IIT JEE , AIEEE Forum - Physics , Mechanics

karthik2007

This question was given by me in a test, which I could not host successfully because my internet connection was off the hook for over 2 weeks. Hence I am posting the only question which was difficult in that test here as a challenge. 5 salutes for the correct answer (with complete solution).

A bead moves on a circular wire of radius r, fixed in the vertical plane. The speed of the bead when it is at the lowest point of the wire is v. Given this, evaluate :
a) The horizontal component of the acceleration of the bead (in the direction of v)
b) If 4gR<v²<5gR, how many points are present on the circle such that the horizontal component of acceleration is zero? (Neglect friction)

feynmann

It is a simple qn .

1) see that there is no force acting on the bead in the horizontal direction at the lowest point .

so horizontal component of accn of the bead at this point = 0 ;

2) v^2 >4gr implies that the bead has sufficient velocity to reach at the top most point where also there is no force acting on the horizontal direction .

So , there are 2 such points .

ganesha1991

1) the horizontal component of acceleration at the lowest point is zero as both the centripetal acceleration and acceleration due to gravity r vertical components
2) 2 points ....in the given range the bead wud complete the semicircle and won't complete the whole circle...therefore acceleration is 0 at topmost and lowermost point bcos of the above reason

correct me if i'm wrong ..........

karthik2007

@feynmann - both your answers are wrong. Your answer is right at the lowermost point, but i want the horizontal component at any point.

I think you solved the question in a hurry. You have to find the velocity of the bead at any point as a function of the angle @ that it makes with the vertical, though I am not supposed to mention it here. You may find it as a function of anything related to the problem for that matter

pramod6990

isiith (v^2*Sin@)/R +gSin@(1-cos@)

horizontal component of a....

karthik2007

No pramod.. gud attempt though..

@feynamn - I want a general expression for acceleration, not the acceleration at the lower most point. Your second soln is wrong though.

akhil_o

let angualr velocity be w
v' is velocity at any point with position angle A...A is angle with vertical
conserving energy
v'²=v²-2rg(1+cosA)
v'=

v²-2rg(1+cosA)
w=v'/r
centripetal acc=v²/r-2g(1+cosA)
and radial acceleration=g
at any point
total a=v²/r-2g(1+cosA)
in direction of horizontal it is
sinA(v²-2rg(1+cosA))/r

karthik2007

ganesha1991

karthik bro check tat rajvarun problem fast

Ksudha_iit

hey ..can we post it tomorrow?? i need some time to solve the pro pls...wait until i post pls

karthik2007

Okay - soln will be posted tomorrow night at 9 pm

pramod6990

isiith (v^2*Sin@)/R -2gSin@(1-cos@)

with correction......

akhil_o

what is wrong in my ans??

karthik2007

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