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I am confused as to how does static friction (analytically) causes rolling motion. As far as i understand For causing a rigid body to rotate about its centre of mass a torque must displace angularly, but the torque of static friction doesn't move angularly. So how does static friction actually cause pure rolling in a rigid body.

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  • $\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$
    – Community Bot
    Sep 4, 2022 at 14:58

2 Answers 2

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I’m not sure what you mean by “move angularly”, but the static friction force, when present, acts perpendicular to the radius of, say, a wheel at the point of contact with the surface producing a torque about the COM of the wheel.

Static friction is necessary to initiate or terminate rolling without slipping (i.e., necessary for acceleration or deceleration). It is only necessary for continued pure rolling if there is a net external force on the object. An example of the latter is rolling down an incline due to the force of gravity as shown in Fig 1 below. Or rolling with a torque applied to the wheel of say a car when it is accelerating or braking on a level surface as shown in FIG 2 below for the case of acceleration.

Static friction is neither produced nor needed for continued pure rolling where there are no external forces acting on the object. That's because static friction only exists to oppose an applied force to prevent relative motion (slipping) between surfaces. An example is rolling on a perfectly horizontal surface with constant angular and linear velocity and no external forces applied to the object (no applied torque, air resistance, rolling resistance, etc.)

Hope this helps.

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  • $\begingroup$ Those are some nice pictures, @BobD. $\endgroup$ Sep 5, 2022 at 0:34
  • $\begingroup$ @DavidWhite Thank you David $\endgroup$
    – Bob D
    Sep 5, 2022 at 0:37
  • $\begingroup$ What i meant by move angularly was that,when static friction acts to intiate rolling it has to do work,both linear and angular and both of which must be opposite in nature,so that the net work done by it is zero and the body is set into pure rolling , however work done by a torque is the integral of torque*angular displacement and since the torque of static friction doesn't displace angularly ,rotational work done by it must be zero,hence it shouldn't be able to rotate the body..and therfore it shouldn't be able to set the body in rolling $\endgroup$
    – Blz
    Sep 5, 2022 at 11:53
  • $\begingroup$ @Blz While static friction enables rotation without slipping, it does not do work giving the object rotational kinetic energy. I will update my answer to explain further in connection with the examples of the figures. $\endgroup$
    – Bob D
    Sep 5, 2022 at 13:29
  • $\begingroup$ Thank you for your quick reply,I meant to say that static friction kinda converts translational kinetic energy into rotational kinetic energy(if it initiates rolling motion),it has to do some equal and opposite rotational and translational work..but its torque doesn't displace angularly..Keenly waiting for your updates.. $\endgroup$
    – Blz
    Sep 5, 2022 at 13:48
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Before you try to understand pure rolling, try to understand what is impure rolling or rolling with slipping. When you give an impulse to a object, the center of mass translates in the direction of impulse provided. In such a scenario, the bottom-most point in contact with the floor it also has some velocity.

Now, analyse the what is the motive of friction - it opposes relative motion. So, the bottom-most point has some velocity and the surface has no velocity (assuming the surface is at rest); so friction will definitely act trying to reduce relative velocity of point of contact to 0 . Analyzing the motion from center of mass frame, the static friction is the only force causing a torque making the body have some angular velocity.

So, friction will keep acting until there is sliding that is until the bottom-most point have some non-zero relative velocity with respect to surface. In absence of sufficient static friction, it is impossible for the bottom-most point to have a zero relative velocity . Note - friction will stop acting the moment relative velocity of the lowermost point becomes zero as there is no relative motion; so no kind of friction acts after a body starts to purely roll

Also, if the rolling was to be done on an inclined plane, friction won't be zero even after relative velocity and relative acceleration of lowermost point becomes zero. This is because friction will help to achieve the relative velocity/acceleration to be zero, but since the gravity causes the body to keep on accelerating and to keep on increasing speed, friction needs to keep on acting to maintain the relative velocity to be zero by the same aforesaid algorithm. In case of a plane horizontal surface, this friction wasn't needed to maintain pure rolling as there was no other force to change relative motion.

Edit - With reference to the comments with other users, I think you what you try to mean is correct. If we just let a sphere roll down a frictionless incline - it will have no angular velocity and the entire potential energy gets converted into translational kinetic energy; however if the incline has a friction sufficient for the object to purely roll, the kinetic energy the sphere finally gets after being rolled down from the same height is identical to the initial case. The difference is initially it had more translational kinetic energy than former and in the former case it has rotational kinetic energy equal to the difference of the translational kinetic energy

It summarizes to the fact that friction does do translational and rotational works while rolling which are opposite in sign but equal in magnitude. A rigorous way to justify that is friction in a small interval does a translational work - F.ds and an rotational work τ.dθ = (r x F).(1/r x ds) = F.ds (all analysis from center of mass frame). Note this is possible only if it is purely rolling otherwise the dθ (angular displacement) wouldn't have been equal to 1/r x ds (linear displacement/r).

Kindly ignore my poor editing skills.

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  • $\begingroup$ Thank you for your inputs could you please also put forward your opinions on my replies to @Bob D ..i would love to read multiple answers😁 $\endgroup$
    – Blz
    Sep 5, 2022 at 13:59
  • $\begingroup$ I tried my best to put forward my ideas in the latest edit. $\endgroup$
    – Sam
    Sep 5, 2022 at 15:15
  • $\begingroup$ Thank you..the edit explained exactly what i was thinking 😁 $\endgroup$
    – Blz
    Sep 5, 2022 at 15:54

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