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what happens to a rolling ball when it enters to a frictionless surface? does the ball keep rolling forward? how to explain the motion of the ball in the BC and CD surfaces?

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  • $\begingroup$ Is there drag from the air? $\endgroup$ – user234190 Jul 7 '19 at 18:38
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It does not keep rolling (because it isn't matching the surface anymore), but it does keep spinning forever.

  • The rough left incline "grabs on" to the contact point of the ball at all times, and causes it to spin. Gravity forces it to move forward (downwards), so the static friction pulls the contact point harder and harder in order to keep it stationary on this rough surface. Thus the spinning speed $\omega$ increases along with the translational speed $v$.
  • When reaching the bottom point, it has reached its top spinning speed $\omega$ and translational speed $v$.
  • Now it starts to move upwards along the right incline. Gravity makes it slow down, so the translational speed $v$ decreases. But the spinning speed $\omega$ stays constant! There is nothing that stops the rotation - no rough surface to grab - so the spinning is never interfered with.
  • When it reaches the top of the right incline, the translational speed become momentarily zero $v=0$, but the spinning speed $\omega$ is still high. So, it spins while just hanging there.
  • And in the next moment, it starts falling downwards again along the right incline, still due to gravity, so the translational speed $v$ increases. Meanwhile, the spinning speed $\omega$ is still unaffected and is still the same as before.

It may feel unintuitive - but you actually see this effect often, for example on bowling balls. The bowling floor is very smooth, although not perfectly smooth, so you can easily make bowling throws that end in the bowling ball translating forward while simultaneously spining, say, sideways or backwards. The "grab" on to the floor is so weak do to so low friction, that the spinning can take place almost independently from the translational motion.

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What you want to keep track of is conservation of energy. Once the ball gets to point B it will keep the same U and the same K up to point C. After that, what will be conserved is the total energy $E=U+K=mgh+\frac{1}{2}mv^2$. So, the ball will go up a height $h=\frac{mv^2}{2mg}$, before coming down again.

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  • $\begingroup$ You forgot the rotational energy component, E=I(omega)^2. $\endgroup$ – PhysicsDave Jul 7 '19 at 19:22

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