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Suppose a cylinder is launched on a horizontal frictional surface such that it has initial translational velocity $v$ and zero angular velocity. the kinetic friction would be applied between the contact points of the cylinder and the surface, opposite to the direction of the translational motion. This kinetic frictional force will simultaneously apply torque on the cylinder (which will increase its angular velocity) and decrease the translational velocity till the cylinder satisfy the condition for pure rolling($V=RW$). when the condition for pure rolling is satisfied, the relative velocity between the contact points and surface would be zero and this there would be static friction between the contact points and the surface.

My confusion is that why is static friction necessary for pure rolling instead of kinetic friction. Static friction is applied when the object satisfies the condition for pure rolling ($V=RW$). The static friction does not increase or decrease, both translational and angular velocity. Kinetic friction on the other hand ensures that the object follows the condition for pure rolling on a horizontal frictional surface.

According to my understanding, if an object is launched on a frictionless surface such that it initially satisfies the condition for pure rolling, then the object would continue to be in pure rolling motion even in the absence of static friction(static friction=0 because µ=0), thus the static friction is not necessary for pure rolling. But if an object is launched on a frictionless surface such that it initially does not satisfy the condition for pure rolling, then the motion of the object would not be transformed into pure rolling due to absence of kinetic friction. Can you explain that why is the static friction is called to be necessary for pure rolling of an object instead of kinetic friction?

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  • $\begingroup$ As the answers below point out, the contact point is not moving with respect to the surface. This is easier to visualize if you think of a rolling tank tread instead of a cylinder. $\endgroup$
    – mmesser314
    Sep 5, 2022 at 16:08

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My confusion is that why is static friction necessary for pure rolling instead of kinetic friction.

The short answer has already been given to you in some of other answers. Pure rolling involves no slipping and kinetic friction only exists where there is slipping.

That said, kinetic friction can play a role in establishing the conditions to initiate pure rolling and does play a role when there is "impure" rolling, i.e., a combination of rolling and slipping. I will attempt to explain by individually addressing a number of statements in your post.

when the condition for pure rolling is satisfied, the relative velocity between the contact points and surface would be zero and this there would be static friction between the contact points and the surface.

Static friction may or may not be needed for the condition of pure rolling, depending on the circumstances, as discussed below.

For the example you provided, static friction would not be needed for continued pure rolling once the condition $v=R\omega$ for pure rolling is satisfied. Exactly what happens will depend on whether or not the horizontal launching force continues to be applied or is removed after motion is initiated, as discussed below:

  1. Launching Force Removed

Once the launching force is removed, in the absence of other external forces (e.g., air drag, rolling resistance), the only horizontal force acting on the cylinder is kinetic friction which reduces $v$ and increases $\omega$ until the condition $v=R\omega$ for pure rolling occurs. At that point there is no kinetic friction (since there is no slipping during pure rolling). There is also no static friction since static friction only exists in opposition to an applied external force, of which there is none. The cylinder will continue to roll without slipping at constant $v$ and $\omega$ due to its inertia.

  1. Continuous Application of Launching Force

Since the horizontal launching force caused slipping, that means the launching force exceeded the maximum possible static friction force of $\mu_{s}mg$ so that friction transitions to kinetic friction only. That leaves only the kinetic friction force and the horizontal launching force $F$ applied to the cylinder. You will now have a condition under which both slipping and rolling occurs, or "impure" rolling if you will, with the angular acceleration determined by the net torque about the COM.

My confusion is that why is static friction necessary for pure rolling instead of kinetic friction. Static friction is applied when the object satisfies the condition for pure rolling ($V=RW$). The static friction does not increase or decrease, both translational and angular velocity.

As already pointed out static friction is not involved in the example you gave because the initial launching force exceeded the maximum possible static friction force. Only kinetic friction is involved during and following launching as long as there are no other external horizontal forces acting on the cylinder.

Now if, instead, the launching force did not exceed the maximum possible static friction force of $\mu_{s}mg$, then no slipping would occur and static friction would apply a torque about the COM. Once again, the angular acceleration would depend on the net torque about the COM. That, in turn, would depend on the location of the applied horizontal force relative to the COM.

Kinetic friction on the other hand ensures that the object follows the condition for pure rolling on a horizontal frictional surface.

Again, kinetic friction never insures pure rolling because kinetic friction only exists when there is slipping. Kinetic friction may insure a combination of rolling and slipping.

According to my understanding, if an object is launched on a frictionless surface such that it initially satisfies the condition for pure rolling, then the object would continue to be in pure rolling motion even in the absence of static friction(static friction=0 because µ=0), thus the static friction is not necessary for pure rolling.

In order to initiate pure rolling of the cylinder on a frictionless surface, the applied horizontal force would need to generate a torque about the COM since there would be no friction force to generate a torque. That is, the applied horizontal force needs to be applied either above the COM to generate pure clockwise rolling, or below the COM to generate pure counter clockwise rolling. If it is applied to the COM the cylinder will simply slide without rolling.

But if an object is launched on a frictionless surface such that it initially does not satisfy the condition for pure rolling, then the motion of the object would not be transformed into pure rolling due to absence of kinetic friction.

The launch on a frictionless surface will not satisfy the condition for pure rolling only if the applied horizontal force is applied to the COM, since there would be no torque about the COM.

But kinetic friction can only initiate pure rolling by satisfying the condition $v=R\omega$. It cannot maintain pure rolling where friction is needed to maintain pure rolling. That's because kinetic friction only continues to exist if slipping is occurring. Only static friction maintains pure rolling where friction is needed to maintain pure rolling.

