Can kinetic and static friction act at the same time for this object rotating inside an accelerating cylinder?

Question

Let say we have a rotating cylinder with constant angular velocity (w). On the inside wall there is an object which is rotating with the cylinder without sliding. And we know that the static friction has its maximum value. This part is the usual high school problem.

Now imagine we give a constant angular acceleration to the cylinder and its angular velocity starts increasing. I'm interested in the motion that object does afterwards.

Here is what I think: Since the friction is at the maximum value, when we increase the angular velocity of the cylinder, sliding will start and because of kinetic friction coefficient being less than that of static friction, the vertical component of friction will be less than the weight of the object and it will start going down as well as increasing its angular velocity (there's a frictional force component in the tangential direction).

The normal force will be more and more because of increasing angular velocity. And that means the force of friction will be increasing. Can this object's vertical velocity ever be zero because of this?

I want to understand what kind of motion the object may do assuming the height of the cylinder is tall enough and the object never reaches the bottom.

Update: Is it possible that once the angular velocity of the cylinder starts to increase and sliding between the cylinder and block occurs, the block does not move vertically because the static friction continues to balance the weight of the block while kinetic friction increases the velocity of the block by acting in the tangential direction?

But this requires static and kinetic friction acting at the same time. I don't think that is the case, but just an idea..

Answer 1

I tried to do it this rainy Sunday ! Please, be indulgent with my poor english (not my native language) and possible miscalculations!

We place ourselves in the frame linked to the cylinder. The position of the point $M$ in this frame of reference is marked by the angle $\theta $ and the vertical position $z$ on a descending vertical axis. We note ${{v}_{z}}=\frac{dz}{dt}$ and ${{v}_{\theta }}=\frac{d(R\theta )}{dt}$

The angular velocity of the cylinder is supposed to increase linearly with time $\omega (t)={{\omega }_{0}}(1+t/\tau )$. The initial angular velocity ${{\omega }_{0}}$ verifies ${{f}_{s}}m{{\omega }_{0}}^{2}R=mg$ with ${{f}_{s}}$ the static coefficient of friction. In addition to the weight and the reaction of the cylinder, one must take into account the inertia force $\overrightarrow{{{f}_{ie}}}=-m\frac{d\overrightarrow{\omega }}{dt}\wedge \overrightarrow{HM}+m{{\overrightarrow{\omega }}^{2}}\overrightarrow{HM}$ and and the Coriolis component $\overrightarrow{{{f}_{ic}}}=-2m\overrightarrow{\omega }\wedge \overrightarrow{v}$. In projection $\overrightarrow{{{f}_{ie}}}=-mR\frac{d\omega }{dt}\overrightarrow{{{e}_{\theta }}}+m{{\omega }^{2}}R\overrightarrow{{{e}_{r}}}$ and $\overrightarrow{{{f}_{ic}}}=+2m\omega \overrightarrow{{{e}_{\theta }}}$

The equations of the movement are :

$\frac{d{{v}_{z}}}{dt}=-g+{{T}_{z}}$

$\frac{d{{v}_{\theta }}}{dt}={{T}_{\theta }}-R\frac{d\omega }{dt}$

$0={{T}_{r}}+2\omega {{v}_{\theta }}+{{\omega }^{2}}R$

With $({{T}_{r}},{{T}_{\theta }},{{T}_{z}})$ the 3 components of the reaction, divided by $m$.

When the mass is slipping, if we write $f$ the coefficient of dynamic friction, the Coulomb law is written $({{T}_{\theta }}^{2}+{{T}_{z}}^{2})={{f}^{2}}{{T}_{r}}^{2}$ and ${{T}_{\theta }}/{{T}_{z}}={{v}_{\theta }}/{{v}_{z}}$. The signs will be fixed after. We deduce ${{T}_{z}}^{2}(1+{{({{v}_{\theta }}/{{v}_{z}})}^{2}})={{f}^{2}}{{T}_{r}}^{2}={{f}^{2}}{{(2\omega {{v}_{\theta }}+{{\omega }^{2}}R)}^{2}}$ or ${{T}_{z}}=-f(2\omega {{v}_{\theta }}+{{\omega }^{2}}R)\frac{{{v}_{z}}}{\sqrt{{{v}_{\theta }}^{2}+{{v}_{z}}^{2}}}$ with a negative sign because the mass begins to descend and therefore the reaction must be positive.

Similarly ${{T}_{\theta }}=-f(2\omega {{v}_{\theta }}+{{\omega }^{2}}R)\frac{{{v}_{\theta }}}{\sqrt{{{v}_{\theta }}^{2}+{{v}_{z}}^{2}}}$

So we have the differential system :

$\frac{d{{v}_{z}}}{dt}=-g-f(2\omega {{v}_{\theta }}+{{\omega }^{2}}R)\frac{{{v}_{z}}}{\sqrt{{{v}_{\theta }}^{2}+{{v}_{z}}^{2}}}$

$\frac{d{{v}_{\theta }}}{dt}=-f(2\omega {{v}_{\theta }}+{{\omega }^{2}}R)\frac{{{v}_{\theta }}}{\sqrt{{{v}_{\theta }}^{2}+{{v}_{z}}^{2}}}-R\frac{d\omega }{dt}$

To simplify a little, we will define dimensionless parameters: ${{V}_{r}}=\frac{{{v}_{r}}}{R{{\omega }_{0}}}$, ${{V}_{\theta }}=\frac{{{v}_{\theta }}}{R{{\omega }_{0}}}$, $t'={{\omega }_{0}}t$, $G=\frac{g}{{{\omega }_{0}}^{2}R}=\frac{1}{{{f}_{s}}}$ and $\beta =\frac{1}{{{\omega }_{0}}\tau }$. Hence the system, with only $f$, $fs$ and $\beta $ as only physical parameters :

