IIT JEE , AIEEE Forum - Physics , Mechanics

tarinbansal

a rock attached to a string swings in a vertical circle then at the topmost point of its path,

a-2 forces act on it, resultant not=0

b- 1 force acts

c-2 forces act on it, resultant =0

d-3 forces act on it

Ans-(a)

My doubt is that there should be 3 forces acting on it- Tension, centrifugal and Mg.

But the answer given is (a).

Again a LOTS question.

karthik2007

This is a simple one mate. The answer is a). There are only two forces acting on the string at the topmost point. The common misconception here is that people consider centripetal force to be another force acting on the particle. See, centripetal force is simply the resultant of the forces already acting, in the direction of the radius vector of the path described by the particle. For a particle to move in a circle, there must be some force holding it, just like how the merry go round cars are held together by steel rods.

Centripetal force never acts by itself, ie, it is NOT a force by itself. It is simply the resultant of the forces already acting on the body, which is directed towards the radius vector. So, at the topmost point, the only forces acting are Tension T and mg, whose resultant has to be T+mg in the downward direction.

tarinbansal

Waah yaar karthik,
What an explanation. This concept was not known to me till now. Thanx yaar. You are simply genius in mechanics.
This has cleared a lot of my previous doubts also.
You deserve a salute.

karthik2007

Now if you have understood, tell me, what are the forces acting on the stone when it is at the bottom, and what is the centripetal force at this point?

tarinbansal

The forces acting at the bottom will be Mg, T.
As Mg>T, therefore net force will be Mg-T which will be equal to the centripetal force.

Am I right?
May I ask you a question? Are you studying ? I mean are you in a college or in ur 11th or 12th?

rooney

At bottom centripetal force is T - Mg. Not mg - T.

tarinbansal

Oh ya.
My foolish mistake.
Mg-T will be centrifugal force in the frame of reference of revolving stone.

But hey,

Mg has to be greater than T. In that case T-Mg will become -ve.

How will you justify that?

karthik2007

No dude, if mg>T, then how will the particle even go around a circle? It is the tension which basically holds the particle together, so that it goes in a circle. So obviously, T should be greater.

To find resultant:

T acts upwards, and mg downwards, so net force radially upwards is T-mg. If T
And I am studying in college :)

tarinbansal

Oh OK.

karthik2007

Here's a tricky one. If you get this one right, then you've almost mastered the concept. Think over it carefully, and then answer. Its not necessary that you have to get the answer instantly.

The Q is : Find the centripetal force acting on the particle of mass m.

rooney

edited

karthik2007

-edited-

tarinbansal

Rooney has given the answer but I want to think on it.
Can you please tell me that is this mass attched to a string or is it rolling on a sphere?

karthik2007

Its rolling on a sphere.

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