0
$\begingroup$

I am trying to solve a problem from the book "Introduction to mechanics, matter and waves" by Uno Ingard.

Its about a weightless piston that pushes with a horizontal force F a uniform rigid cylinder of mass M and radius R, it also says that there is friction between the piston and the cylinder, and the cylinder is rolling without slipping. I'm having trouble writing the equations of motion.

Given that the cylinder is rolling without slipping, there is a friction force between the cylinder and the horizontal plane, and because of the friction between the piston and the cylinder we have two frictions forces, a horizontal (horizontal plane) and a vertical one (piston).

My question is, does the friction between the piston and the cylinder affect the friction between the cylinder on the horizontal plane?

If I wanted to write the equations of motion from a frame of reference Q at the point of contact between the cylinder and the horizontal plane, which directions should the friction forces have and how many forces would I have?

An image of the problem

$\endgroup$
1
  • $\begingroup$ can we assume there is no friction between the piston and the "cylinder" within which it moves? $\endgroup$
    – Bob D
    Commented Jul 14, 2023 at 19:14

3 Answers 3

1
$\begingroup$

My question is, does the friction between the piston and the cylinder affect the friction between the cylinder on the horizontal plane?

Consider the following, then come to your own conclusion as to whether or not the friction forces affect one another.

The figure below is a free body diagram (FBD) of the cylinder which assumes there is no kinetic friction between the piston shaft and the "cylinder" housing it, so that the applied force $F$ is the normal force between the piston face and cylinder surface. $f_B$ is the kinetic friction force acting downward providing a counter clockwise torque on the cylinder, whereas $f_B$ is the static friction force between the cylinder and the horizontal plane provides a clockwise torque on the cylinder.

The kinetic friction force depends only on the applied force $F$ which is the normal force between the piston and cylinder and the coefficient of kinetic friction between those surfaces. The static friction force is only needed to meet the non slip conditions (discussed below). The maximum possible static friction force depends on the weight of the cylinder and the coefficient of static friction between the surfaces.

It should be noted that the two friction forces are related to the extent that that the maximum possible static friction force between the cylinder and horizontal plane, which is $\mu_{s}Mg$ where $\mu_s$ is the coefficient of static friction between the surfaces, cannot be exceeded by the applied force $F$. If the applied force $F$ exceeds that, friction becomes kinetic and the cylinder will slip.

As indicated above, the magnitude of the static friction force $f_A$ is only that needed for the non slip condition which is

$$a=\alpha R$$

Where $a$ is the linear horizontal acceleration of the cylinder and $\alpha$ its angular acceleration. Thus we have, applying Newton's 2nd law for linear and angular acceleration:

$$\frac{F-f_{A}}{M}= \frac{(f_{A}-\mu_{k}F)}{I}R^2$$

Where $I$ is the moment of inertia of the cylinder.

Hope this helps.

enter image description here

$\endgroup$
10
  • $\begingroup$ $I\propto MR^2$ and so your equation is dimensionally inconsistent. Missing a factor of R above. $\endgroup$ Commented Jul 15, 2023 at 2:18
  • $\begingroup$ Also, you should not be basically directly solving the question. $\endgroup$ Commented Jul 15, 2023 at 2:24
  • $\begingroup$ Now your torque and angular acceleration are correct, but this is not linear acceleration. If you substitute into the correct relation that you had in your original answer, you will see that there will be $\frac1IR^2$, not $\frac1IR$ that you originally had in the answer. As for the friction forces, I already told you. You can vary $f_B$ any time by putting lubricants or something, and that will cause a change in $\mu_k$. The equation you derived shows that $f_A$ explicitly depends upon $\mu_k$, i.e. depends upon $f_B$, even though $f_B$ does not depend upon $f_A$. You are just wrong. $\endgroup$ Commented Jul 15, 2023 at 3:09
  • $\begingroup$ @naturallyInconsistent “but this is not linear acceleration”. Not sure what you mean. I didn’t say torque and angular acceleration are liner acceleration. The linear acceleration is the horizontal acceleration of the COM and is on the left side of the last equation in my answer. $\endgroup$
    – Bob D
    Commented Jul 15, 2023 at 15:00
  • $\begingroup$ “If you substitute into the correct relation that you had in your original answer, you will see that there will be $\frac{1}{I}R^2$, not $\frac{1}{I}R$ that you originally had in the answer.” Yup. I inadvertently dropped the $R$ from the equation $a=\alpha R$. Corrected. Thanks. $\endgroup$
    – Bob D
    Commented Jul 15, 2023 at 15:01
1
$\begingroup$

My question is, does the friction between the piston and the cylinder affect the friction between the cylinder on the horizontal plane?

Yes, the friction between the piston and the cylinder will change the magnitude of the friction between the cylinder and ground. To see this, you have to consider rotational behaviour.

which directions should the friction forces have and how many forces would I have?

What you should do, is that you actually do know how the cylinder will be moving, and thus you can directly work out which slippage directions the cylinder will be having, and thus work out the directions of the frictions involved.

$\endgroup$
2
  • 1
    $\begingroup$ "Yes, the friction between the piston and the cylinder will change the magnitude of the friction between the cylinder and ground. To see this, you have to consider rotational behaviour." How is that so? I could be wrong, but it seems to me the kinetic friction force between the piston and cylinder is independent of the static friction force between the cylinder and horizontal plane. The only connection between the two that I see is that the applied force F cannot exceed the maximum possible static friction force between the cylinder and horizontal plane. $\endgroup$
    – Bob D
    Commented Jul 14, 2023 at 20:30
  • $\begingroup$ @Alv in BobD's answer below, he himself derived that the magnitude of $f_A(\leqslant\mu_sMg,$ not always equal) depended upon $\mu_k$ for fixed $F$; I was careful to state it in a way that conferred the directionality of the dependence too, so I am lost as to why both of you got it wrong. $\endgroup$ Commented Jul 15, 2023 at 2:23
0
$\begingroup$

enter image description here

From the FBD you obtain two equations

$$\sum F_x=0=m\,a+F_c+F_{\mu H}-F$$ where $~F_c~$ is the constraint force at the contact point A. $$\sum \tau_z=0=I_z\,\alpha+\left(F_{\mu V}-F_{\mu H}-F_c\right)$$

with the rolling condition

$$a-R\,\alpha=0$$

you obtain 3 equations for the unknows $~a,\alpha\,,F_c~$

$$\alpha=R\,\frac{F-F_{\mu V}}{I_z+m\,R^2}\quad ,a=R\,\alpha$$

the vertical friction force $~F_{\mu V}$ can taken as a function of the cylinder rotation $~\varphi~$

Remark

because the rolling condition the horizontal friction force $~F_{\mu H}~$ doesn't affected the rotation (or the translation) of the cylinder

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.