Let a ball of mass $m$ is given a velocity over the the top of a smooth sphere of radius $r$. The equation of motion at the topmost point will be $$mg – N = m\dfrac{v^2}{r}.$$ As $v$ increases, $N$ decreases to zero. Hence, $v$ at that case will be $\sqrt{rg}$. The ball will lose contact with the sphere if it has velocity at the topmost point at the beginning greater than $\sqrt{rg}$. When it has velocity less than this , it will still lose contact but after moving certain distance.
My questions:
$\bullet$ Why does the particle lose contact immediately, when traveling at $v > \sqrt{rg}$?
$\bullet$ Why does it lose contact after certain time , when traveling at a velocity $v < \sqrt{rg}$?
[Note that the sphere is smooth & provides no friction.]
Best Answer
Assumptions :
1. Sphere and ball are perfectly rigid bodies.
2. radius of ball negligible in comparison to sphere (in other words equivalent to a point mass)
3. friction between ball and sphere is sufficiently large ** (remove) 4. surface between sphere and table (or whatever the sphere is on) is frictionless
Initially due to friction between ball and sphere, the point on sphere in contact with the ball tries to move with velocity v. For this the movement of the ball w.r.t sphere is moving in a circular path about the center of the sphere. Hence sufficient centripetal acceleration must be provided by the normal force exerted by the sphere on ball.
Hence the equation $mg - N = m\dfrac{v^2}{r}$.
clearly (sphere cannot pull ball towards it!)$$N>0 \implies v<\sqrt{rg} $$
so when the above happens the ball moves in circular path w.r.t the center of sphere.
When $N=0$ the ball loses contact with the sphere. There is no longer any friction force to support the movement of the ball along the sphere as $f=\mu N$.
This happens when the ball makes an angle $\theta $ with vertical such that $\cos\theta = \dfrac{v'^2}{rg}$ (Use free-body-diagram and equate net tangential force = $m\dfrac{v'^2}{r}$). where $v'$ is velocity of ball at that instant. To find $v'$ us energy conservation.
$$K.E.(\text{ball})+K.E.(\text{sphere})+P.E.(\text{ball}) = \text{constant} \qquad \& \qquad (v\cos\theta - v_{cm})^2 + (v\sin\theta)^2 = v_{cm} ^2$$ (for pure rolling).
After this the ball behaves like a projectile.
Edit :
Even if there exists no frictional force, little changes in the above solution.
Although we considered the dimensions of the ball to be small, a slight displacement in the x-direction does not remove contact from sphere (had it been a point mass it would have left the sphere (didn't notice it before :) ) ), rather the point of contact changes, bringing about change in direction of $N$ and the required centripetal acceleration to keep it moving in a circular path w.r.t center of sphere. Note that radius of ball << radius of sphere results in zero R.K.E. (about its own axis)(almost) so you can treat it like a point mass here.