Direction of static friction in pure rolling up an incline?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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)

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

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@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 !!!