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Consider a simple situation where a round body is rolling up an incline plane and hence it is under the influence of gravity...

My physics teacher says that to identify the direction of friction, we should imagine a situation where there is no friction, identify where the slipping occurs and the friction will act oppose the slipping.

This works when the body is slipping down an incline because then the friction acts up the incline which makes sense because its torque supports the increasing angular velocity due to gravity.

But in case of pure rolling up an incline it is seen that friction again acts up the incline. But if we imagine a situation without friction it is seen that the bottommost point slips up the incline and to ensure that pure rolling occurs, the friction must act down the incline so that the bottommost point is at rest w.r.t the incline.

So why does friction act up the incline if a body is pure rolling up an incline under the influence of gravity?

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  • $\begingroup$ The friction force act always opposite to the velocity of the body $\endgroup$
    – Eli
    Nov 30, 2022 at 8:32
  • $\begingroup$ If an object rolls without slipping, then the translational speed is tied to the angular velocity about the center of mass, and the the translational acceleration is tied to angular acceleration about the center of mass. If you draw a free-body diagram of the object on the incline you will see that neither gravity or normal can provide a torque about the center of mass. So, the angular acceleration must be caused by the torque of static friction. If acceleration points down the incline (it always does, whether rolling up or down), then $\vec{F}_{fs}$ must point up the incline. $\endgroup$
    – Ben H
    Nov 30, 2022 at 13:13
  • $\begingroup$ @BenH I understand this much... But if we imagine a situation without friction it is seen that the bottommost point slips up the incline and to ensure that pure rolling occurs, the friction must act down the incline so that the bottommost point is at rest w.r.t the incline. and if the friction acts up the incline then how does the bottommost point remain at rest wrt incline?? $\endgroup$ Nov 30, 2022 at 14:24
  • $\begingroup$ The only case I can think of with a down-the-incline friction force is if the incline is half frictionless and half frictional. If a ball slides up halfway (without any rotation) and then encounters friction, the angular acceleration will have to be such that the rotation "catches up" to the translation motion of the ball. Then the frictional force will point down the incline, until the ball begins to roll without slipping (but I think it will be a kinetic frictional force that accomplishes this transition, since the ball will slide on the frictional surface before r-wo-slipping). $\endgroup$
    – Ben H
    Nov 30, 2022 at 15:30
  • $\begingroup$ imagine a situation where there is no friction, identify where the slipping occurs and the friction will act oppose the slipping ... This is often a useful approach for static friction, but I don't find it very helpful for rolling. I think the approach I described above (and in my answer) is easier to see: the frictional force directed such that the necessary angular acceleration is accomplished. $\endgroup$
    – Ben H
    Nov 30, 2022 at 15:34

4 Answers 4

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Always start from principles of dynamics, when you have some doubts.

Let's write the equations of motion of the round rigid body rolling on the inclined surface with angle $\theta$ w.r.t. the horizontal direction,

  • translation: $m \dot{\mathbf{v}} = \mathbf{F}^{ext}$, being $\mathbf{v}$ the velocity of the center of mass;
  • rotation: $\dot{\boldsymbol{\Gamma}}_H = - \dot{\mathbf{x}}_H \times \mathbf{Q} + \mathbf{M}_H^{ext}$, or using the center of mass as the pole $H \equiv G$, $\dot{\boldsymbol{\Gamma}}_G = \mathbf{M}_H^{ext}$.

We can use a reference frame with $x$-axis aligned with the surface of the inclined plane, $y$-axis orthogonal to it, and $z$-axis pointing out of the plane being the axis of rotation of the body.

