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`
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)`.
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:
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^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)`
So, `v^2=2agcostheta+ u^2-2ag`
Hence the normal reaction, from eqn. (ii) we have
`= m/a(ag(2costheta+ n-2)+agcostheta)`
Therefore, the normal reaction of the surface on the particle while it is still in contact with the surface is given by