3
$\begingroup$

It is a problem from Kleppner mechanics:

A bead of mass $m$ slides without friction on a rod that is made to rotate at a constant angular speed $\omega$. Neglect gravity.

(a) Show that $r = r_0 e^{\omega t}$ is a possible motion of the bead, where $r_0$ is the initial distance of the bead from the pivot.

(b) For the motion described in part (a), find the force exerted on the bead by the rod.

(c) For the motion described above, find the power exerted by the agency that is turning the rod and show by direct calculation that this power equals the rate of change of kinetic energy of the bead.

I am confused about the fact that how there can be radial acceleration without any radial force. Actually I can see it from equation but how can I realize the physical meaning of this?

$\endgroup$
3

2 Answers 2

2
$\begingroup$

To move in a circle, there needs to be cetripetal acceleration, and for that, there needs to be a centripetal force. As you yourself pointed out, there is no force along radial direction, which means that the bead cannot move in a circle.

Another way to see this is that the rod exerts a tangential force on the bead, so it gains some velocity in the tangential direction. However at the next instant, as the rod rotates through some angle, the earlier tangential direction is no longer the tangential direction; there is some radial component to it. Thus the bead travels in the radial direction due to the forces exerted by the sides of the rod.

$\endgroup$
1
  • $\begingroup$ Thank you ! Today I also imagined it like that and satisfied now by your answer! $\endgroup$ Commented Oct 6, 2020 at 6:23
1
$\begingroup$

I am confused about the fact that how there can be radial acceleration without any radial force. Actually I can see it from equation but how can I realize the physical meaning of this?

For completeness, let's first include the math here.

For Newton's second law in polar coordinates we have $$\mathbf F=m\mathbf a=m(\ddot r-r\dot\theta^2)\,\hat r+m(r\ddot\theta+2\dot r\dot\theta)\,\hat\theta$$

Now, one should be careful about what they mean by "radial acceleration". If by radial acceleration you mean $a_r=F_r/m$, then of course if there is no radial force then there is no radial acceleration. However, you seem to be more interested in $\ddot r$ as the "radial acceleration". And of course as you can see, if $F_r=\ddot r-r\dot\theta^2=0$, this does not mean that $\ddot r=0$ unless $r$ or $\dot\theta$ are $0$.

But what is happening physically? The issue here is that $\hat r$ and $\hat\theta$ change directions in space. This is different from the intuition we develop in introductory physics in Cartesian coordinates where the unit vectors are constant. Therefore, you cannot equate motion in some "direction" with acceleration in some "direction". This is because "radial" and "tangential" are not unique, constant directions; my radial could be your tangential. Indeed, as @dnaik has already pointed out more generally, in uniform circular motion the acceleration is entirely radial, and yet there is no motion in the radial direction.

If you want to get back to this intuition, then move back to Cartesian coordinates. Of course it is harder to keep track of the forces, but it will work.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.