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enter image description here

This is a picture taken from wikipedia. Here the rolling friction is explained.

W is the weight, F is the driving force( neglecting friction in the bearing etc,) r is the radius of the wheel, R is the reaction force which may add to frictional force while rolling.

My question is that, what does the reaction force from the other side for ? (Consider the total arc of contact and take normal reactions ? seems like they cancel each other !!

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There are forces on the wheel all the way along the arc of contact. Some of these forces point forwards, some point backwards. The forces pointing backwards are the larger because the wheel is moving left and is pushing more against the ground ahead of it than against the ground behind.

$R$ is the resultant of all these forces. The "cancelling out" has already been done. $R$ is what is left over, due to the forces at the front being bigger.

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If we assume that the object is not completely rigid then there will be an arc of contact normal reaction acts along the normal at the point of contact (here points). There is no motion vertically hence the forces in vertical direction must balance each other hence the weight must be balanced by the vertical component of reaction force, hence reaction force is not zero(they don't cancel each other completely) however the horizontal component of reaction force may or may not cancel each other depending upon the exact shape of arc and mass distribution of the wheel. If the object is completely rigid and the shape of wheel is a perfect circle then there will be only one point of contact and the reaction force will be perpendicular to the ground.(no horizontal component) hope this helps.

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  • $\begingroup$ In that consideration, you are saying area doesnt matter the rolling frictional force. But again, we may not be able to answer " why deflated wheels come to rest quicker than inflated ones " ? $\endgroup$ – Vinayak Nov 8 '18 at 16:11
  • $\begingroup$ I am not saying anything about rolling friction just telling about the reaction force. $\endgroup$ – Anmol Rastogi Nov 9 '18 at 18:25
  • $\begingroup$ And by reaction force here I specifically meant normal reaction $\endgroup$ – Anmol Rastogi Nov 9 '18 at 18:26
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My question is that, what does the reaction force from the other side for ? (Consider the total arc of contact and take normal reactions ? seems like they cancel each other !!

I can't precisely understand what you are asking,but I think you are questioning why should there be an R and wouldn't it get cancelled if we resolve it into components?

In the case of a rolling body, surface deformations give rise to tangential reaction force components acting at the surface. They also give rise to force components which, though normal to the deformed surface where the bodies contact, are not necessarily normal to the undeformed surfaces well away from the contact region. These deformations, though often small, are still larger than the microscopic processes responsible for friction phenomena. Their effect on a body rolling without slipping is called "rolling resistance" in the engineering literature. Introductory physics books often lump this together with resistance, or even call it "rolling friction", a misleading term. Some even lump the two together and call it "friction". In some simple, or idealized, problems one can get away with that carelessness

If the ball were not rolling we would expect a symmetric deformation unbiased in either direction. The dynamics of the initiation of motion would need to be considered, but we need not get bogged down in the details. Once motion is established, the front side of the ball is compressing itself and the surface below, while at the back side, the surfaces are both relaxing back to their undeformed condition. Typically, elastic materials display hysteresis, that is, the compression and relaxation forces are unequal functions of the amount of deformation

Reference:

https://lockhaven.edu/~dsimanek/scenario/rolling.htm

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