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.

If P leave the surface when it is at a height a/2 above the level O show that n = 7/2.

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Let P be the position of the particle at any time t where arc AP = s and angle AOP = `theta` (Refer to the attached image). 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^2 theta)/(dt^2)*(d theta)/(dt)=2agintsintheta(d theta)/(dt)` [for the LHS integral, put `x=(d theta)/dt, dx=(d^2theta)/(dt^2)` and it reduces to `intxdx=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, put

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)`

Again considering 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)`

When the particle leaves the sphere, R=0

`mg(3costheta+n-2)=0`

`rArr costheta=-(n-2)/3`

If angle VOQ=`alpha` ,

Then, `alpha=(pi-theta_1)`

Therefore, `cosalpha=cos(pi-theta_1)=-costheta_1=(n-2)/3`

Then AT=OA+OT`=a+acosalpha=a+a*(n-2)/3=a/3(n+1)`

By condition, the particle leaves the inner surface of the sphere at a distance `a/2` above the level O, i.e. AT=`3/2a`

So, `a/3(n+1)=3/2a`

`rArr n=7/2`

Hence the proof.

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