# A particle P, free to move on the smooth inner surface of a fixed hollow sphere of centre O and radius a is placed at the lowest point of A of this surface. It is then projected horizontally with...

A particle P, free to move on the smooth inner surface of a fixed hollow sphere of centre O and radius a is placed at the lowest point of A of this surface. It is then projected horizontally with initial speed `sqrt(nga)` where n>0.

While particle is still in contact with the surface find the reaction on the particle from the surface when the angle turned by OP is `theta`

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### 2 Answers

We can use energy conservation law here assuming that there is no energy losses during the motion.

Let us define the horizontal line that passes through A as datum and the mass of P as m.

When P is a A;

Energy `(E) = 1/2xxmxx(nga)-mga`

When particle is at an theta angle;

`E = 1/2xxmxxv^2-mg(a-acostheta)`

`1/2xxmxx(nga) = 1/2xxmxxv^2+mg(a-acostheta)`

`nga = v^2+2ga(1-costheta)`

`v^2 = ga(n-2+2costheta)`

Using Newtons second law when P is at an angle `theta` towards direction of the centre of the circle which is OP;

`F = ma`

`R-mgcostheta = mv^2/a`

`R = mgcostheta+m(ga(n-2+2costheta))/a`

`R = mg(n-2+3costheta)`

** So reaction(R) on the particle from the surface when the angle turned by OP is theta is** `mg(n-2+3costheta)`.

**Sources:**

The particle, of mass m, is projected from the lowest point A of a sphere with velocity `u=sqrt(nga)` to move along the inside of the hollow sphere (refer to the attached image).

Let P be the position of the particle at any time t where arc AP = s and angle AOP = `theta` . If v be the velocity of the particle at P, the equations of motion along the tangent and normal are:

`m*(d^2s)/dt^2=-mgsintheta` ..........(i)

and `(mv^2)/a=(R-mgcostheta)` ..........(ii)

also, `s=a*theta` ........(iii)

From (i) and (iii),

`a*(d^2theta)/(dt^2)=-gsintheta`

Multiplying both sides by `2a(d theta)/(dt)` and integrating,

`2a^2*int((d^2theta)/(dt^2)*(d theta)/(dt))=-2agint(sintheta*(d theta)/(dt))` [for LHS integral, put `x=(d theta)/dt, dx=(d^2theta)/(dt^2)` and it reduces to `int(xdx)=1/2 x^2` ]

So, `(a(d theta)/(dt))^2=2ag*costheta+C`

`rArr v^2=2ag*costheta+C` (since `v=a*(d theta)/(dt)` )

To evaluate C, the integration constant,

At A, when `theta=0, v=u=sqrt(nga)`

`u^2=2ag*1+c`

`rArr c=(u^2-2ag)`

So, `v^2=2agcostheta+ u^2-2ag`

`=2agcostheta+ nag-2ag`

`=ag(2costheta+ n-2)`

Hence the normal reaction, from eqn. (ii) we have

`R=m/a(v^2+agcostheta)`

`= m/a(ag(2costheta+ n-2)+agcostheta)`

`=mg(3costheta+n-2)`

Therefore, the normal reaction of the surface on the particle while it is still in contact with the surface is given by

`R= mg(3costheta+n-2)`

**Sources:**