A ball rolling on a horizontal surface without slipping encounters a curved incline as shown in diagram below. The curved incline is smooth and frictionless.


What will happen to the rolling ball on the curved incline and why? Will it continue to roll without slipping or will it move up the incline without rolling?

My take on this situation

I think the ball will continue rolling but with decreasing angular velocity and decreasing linear velocity of it's center. The angular velocity will decrease because the gravity force mg acting vertically down will exert a clockwise torque about an axis passing through the point of contact of ball with incline and perpendicular to the incline plane (i.e. axis is perpendicular to the plane of paper). I am not sure about this explanation and observation.

Rolling ball moving up a smooth incline


closed as off-topic by John Rennie, Kyle Kanos, Yashas, Jon Custer, David Hammen Mar 9 '17 at 18:03

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  • $\begingroup$ If the curved surface is frictionless, how can it exert a torque? $\endgroup$ – sammy gerbil Mar 9 '17 at 6:32
  • $\begingroup$ I meant torque due to the gravity force about the axis through point of contact between ball and incline. When a ball rolls without slipping then we can consider the ball to be rotating about the point of contact. $\endgroup$ – Sunil Mar 9 '17 at 6:36
  • 1
    $\begingroup$ When the ball is on the frictionless surface it will roll with slipping. $\endgroup$ – sammy gerbil Mar 9 '17 at 6:39
  • $\begingroup$ Why would it slip on the incline? The point of contact has zero velocity when on horizontal surface since it has w * r velocity in backward direction and velocity v in forward direction, which are equal and opposite resulting in net zero velocity for point of contact. $\endgroup$ – Sunil Mar 9 '17 at 6:52
  • $\begingroup$ As the ball rises its linear velocity $v$ falls below $r\omega$, so it no longer fulfills the no slip condition. $\endgroup$ – sammy gerbil Mar 9 '17 at 6:57

Without friction, $\omega$ and $v$ are decoupled, and the ball will climb up to $h_{2}=\frac{v^{2}}{2g}$ (when all translational kinetic energy is transformed into potential energy) and then fall back the curve. During this process, the rotation will be what it was when entering the curve.

When the it reaches the horizontal part again, there is necessarily slipping, since the translational velocity is still $v$ but in the other direction, and the angular velocity is $w$ as before.

  • $\begingroup$ Are you saying that angular velocity will not change as the ball goes up the incline? If yes, why would the angular velocity not change? $\endgroup$ – Sunil Mar 9 '17 at 6:46
  • $\begingroup$ sammy told you already : without friction there is no torque, because there is no force applied. And up the curve there is no friction. $\endgroup$ – Naptzer Mar 9 '17 at 6:53
  • $\begingroup$ Ok. So, the ball will be rotating and at the same time sliding up the incline. Then, when it starts its downward journey it will again rotate and slide down at the same time. $\endgroup$ – Sunil Mar 9 '17 at 7:06

I think the ball has to continue rolling till a certain height (i.e till all the Kinetic energy is transferred to its potential energy) and then roll down without slipping for the reason that it has an angular momentum and it has to be conserved as it rolls over the frictionless surface. There is no torque present about the centre due to gravity as all of its component acts at the centre. i.e r=0 hence Torque = 0. So in short one can say there is no external force so as to alter the angular momentum. Hence rotation/rolling remains.

Any correction would be highly appreciated.


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