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.
If P leave the surface when it is at a height a/2 above the level O show that n = 7/2.
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:
and `(mv^2)/a=(R-mgcostheta)` ........(ii)
also, `s=a*theta` ........(iii)
From (i) and (iii),
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)`
So, `v^2=2agcostheta+ u^2-2ag`
Again considering the normal reaction, from eqn. (ii) we have
`= m/a(ag(2costheta+ n-2)+agcostheta)`
When the particle leaves the sphere, R=0
If angle VOQ=`alpha` ,
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`
Hence the proof.