Can you explain that why is the static friction is called to be necessary for pure rolling of an object instead of kinetic friction?

Hopefully my comments above taken together have answered your question.

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Pure rolling needs that there is no relative motion between the contact points, and thus that they have the same velocity. Since there is no relative motion between these two points, this is provided by static friction.

Static friction acting on the cylinder, generated as the reaction between the cylinder and the surface where it is rolling, introduces on the cylinder the external moment that is required to change its angular velocity, following the equation

$I_G \dot{\omega} = M_G^{ext} = R F^{stat. fr.}$

being $I_G$ the moment of inertia of the cylinder around its axis, $\omega$ the components of the angular velocity "around" the axis, $R$ the radius of the cylinder.

As you can see from the equation above, if the angular velocity of the cylinder is constant, the static friction is equal to zero. As an example, a wheel of a car (approximately, since we are in real life) experiences net static friction only during acceleration or braking along the trajectory.

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  • $\begingroup$ " If an object is launched on a frictionless surface such that it initially satisfies the condition for pure rolling, then the object would continue to be in pure rolling motion even in the absence of static friction(static friction=0 because µ=0), thus the static friction is not necessary for pure rolling" is this assumption wrong? $\endgroup$
    – hsdfasd
    Sep 5, 2022 at 13:55
  • $\begingroup$ this is true if the angular velocity of the cylinder and the linear velocity of its center of gravity don't change $\endgroup$
    – basics
    Sep 5, 2022 at 13:58
  • $\begingroup$ if you don't apply any external forces along the direction of motion, the velocity of the center of mass is constant. if you don't apply any external moment, the angular velocity is constant. This is the case of a cylinder rolling on a horizontal frictionless surface. This is not the case of a cylinder rolling on an inclined surface, since the weight has a component on the direction of motion that makes the velocity of the center of mass increase, while the angular velocity remains constant, and thus there is some sliding between contact points an no more pure rolling $\endgroup$
    – basics
    Sep 5, 2022 at 14:02
  • $\begingroup$ "A wheel of a car (approximately, since we are in real life) experiences net static friction only during acceleration or braking along the trajectory." wouldn't the friction experienced by the wheel be kinetic instead of static friction. I am saying this because when the wheel is accelerating or braking along the trajectory, there is a relative velocity(non-zero) between the contact points and the surface. Thus, I think the friction which is applied between the wheel and the surface should be kinetic friction instead of static friction. This was my confusion in my original post as well. $\endgroup$
    – hsdfasd
    Sep 5, 2022 at 14:11
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    $\begingroup$ You experience dynamic friction if the wheel slips on the ground: this is a very inefficient way of braking, and in car racing you see smoke coming from the wheels and flat spots on the tyre. When you brake efficiently, the point of the wheel in contact with the ground has zero velocity: you rely on pure rolling between the tyre and the ground and dynamic friction in your brakes (Brembo, or whatever), between the disk and the pads (pushed to the disk by pistons). You can use flat spots on tyres as a proof of sliding and thus dynamic friction that produces extreme and localized tyre wear $\endgroup$
    – basics
    Sep 5, 2022 at 14:18
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There's a lot of confusion around the role of friction in rolling motion, myself included, but here is my current understanding. If the cylinder is undergoing pure rolling and the translational speed is constant, then there is no static friction at the point of contact. If there were, the translational speed would decrease due to a net force opposite the motion and the angular velocity would increase due to a net torque about the center of mass. Under the assumption that neither of these is occurring, there is no friction force acting.

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  • $\begingroup$ If someone downvotes an answer, it would be helpful to explain the reason. $\endgroup$ Sep 5, 2022 at 20:51
  • $\begingroup$ I totally agree, but there is no obligation. Someone once compared downvoting as an example of "hit and run". $\endgroup$
    – Bob D
    Sep 6, 2022 at 21:23
  • $\begingroup$ In reviewing my answer, I'm still a bit confused about something. I argue that there is no static friction, but since the point of contact is not moving there could be static friction but it doesn't do any work hence doesn't change the translational speed. So is there a static friction force or not ? $\endgroup$ Sep 7, 2022 at 12:39
  • $\begingroup$ Just to be clear before responding, are you talking about the case where the launching force is removed after causing sliding (my case 1) or where it continues to be applied (my case 2). $\endgroup$
    – Bob D
    Sep 7, 2022 at 13:43
  • $\begingroup$ Your case 1 - we have an object already in pure rolling. You state that there is no static friction as there is no external force to oppose, which seems right and answers my question. Thanks. $\endgroup$ Sep 7, 2022 at 15:21
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If you assume that your body is in pure rolling motion, the contact point is not moving relatively to the surface. You apply static friction for this reason.

Look at the tires of the cars on a street: the point of contact is still with respect to the asphalt (unless the car is drifting of course)

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So this is one of those cases where you are right but you are stating something that on reflection becomes kind of meaningless.

You are stating that if the friction is zero then there is no difference between kinetic or static friction, and why would I use one over the other? And this is correct: if they are both zero and then who cares.

There is a weak argument that you should instead use static friction because it is the convention. So, this weak argument says you should treat a frictionless surface as a surface of arbitrarily low but non-zero friction, and then there is no problem. I call this argument weak because in practice, we don't do it with other things. For example when you learned about parallel lines you learned that they do not cross. You did not learn “they cross at the point at infinity.” The latter is a valid viewpoint, we use it all the time in complex analysis where we add a point at infinity and connect the entire complex plane into the Riemann sphere, but it's not how we teach geometry.

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