Question

A bead rests in a smooth vertical ring of radius R Bead is given a slight push so that it slides down around the ring.At some instant bead is at angular position theta with vertical then choose correct option for the following
1).Acceleration of bead is horizontal at theta=π-arc cos1/3
2).Acceleration of bead is horizontal at theta=arc cos 1/3

Answer 1

@ganeshie8

Answer 2

Oops it was ..
A bead rests on a smooth vertical fixed ring

Answer 3

You may try posting it regular physics section for better response. Thanks.

Answer 4

Sir i have posted questions there!! And i were left with no choice so did it

Answer 5

@michele_laino

Answer 6

@Michele_Laino sir I came up with this equation
2cos^2theta+2costheta-1=0
And this gives us theta as
arc cos(√3-1)
And in the option I gave for the question none satisfy this...

Answer 7

@Michele_laino

Answer 8

is the initial position of the bead like this:

Answer 9

Yeah!! Exactly

Answer 10

at generic position, we can write this:

the acceleration is:
\[a = \left( { - \frac{{{v^2}}}{R}\cos \theta ,\quad \frac{{{v^2}}}{R}\sin \theta - g} \right)\]