The momentum of the body reads $m \mathbf{v} = m \dot{x} \mathbf{\hat{x}}$, its derivative $m \dot{\mathbf{v}} = m \ddot{x} \mathbf{\hat{x}}$; its angular momentum reads $\boldsymbol{\Gamma}_G = I \dot{\theta} \mathbf{\hat{z}}$ and its time derivative $\dot{\boldsymbol{\Gamma}}_G = I \ddot{\theta} \mathbf{\hat{z}}$; external forces acting on the body are due to the gravitational field, namely its weight $m \mathbf{g} = -m g \sin \alpha \mathbf{\hat{x}} - m g \cos \alpha \mathbf{\hat{y}}$, and the reaction with the rolling surface that can be written as the sum of the normal and parallel components, namely $\mathbf{F}_R = F^f \mathbf{\mathbf{x}} + N \mathbf{\hat{y}}$, namely the normal reaction $N$ taken as positive if pointing in $\mathbf{\hat{y}}$-direction towards the center of the rolling body, and $F^f$ taken positive if pointing in $\mathbf{\hat{x}}$-direction; the weight acts at the center of mass, the reaction at the point of contact.

The kinematic constraint of pure rolling gives us $\dot{x} = R \dot{\theta}$, being $R$ the radius of the rolling body

Now, writing the components of the equation of motions

  • $x$-momentum: $ m R \ddot{\theta} = -mg \sin \alpha + F^f$
  • $y$-momentum: $ 0 = N - m g \cos \alpha$
  • rotation: $I \ddot{\theta} = - F^f R$,

so that we get

  • $( m R^2 + I) \ddot \theta = - mg R \sin \alpha$, and thus $\theta(t) = - \dfrac{1}{2} \dfrac{mg}{mR^2 + I} \sin \alpha \cdot t^2 + \Omega_{0} t + \theta_{0}$
  • $N = m g \cos \alpha$,
  • $F^f = - \dfrac{I}{R} \ddot \theta = \dfrac{I}{mR^2 + I} mg \sin \alpha$.

As you can see, the component of friction in $x$-direction, $F^f$, has the same sign of $\sin \alpha$, i.e.

  • if the body is "climbing up to a hill" $\sin \alpha > 0 $ and thus $F^f > 0$ points towards $x$-direction;
  • if the body is "rolling down" along direction $x$, $\sin \alpha < 0 $ and thus $F^f < 0$ points towards $-x$-direction.
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  • $\begingroup$ It looks like you conclude that the direction of $\vec{F}_{fs}$ depends on whether the object is rolling up the incline or rolling down the incline. This is not the case. $\endgroup$
    – Ben H
    Nov 30, 2022 at 13:20
  • $\begingroup$ it looks like you have to read the answer without extrapolating. Friction acts like this IN THIS VERY PROBLEM $\endgroup$
    – basics
    Nov 30, 2022 at 13:30
  • $\begingroup$ Sorry, I just cannot understand your answer. I wasn't extrapolating, just seeing "up the hill ... points towards $x$-direction" and "rolling down ... points towards $-x$ direction". Is that not what you conclude? Or is your $x$ direction changing depending on the scenario? $\endgroup$
    – Ben H
    Nov 30, 2022 at 13:32
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To make it concrete, imagine the incline as a wedge with vertical side to the right. If an object rolls down the wedge then the translational acceleration points down the ramp to the left. Both the angular velocity and the angular accleration, about the center of mass, are CCW, since it is rolling that way and speeding up. This requires a "CCW torque" about the center of mass. So, static friction, acting at the base of the object, points up the ramp (to the right). If the object is, instead, rolling up the ramp, the acceleration vector still points down the ramp (the object is slowing). Angular velocity is now CW, but the angular acceleration is still CCW (the object is slowing so angular acceleration is opposite to its velocity). So, again to give that CCW torque, the force of static friction must still point up the ramp.

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My physics teacher says that to identify the direction of friction, we should imagine a situation where there is no friction, identify where the slipping occurs and the friction will act oppose the slipping.

I think it's best just to assume this is incorrect or incomplete. Friction will be directed up the incline in this case.

Perhaps your teacher's logic still holds if you consider what the cylinder would do if there was no initial velocity.

The frictional force pushing up the ramp is what allows the cylinder to reach higher than it could on the influence of its linear kinetic energy alone.

