Let's say a solid cylinder of radius r is rolling on a stationary horizontal surface with linear velocity v, angular velocity $\omega$, linear acceleration a and angular acceleration $\alpha$

Bottom most point is P


Now I require to find the acceleration of the point P (on the cylinder).

To do this first I could see it in the COM( Centre of Mass ) Frame. I observe it to have an $\omega ^2 r$ centripetal acceleration and a leftward $r\alpha$ tangential acceleration.

Coming to the ground frame I would have to add the acceleration of COM to it which cancels the $r\alpha$ component (as it is rolling) leaving $\omega ^2 r$ upward acceleration.

However I can't get the intuitive understanding as to why there exists a vertical component.

In the ground frame the velocity of P is 0 and so is the tangential acceleration. Due to no velocity I can't directly assume it to have a radial acceleration.

Could someone explain intuitively why it does? Or maybe how I could derive the same by completely solving it from the ground frame?


Consider just a simple wheel that is spinning with constant angular speed $\omega$. Imagine there is a particle on the edge. Recall that velocity is speed with a direction.

As the wheel spins, the velocity direction must change, but the speed remains constant (because the wheel is said to be spinning with constant angular speed). What kind of forces can affect velocity direction but not speed? The answer is: forces that act perpendicular to the velocity (can be shown by Work-KE theorem). Since the velocity is tangent to the circle, the perpendicular force points inwards, towards the center of the circle.

In summary, to change the velocity direction of the particle on the edge of the wheel, a force is required.

Another good example is taking a string and attaching it to a small object and then spinning it around in a circle. The tension force that ensures the object spins in a circle is directed along the string, which points to the center of your circle.

  • $\begingroup$ Amazing I thought about this too after reading it and I got perfect clarity on this!(+1) $\endgroup$ Jan 13 at 19:11

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