Why Does a Block Lose Contact with a Frictionless Sphere?

[MathematicalPhysicist]

a small block slides from rest from the top of a frictionless sphere of radius R, how far below the top x does it lose contact with the sphere? the sphere doesn't move.
the question is from kleppner's in troduction to mechanics page 196 problem 4.6. (my scanner doesn't work so i can't scan the picture).

anyway, x is the displacement from when the block was on top the shpere up until where it loses contact with the sphere.

what i got so far is:
i calculated the tan of the angle of velocity, i got that tan(a)=sqrt(2Rx-x^2)/x and i got by energies that v^2=2xg and i know that v_y/v_x=tg(a) and v^2=v_y^2+v_x^2
and that v_y^2=xg, but after all that i didnt get the answer in the book, which is R/3.
can someone help me here?
thanks in advance.

 

[radou]

Loss of contact implies that the normal force N equals 0. Since N = mgcosA - mv^2/R, we have cosA = v^2/(Rg). Since you already found the velocity v for a displacement x, your problem is solved.

 

[MathematicalPhysicist]

im having a problem here:
iv'e done the calclulation for v, v^2=2xg
and cos(a)=x/sqrt(2xR)
but then:
x/sqrt(2xR)=v^2/Rg=2xg/Rg=2x/R
x/2R=4x^2/R^2
Rx=8x^2
x=R/8
and according to my book the answer is R/3, where did i go wrong here?

 

[MathematicalPhysicist]

someone can help me here?
any help is appreciated.

 

[neutrino]

Find cos(a) in terms of x and R from what you've already found out[cos(a) = v^2/Rg] and then find what cos(a) is from the diagram (right triangle).Equate the two.

 

[MathematicalPhysicist]

a is the angle of the velocity of the block, i found that:
cos(a)=x/sqrt(2xR)
bus as i said i don't get in the end the right answer, which is R/3.

 

[neutrino]

angle of the velocity?

Take a look at radou's post. A is the angle bet. the vertical and the position at which the block falls off.

 

[MathematicalPhysicist]

so how should i calculate the angle, i thought that my calclualtion was correct i.e cos(a)=x/sqrt(2xR)
how should i calculate the angle?

 

[MathematicalPhysicist]

what did i do wrong here?

 

[neutrino]

$$\cos{a} = \frac{v^2}{Rg}$$

$$v^2 = 2gx$$ - conservation of energy

Also, $$\cos{a} = \frac{R-x}{R}$$ (from the right triangle)

Angle a is measured clockwise from the vertical two the point where the block loses contact.