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In the question it is stated that there is a ring which can move along a smooth rod. The rod itself is rotating in a horizontal plane with one fixed end with uniform angular velocity. Initially the ring is kept near to the fixed end. We have to calculate the velocity of the ring after it leaves the rod, given the angular velocity and the length of the rod.

According to me, when the ring leaves the rod there should be just a tangential velocity which can be calculated by integrating the centrifugal force with respect to the radius.

There seems to be no point of a radial velocity. Am I correct?

enter image description here

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  • $\begingroup$ This is really crying out for a drawing!! $\endgroup$ – Gert Sep 3 '19 at 15:21
  • $\begingroup$ @Gert Not really. A ring sliding on a rod is a pretty common system used to analyse rotating reference frames and is easily envisioned without a picture. $\endgroup$ – Aaron Stevens Sep 3 '19 at 15:35
  • $\begingroup$ Note: The included sketch is a top view, looking down along a vertical axis at the position indicated by an x. $\endgroup$ – R.W. Bird Sep 3 '19 at 18:56
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No, this is not correct

Certainly the ring will move tangent to what would have been its trajectory at the point of release had it not been released. But as the ring moves outward it will not be maintaining a constant distance from the fixed point of the rod. Therefore, its radial position $r$ will be changing, i.e. it has a non-zero radial velocity still.

You can determine the equation of motion by looking at Newton's second law for planar motion in polar coordinates: $$\mathbf F=m\mathbf a=m(\ddot r-r\dot\theta^2)\hat r+m(r\ddot\theta+2\dot r\dot\theta)\hat\theta$$ Since there is no force acting on the ring after it leaves the rod, it must be that each component of the acceleration is $0$: $$\ddot r=r\dot\theta^2$$ $$r\ddot\theta=-2\dot r\dot\theta$$

These equations, along with the initial conditions you describe, determine the motion of the ring after it leaves the rod.

You have already convinced yourself that after the ring leaves the rod, $\dot\theta\neq0$. And we know that $r\neq0$. Therefore $\ddot r\neq0$, and so it must be that there is a non-zero $\dot r$.

Therefore, to answer your title question

Will there be any radial velocity in absence of centripetal force and angular acceleration?

Yes, there will be. It is determined by the motion of the ring once it reaches the end of the rod.

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Why do you think there will be no radial velocity? The radial velocity is what keeps the ring moving outward along the rod, it will not be zero (except possibly at the beginning).

In terms of fictitious forces in the frame moving with the rod, the centrifugal force is outward and what provides the radial velocity.

The tangential/azimuthal velocity is dictated by angular velocity of the rod. For instance, at the time the ring leaves the rod, its tangential velocity will be equal to that of the free end of the rod.

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  • $\begingroup$ I see that, but the velocity of the ring after it leaves the rod is the same as its velocity immediately before it leaves the rod (at least in the inertial frame). The radial component of the final velocity can be calculated by analyzing the motion along the rod. $\endgroup$ – Puk Sep 3 '19 at 15:32
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At each position of the ring on the rod, a force would be required to keep the ring in that position with a centripetal acceleration of r(w^2). In the absence of that force the ring will accelerate outward along the rod at a rate of r(w^2). That suggests that r = ro e^(wt). Then the radial velocity would be v = ro w e^(wt) giving a radial acceleration of a = ro (w^2) e^(wt) = r(w^2) which matches our basic assumption. When the ring reaches the end of the rod r = L = ro e^(w tf). (The ring leaves the rod at time tf =(1/w) ln(L/ro). The radial velocity at that point will be v = ro w e^ln(L/ro) = wL which is equal to the tangential velocity at that point. The rotation angle of the rod will be A = w tf = ln(L/ro). When the ring leaves the rod, its velocity vector will form an angle of 45 degrees with the rod.

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  • $\begingroup$ After further consideration, it occurs to me that this analysis applies only if ro is very small, since the velocity equation does not allow the initial radial velocity to be zero. $\endgroup$ – R.W. Bird Sep 8 '19 at 18:53

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