$\frac{d{{V}_{z}}}{dt'}=-\frac{1}{{{f}_{s}}}-f(1+\beta t')(1+\beta t'+2{{V}_{\theta }})\frac{{{V}_{z}}}{\sqrt{{{V}_{\theta }}^{2}+{{V}_{z}}^{2}}}$

$\frac{d{{V}_{\theta }}}{dt'}=-f(1+\beta t')(1+\beta t'+2{{V}_{\theta }})\frac{{{V}_{\theta }}}{\sqrt{{{V}_{\theta }}^{2}+{{V}_{z}}^{2}}}-\beta $

For the rest, I used Python to solve this system. I don't want to be too long. I can give you the script if you need it. I joined the curves with ${{f}_{s}}=0.8$, ${{f}}=0.7$ and $\beta =0.1$

We find that the vertical and horizontal speeds increase and then decrease and the mass stops again.

By writing the static equilibrium condition, it is easy to verify that at this moment the static equilibrium conditions are satisfied.

Answer 2

I do not have the solution to your problem, which does not seem to be easy. But in Coulomb's law, I think it is not possible to suppose a "partial slip" in one direction. Ther is only one friction coefficient which can depend of the Sliding speed. If the mass slips, the norm of the tangential component $\sqrt{{{T}_{\theta }}^{2}+{{T}_{z}}^{2}}$ is $fN$. And the direction of $\overrightarrow{T}$ is opposite to the sliding speed $\frac{{{T}_{\theta }}}{{{T}_{z}}}=\frac{R\overset{\centerdot }{\mathop{\theta }}\,}{\overset{\centerdot }{\mathop{z}}\,}$ in the frame of the rotating cylindrer. Normally, it must make the problem mathematically complete.

There is also a problem of stability by putting oneself at the limit of the slip. Even if the angular velocity remains constant, a small negative vertical velocity destroys the initial equilibrium which is therefore an unstable one ?

Answer 3

Both cases are possible depending on the physical origin of the static friction, which is actually not a simple matter.

For example, a car wheel which slides sideways will have difficult time sticking to the ground in the froward direction as well. That means, sliding in any direction makes the static friction force 'switch' to the kinetic one.

However, one might carefully construct a surface that is so anisotopic that an object glides easily in the x direction while very hard to move in the y direction. In such a case, this object can experience kinetic and static friction in the same time.

Answer 4

I change this model ,so it is more "realistic" model , but i try to answer the question "Can kinetic (dynamic) and static friction forces act at the same time?"

We have to distinguish between two states , static state and dynamic state.

Static State

As long as the weight $mg$ is less then the static friction force , the mass will stick to the cylinder wall. The mass will start to move when:

$$m\,g > \mu_s\,F_s+\mu_s\,m\,\omega^2\,R\tag 1$$

with: $\mu_s$ static friction coefficient,$F_s$ spring force and $\omega$ the angular velocity of the cylinder.

With equation (1) we can calculate the angular velocity ,$\omega_0$,that required so the mass start to move.

$$\omega_0=\sqrt {{\frac {m\,g-\mu_{{s}}F_{{s}}}{\mu_{{s}}m\,R}}}\tag 2$$

$\omega_0$ must be greater then zero.

$\Rightarrow\quad $ $$m > \frac{\mu_s\,F_s}{g}\tag 3$$

Dynamic State

We can write the equation of motion for this state.

$$m\,\ddot{z}=-m\,g+\mu_d\left(\,m\,\omega(t)^2\,R+F_s\right)\tag 4$$

with $\mu_d$ the dynamic friction coefficient .

As long as $\omega(t) < \omega_0 $ the mass acceleration, velocity and deflection will be zero ( $\ddot{z}=0,\dot{z}=0\,z(t)=0)$

Simulation result $z(t) $

Conclusion:

The static friction force don't act together with the dynamic friction force, we have two separate states "static state" and "dynamic state" these described the equation of motion of the mass.

Remark

If the spring force is equal to zero (model without spring) equation (2) for $\omega_0$ is still valid, but there is no more restriction on the mass ,equation (3), so we get the same conclusion.

Edit:

The EOM's with applied torque $\tau$ on the cylinder

Start with the position vector $R_p$ of the mass :

$$ \vec{R_p}= \left[ \begin {array}{c} R\cos \left( \varphi \right) \\ R\sin \left( \varphi \right) \\ z\end {array} \right]$$

The velocity:$\quad \vec{v}=\vec{\dot{R}}_p$

so we can calculate the kinetic energy of the mass:

$T_m=\frac{1}{2}\,m\,\vec{v}^T\,\vec{v}$ and the potential energy of the mass
$V=m\,g\,z$

The kinetic energy of the cylinder is :

$T_c=\frac{1}{2}\,\Theta\,\dot{\varphi}^2$

The Lagrange function $L$ :

$$L=T_c+T_m-V=\frac{1}{2}\,m\,R^2\dot{\varphi}^2+\frac{1}{2}\,\Theta\,\dot{\varphi}^2+\frac{1}{2}\,m\,\dot{z}^2-\,m\,g\,z$$

The applied forces on the mass are:

$f_x=0$

$f_y=0$

$f_x=F_R$

with $F_R$ the friction force

$$F_R=\mu\,m\,R\,\dot{\varphi}^2$$

The Euler-Lagrange equations :

$$\ddot{\varphi}=\frac{\tau}{m\,R^2+\Theta}$$

$$\ddot{z}=\frac{-m\,g+\mu\,F_R}{m}$$

This give us the same conclusion as mention above