Question

A particle of mass m is projected with speed √rg/4 from top of a smooth hemisphere as shown in figure. if the particle starts slipping from the highest point, then the horizontal distance between the point...

Answer 1

let particle leaves contact at an angle θ with the vertical line as shown in figure.

so, h = R - Rcosθ

use formula, v² = u² + 2as

here, u = √{Rg/4}, a = g , and s = h = R(1 - cosθ)

so, v² = Rg/4 + 2gR(1 - cosθ)

= 9Rg/4 - 2Rgcosθ......(1)

at equilibrium,

mv²/R = mgcosθ - N

or, m(9Rg/4 - 2Rgcosθ)/R = mgcosθ - N

when particle leaves the contact from hemisphere. N = 0

so, m(9g/4 - 2gcosθ) = mgcosθ

or, 9mg/4 = 3mgcosθ

or, cosθ = 3/4 .....(2)

now, horizontal distance = Rsinθ

from equation (2),

cosθ = 3/4 ⇒sinθ = √7/4

hence, horizontal distance between the point where it leaves the contact from teh sphere and the point at which the particle was placed = R√7/4 [Ans]

Answer 2

Answer:

Explanation: i hope it's better