2
$\begingroup$

Centripetal acceleration of a point with velocity $v$ and moving on a path of radius of curvature $r$ is given by $v^2/r$. If $v= rw$ where $w$ is angular velocity of a body , then centripetal acceleration is given by $w^2r$.

So, my doubts are as follows:

Doubt 1: in pure rolling motion as velocity of lowermost point is zero so it’s centripetal acceleration should be equal to zero but it is traversing a path of radius of curvature r about the centre of object so it seems contradictory to me . Also, I was taught that acceleration of lowermost point is $w^2r$ and hence it has an acceleration towards the centre of the object which also confuses me.

Doubt 2 : Also, if the lowermost point has a centripetal acceleration and we solve any problem from the frame of reference of the lowermost point in a purely rolling object, then is a pseudo force $mw^2r$ applied on the centre of mass towards the lowermost point ? Or does the centre of mass have a centripetal acceleration towards the lowermost point that means a centrifugal force away from it?

Any help would be greatly appreciated!

$\endgroup$

1 Answer 1

2
$\begingroup$

Suppose we have a nice rolling wheel, with its center of mass (CoM) at its geometric center. The wheel is rolling on flat ground (so CoM moving at constant velocity).

For doubt 1: there are two types of motion going on. The wheel is rolling, and also translating.

The speed given by $v=\omega r$ is for rolling, and is measured in the center of mass frame (i.e. frame that is moving with the CoM). If you want the velocity with respect to an inertial frame (e.g. you standing still, observing the motion), you also have to account for the velocity of the center of mass.

The acceleration vector in the CoM frame will be centripetal acceleration, $\omega^2 r$. However, if $\alpha\neq0$, within the CoM frame, you will also have tangential acceleration, $r\alpha$ (such as rolling down an incline). If you want the acceleration with respect to an inertial frame, you'd have to account for the acceleration of the CoM.

For doubt 2: depends which frame you're considering.

If you're in an inertial frame (e.g. you, observing a rolling wheel) then there is no pseudo force.

If your frame is moving with a point (i.e. the frame follows a point on the rim), then there will be an outwards pseudo force.

If you're considering a frame that tracks the point of contact on the ground (I think this is the one you're talking about), then there may or may not be a pseudo force.

  • Case 1: the wheel is rolling on flat ground and CoM has constant velocity. In this case, there are no pseudo forces, since the frame of reference is inertial. There will be a net centripetal force of $\omega^2 r$ applied to the CoM, directed towards the point on the ground.
  • Case 2: the wheel is rolling down an incline. In this case, you're in an accelerating frame of reference (since $\bf \vec a_{\rm CoM}\neq 0$, so there will be a pseudo force (it will actually be a force that opposes static friction such that $\sum \bf \vec F= 0$).

Hope this clears your doubts.

$\endgroup$
5
  • $\begingroup$ Thanks ! Doubt 1 is cleared but I still don’t understand that if I am considering the frame of the lowermost point of the rolling object say a disc then as the lowermost point has an acceleration $w^2r$ upwards so if we wish to consider this frame, why do we give the COM an acceleration of $w^2r$ towards the point ( as I have seen being done in solutions to many problems) ( and please clarify if this what we call the radial acceleration of COM about that point) instead of applying a pseudo force ( mass of object * negative of acceleration of observer) on the COM towards the point? $\endgroup$
    – Nil
    Commented May 15, 2021 at 8:24
  • 1
    $\begingroup$ @Nil Because the contact point is an "instantaneous center of zero velocity" (look it up). Essentially, the body will exhibit pure rotation about the center of zero velocity (which, in this case, happens to be the contact point with the ground). So you can imagine the disc rotating around the contact point, because the contact point is the center of rotation. Now, because the centripetal force is directed towards the center of rotation, it is directed towards the contact point. $\endgroup$
    – user256872
    Commented May 15, 2021 at 8:39
  • 1
    $\begingroup$ As for your second question, this is just the centripetal force (aka radial force) except viewed from a different frame. $\endgroup$
    – user256872
    Commented May 15, 2021 at 8:39
  • 1
    $\begingroup$ (case 1) if the wheel's CoM moves at a constant velocity, then the bottommost point also moves at a constant velocity, so the frame of reference of the contact (or bottommost) point will be inertial, and no pseudoforces come up. However, a pseudoforce would occur if you consider the frame of the contact point rotating about the CoM of the disc. That would be your fictitious centrifugal force, pointing the opposite way. (analogous to watching a car making a turn vs. being in that car) $\endgroup$
    – user256872
    Commented May 15, 2021 at 8:42
  • $\begingroup$ Also, if this answers your question, please mark my answer as "accepted." $\endgroup$
    – user256872
    Commented May 15, 2021 at 8:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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