Question

(2) /
(1) zva
(3) Ź
(4)
20.
nner surface of a smooth vertical circle of radius R, its velocity at the lowest
A particle is projected along the inner surface of a smooth vertical circ
It will leave the circle at an angular distance 'O' from the upward vertical. The value of
point being
o is
(2) 53°
(3) 60°
(4) 30°
(1) 37°​

Answer 1

Answer:

By using Centeroid Accleration Theory

R-mg cos (180 -Tita)=R+mg Cos tita=mv^2/r

now using motion Fromluas

V^2=2gh

h=r-r cos tita

now R+mg cos tita=m(2gr(1-cos tita))/r

R+mg cos tita=2mg-2mg cos tita

when leave the Circle R=0

so 3 cos tita=2

so Tita (o)=cos-1 (2/3)=53 degress