(R+R sinΘ) + 1/2 v^2
2gR - gR - gR sinΘ = 1/2 v^2
gR (1-sin Θ) = 1/2 v^2
v = √(2 gR(1-sinΘ))

Is it right ?

Right. So v_y is?

 

[terryds]

_y
Right. So v_y is?

v_y = v cos Θ
v_y = cos Θ √(2 gR(1-sinΘ))

The sin of angle θ is 2/3
So, the cos of angle Θ is √5 / 3
v_y = √5 / 3 √(2 gR(1-(2/3)))
v_y = √((10/27) gR)

Then,
V_y(t) ^2 = V_y(0) ^2 - 2 g Δy
0 = (10/27) gR - 2g Δy
Δy = 5/27 R

Then, the maximum height is R+(2/3)R+(5/27)R = (50/27) R
Right ?
Please tell me if I missed something

 

[haruspex]

v_y = v cos Θ
v_y = cos Θ √(2 gR(1-sinΘ))

The sin of angle θ is 2/3
So, the cos of angle Θ is √5 / 3
v_y = √5 / 3 √(2 gR(1-(2/3)))
v_y = √((10/27) gR)

Then,
V_y(t) ^2 = V_y(0) ^2 - 2 g Δy
0 = (10/27) gR - 2g Δy
Δy = 5/27 R

Then, the maximum height is R+(2/3)R+(5/27)R = (50/27) R
Right ?
Please tell me if I missed something

That's what I get.