IIT JEE , AIEEE Forum - Physics , Mechanics

rini

A particle originally at rest at the highest point of a smooth vertical circle of radius R is slightly displaced.

it will leave the circle at a vertical distance h below the highest point.

find h (in tearms of R).

ans: h= R/3

rooney

Let the object make an angle of

with vertical moving with velocity V
Forces acting on the body are weight Mg and Normal force N
Forming equations~
Mgcos

- N =Mv²/R
By energy conservation ~
(1/2)Mv²=MgR(1-cos

)

When body leaves circle, N = 0. Put N= 0 in 1st equation and solve wih 2nd equation.
you get cos

=2/3

Distance from top = R(1 - cos

) = R/3

shubham_sachdeva

Let it loose contact after making angle

equating forces, we get

mv^2/R=mgcos

v^2=Rgcos

;

Again 1/2mv^2=mg(R-Rcos

) ;v^2=2gR(1-cos

)

so

=cos^-1(2/3)

so R - h = 2R/3

which implies

h = R/3

rini

thanks... a lot.