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.
Show that if 2<n<5 the particle will leave the surface with a speed of `sqrt(((n-2)ga)/3)`
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The particle is projected from the lowest point A of a sphere with velocity `u=sqrt(nga)` to move along the inside of the hollow sphere.
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, respectively, 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)=-2agintsintheta(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`
Again considering normal reaction, from eqn. (ii) we have
`= m/a(ag(2costheta+ n-2)+agcostheta)`
When the particle leaves the sphere, normal reaction, `theta=theta_1` , and R=0, then
If `v_1` is the velocity of the particle at point Q, where it leaves the surface, then putting `v=v_1` , R=0 and `theta=theta_1` , then according to eqn. (ii),
`rArr v_1=sqrt(((n-2)ga)/3)` (Note, for real values of `v_1` , n must be greater than 2. Again, for n=5, `costheta=-1` , `rArr theta=pi` , inother words, the particle does not leave the surface, so for falling off, 2<n<5)
Therefore, the particle will leave the surface with a speed of `sqrt((n-2)ga)/3` .
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