IIT JEE , AIEEE Forum - Physics , Mechanics - Angle traced when particle slides down and leaves the contact with a frictionless sphere.

tarinbansal

A particle slides on the surface of a fixed smooth sphere starting from the topmost point. Find the angle rotated by the radius through the particle, when it leaves contact with the sphere.

Ans- cos^-1(2/3)

Its a common question of HCV. I have a doubt in this question which I will put forward when someone will give me a proper solution to this.

Rates assured.

rooney

Tell the doubt.

tarinbansal

Well I dont know how to solve the question even but I am telling the doubt.

Let the angle be

from vertical. Then at the point it loses contact with sphere, normal N=0.

N=mgcos

+ mv^2/R

But as N=0.

Hence, mgcos

= - mv^2/R

Or cos

= -v^2/Rg

as cos

is coming out to be negative,

it means that 90<

<270

But that is not possible, as the particle leaves contact before 90 degrees.

SO my doubt is that when such a condition is not possible then why is it coming?

rooney

Here is the error

N=mgcos

+ mv^2/R

It should be mgcos

- N =mv^2/R

coz mv^2/R is directed towards the center. and mgcos

is also towards center. So, mgcos

> N.

tarinbansal

But as the net force in the direction normal to the sphere is zero hence it should be
N=mgcos(theta) + mv^2/R

rooney

Accd to your equation, magnitude of mgCos(theta) is less than that of N which is never true. It is mgcos(theta) - N =mv^2/r

biki

let angle made =

(as shown in figure)

let v = velocity just before falling

So at an instant just before falling---

mv²/r = mg.cos

_____(1).........where m = mass, r = radius of the sphere

or v²/r = g.cos

_____(1)

again change in K.E. = (1/2)mv² - 0 ......as initial velocity = 0

and loss in P.E. mg(r - r.cos

)

So (1/2)mv²= mg(r - r.cos

)

or v²/2 = gr (1 - cos

)

or v²/r = 2g(1 - cos

)

or g.cos

= 2g - 2g.cos

or 3g.cos

= 2g

or cos

= 2/3

or

= cos^-1(2/3)

ramyani

fatafati.

tarinbansal

Hey biki,
Let me tell you that when we are standing in an inertial frame, and a particle is doing circular motion, Then centripetal force acts on it not centrifugal.
In your soulution, you have applied centrifugal force to the particle, it should be towarrds centre.

tarinbansal

rooney how can we say that the magnitude of mgcos(theta) can never be less than Normal?
If you see I have just balanced all the forces, in the direction normal to the sphere as net acceleration and displacement in that direction is zero.
Then why I am getting the wrong conclusion?

rooney

Tarin,

See . what is the net force ?
mgcos(theta) - N

What is net acceleration BEFORE the mass m leaves the sphere?
v^2/R in direction of mgcos(theta) [radial direction]

So, mgcos (theta) - N =mv^2/R

This is valid only when the body has not left the contact from the sphere's surface.

tarinbansal

Well Rooney,
I think I have understood ur point.
One more request, check biki's solution. Don't you think he has solved it wrong by taking mv^2/r outwards as centrifugal force?

rooney

No. When you consider the pseudo force( or centrifugal force), you no longer consider the net acceleration to be v^2/R. You assume all the forces to be balanced. You get the same equation by both ways.

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