Block slides from rest from the top of a fixed frictionless sphere

[leyyee]

can anyone help me to solve this question? please and thank you :D

A small block slides from rest from the top of a fixed frictionless sphere of
radius R.

Determine the vertical distance traveled by the block, x , where it loses
contact with the surface of the sphere and its speed at this point.

 

[Fellow]

let m be the mass of the block.
let y be the angular displacement of the block between its starting point and the point in which it loses contact with the sphere.

we can see that cos(y) = (R-x)/R, x = R[1 - cos(y)]

since there is conservation of mechanical energy, loss of gravitational potential energy will result in a gain in kinetic energy of the block, mgx = (1/2)mv²
hence, v² = 2gx = 2gR[1 - cos(y)]

next, as the block moves down the sphere, a portion of its weight must act radially into the sphere so as to provide the required centripetal force for the motion of the block.
hence, required centripetal force = mg*cos(y) = mv²/R
substituting the value of v² found previously into the equation, mg*cos(y) = 2mgR[1 - cos(y)]/ R
hence, rearranging the terms, we get cos(y) = 2/3

since x = R[1 - cos(y)], x = R(1 - 2/3) = R/3

since v² = 2gx = 2gR[1 - cos(y)], v² = 2gR/3, v = (2gr/3)^1/2

is it correct?

 

[leyyee]

YEAh. THANK YOU!