Question

I need a help frnd...

A particle at rest starts sliding down from the top of a large frictionless sphere of radius R. The sphere is fixed on the ground. Calculate the height from the ground at which the particle leaves the surface of the sphere.

##follow me, I will follow back..​

Answer 1

↪️$\boxed{\blue{you\: are\: connected\:to\: boAtstone}}$↩️

$\huge\boxed{ANSWER:-}$

Mg cos theta -N = mv²/R

➡️The point where particle leave the sphere, normal force must be zero

Mg cos theta = mv²/R

cos theta = v²/Rg................(1)eqn

now,

Applying law of conservation of energy

∆E total = ∆Ek + ∆Ep

∆E total = (K.Ef - K.Ei) + (P.E f - P.E i)

∆E = (1/2mv²-0 ) + (0 - mgh)

∆E = 1/2 mv²-mgh

since, there is no friction...∆E=0(conserved)

1/2mv² - mgh = 0

v² = 2gh...........(2)eqn

putting the value of v² in equation (1) we get,

cos theta = 2gh/Rg = 2h/R

(from attachment)

cos theta = R-h/R

R-h/R = 2h/R

h= R/3

In Question it is asked to find distance from ground

[R +(R -h)]

=>[R + ( R - R/3)]

=>5R/3✅✅

$\boxed{refer\:to\:attachment}$