IIT JEE , AIEEE Forum - Physics , Mechanics

karthik2007

Refer figure:

Consider an arbitrary point A.

Conserving energy, we get:

1/2mv^'2 + mgR(1+cos@)= 1/2mv²

Which gives v'² = v²-2gR(1+cos@)

Now, from newton's laws, we get:

N+mgcos@ = mv'²/R

Which gives N = mv²/R - mg(2+3cos@)

We have to find the horizontal acceleration of the bead. The horizontal force acting on the bead in the direction of v is the component of the Normal force -Nsin@ (Nsin@ is towards the left)

Hence we get -Nsin@ = [mg(2+3cos@) - mv²/R]sin@

Hence, horizontal acceleration = -Nsin@/m = [g(2+3cos@) - v²/R]sin@

For the second part:

horizontal acceleration = 0 is

sin@ = 0, or the term in the bracket = zero.

hence, sin@ = 0 implies @ = 0, pi, 2 pi, 3 pi... etc.

g(2+3cos@) - v²/R = -

Or cos@ = u²-2Rg/3Rg

this means @ will have two values as 4Rg<v²<5Rg

@ = cos^-1(u²-2Rg/3Rg) = t , where @<2pi< t.

Which implies four such points exist.

sachin_gupta1991

Is the answer {v²- 2gr(1-cos

)}sin

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r

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