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  • $\begingroup$ But still...if friction opposes slipping and the body would have slipped up the incline then shouldn't friction act down the incline? $\endgroup$ Nov 30, 2022 at 7:09
  • $\begingroup$ -1; your answer is wrong $\endgroup$
    – basics
    Nov 30, 2022 at 7:18
  • $\begingroup$ Doesn't it depend if the mass is pulled up the incline by an external force or if it has something that drives it upwards (imagine on a two whel drive car the front and rear axle, they have opposite directions of static friction in my opinion) $\endgroup$
    – kruemi
    Nov 30, 2022 at 8:27
  • $\begingroup$ @basics no, it's not, you're just wrong. Friction is directed up the incline for a rolling cylinder slowing down. Friction is what converts the rotational energy into translational KE. $\endgroup$
    – Señor O
    Nov 30, 2022 at 18:12
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enter image description here

First choose the direction of the velocity $~v~$. the direction of the friction force $~F_\mu~$ is always opposite to the velocity direction.

from the free body diagram you obtain

$$\sum F_{\rm CM}=m\,\dot v+F_\mu-F+m\,g\,\sin(\alpha)=0\tag 1$$ $$\sum \tau_{\rm CM}=I\,\ddot\phi-F_\mu\,R=0\tag 2$$

three cases

  • rolling condition $~\Rightarrow~ \dot v=R\,\ddot\phi$
  • static friction force $~\Rightarrow~ F_\mu=\mu\,m\,g\,\cos(\alpha)$
  • dynamic force (tire force)$~\Rightarrow~ F_\mu=-C_s\,\frac{\dot\phi\,R-v}{|\dot\phi\,R|}$

for any case , you obtain 3 equations for the 3 unknows $~\dot v~,\ddot\phi~,F_\mu$

to make the equation "work" for inclined up or down (static friction)

$$F_\mu\mapsto F_\mu\,\rm {signum}(v)$$


@Ben H

solve my equation (1),(2) and the rolling condition $~\dot v=R\,\ddot\phi$

you obtain

$$\dot v={\frac {{R}^{2} \left( F-mg\sin \left( \alpha \right) \right) }{m{R}^ {2}+I}} $$ $$\ddot\phi={\frac { \left( F-mg\sin \left( \alpha \right) \right) R}{m{R}^{2}+I} } $$

your equation (2) is $$\sum \tau_{\rm CM}=I\,\ddot\phi+F_\mu\,R=0$$

thus $$\dot v={\frac {{R}^{2} \left( F-mg\sin \left( \alpha \right) \right) }{m{R}^ {2}-I}} $$

$$\ddot\phi={\frac { \left( F-mg\sin \left( \alpha \right) \right) R}{m{R}^{2}-I} } $$

which is wrong !!!

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  • $\begingroup$ It is true that the force of kinetic friction is opposite to the velocity vector (the velocity of the object with respect to the surface on which it slides). But the same rule cannot be used for static friction: there is no sliding of surfaces. Here static friction points up the incline, as shown in my answer. $\endgroup$
    – Ben H
    Dec 1, 2022 at 23:38
  • $\begingroup$ (regardless of the direction of motion) $\endgroup$
    – Ben H
    Dec 1, 2022 at 23:45
  • $\begingroup$ @BenH I think that this EOM is correct $\dot v=\frac 1m(F-F_\mu-m\,g\,\cos(\alpha))$ there is no difference sign for static friction. the direction of the friction force is always opposite to the velocity direction. with friction force you obtain less acceleration. your unter vote my answer is wrong!! $\endgroup$
    – Eli
    Dec 2, 2022 at 12:31
  • $\begingroup$ Sorry, it's just not true that the direction of friction is always opposite to the velocity direction. For static friction there is no velocity, so it can't be true. If you need a more authoritative source for the static friction pointing up the ramp, here is an image of the example from the Young and Freedman textbook. $\endgroup$
    – Ben H
    Dec 2, 2022 at 15:14
  • $\begingroup$ So What is wrong with my equation?. You dont need velocity for the friction force, just die direction. Can you please write your equation, so we can discuss about it? $\endgroup$
    – Eli
    Dec 2, 2022 at 15:24

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