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In school we solve a lot of physics problems involving rolling cylinders/hoops/wheels. Often it is specified in the question statement that "rolling can be assumed to occur without slipping". Then we know we can use the relationship between translational acceleration and rotational acceleration: $a = \alpha r$.

But can we know from a particular set up when the object can be assumed not to slip when rolling? Assume we know the usual macroscopic details of the set up such as the mass and radius of the rolling object, the angle of the plane, friction coefficients etc.. without deep insight into specifics of the materials of the wheel and the plane.

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  • $\begingroup$ "Assume we know the usual macroscopic details of the set up such as the mass and radius of the rolling object, the angle of the plane, friction coefficients etc.. without deep insight into specifics of the materials of the wheel and the plane." Yes. Then all accelerations can be calculated, using Newton's 2nd, basically. $\endgroup$ – Gert Apr 4 at 18:09
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Let's first look at an example you are probably familiar with: a block on an incline with friction. If we assume the block is stationary due to static friction, then it is easy to show that $$f=mg\sin\theta$$ where $m$ is the mass of the block, $g$ is the acceleration due to gravity near the Earth's surface, $\theta$ is the angle the incline makes with the horizontal, and $f$ is the static friction force.

Now, if we could imagine slowly increasing the angle of the incline $\theta$ we could find that eventually we hit a threshold where static friction is no longer strong enough to hold the block up, and the block will then accelerate down the incline. In other words, there exists some critical angle $\theta_{crit}$ such that $$mg\sin\theta_{crit}=f_{max}$$ where $f_{max}$ is the largest the static friction force can be.$^*$

Now, let's look at your question about rolling without slipping. Let's say we have a round object (could be a ball, cylinder, etc.) on a surface with friction and we apply a force $F$ tangentially to the top of it to the right. Of course, if our applied force is small enough the object will roll without slipping, with friction acting on the bottom of the object to the left.

Looking at this scenario, we can see that the linear acceleration of the object will be given by $$\sum F=ma=F-f$$ and the angular acceleration is given by $$\sum\tau=I\alpha=(F+f)R$$ where $I$ is the moment of inertia of our object and $\alpha$ is its angular acceleration, both about the center of the object.

Now, if we want rolling without slipping, then as you said $a=R\alpha$, and so solving the system of equations (left to you) we have $$f=\frac{mR^2-I}{mR^2+I}F$$ Furthermore, if we have nice objects like cylinders or spheres, we also have $I=\gamma mR^2$, therefore $$f=\frac{1-\gamma}{1+\gamma}F$$

What this means is that the friction force has to be equal to this value in oder for rolling without slipping to occur. Now, just like in the incline example, we can think about increasing the applied force $F$ until slipping occurs. In other words, there exists some force $F_{crit}$ such that $$f_{max}=\frac{1-\gamma}{1+\gamma}F_{crit}$$

In comparison to the incline example, $\frac{1-\gamma}{1+\gamma}F$ is analagous to $mg\sin\theta$. So, we know that there will be rolling without slipping if the frictional force is not required to be too large in order to have rolling without slipping. If the required frictional force is too large, then the object is going to slip.$^{**}$


$^*$ Note that sometimes you see people assume the maximum static friction force is proportional to the normal force between the block and the incline, i.e. $f_{max}=\mu N=\mu mg\cos\theta_{crit}$, in which case $$\mu=\tan\theta_{crit}$$ but assumptions about how the friction force actually works is not important for our discussion here. All we need to understand is that there is some maximum value it can have before "failing".

$^{**}$ This is somewhat of a specific example. You can show that if the force is instead applied at some distance $\beta R$ above center of the object ($\beta>0$), then the required friction force to have rolling without slipping is $$f=\frac{\beta-\gamma}{1+\gamma}F$$

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  • $\begingroup$ May I ask what is $\gamma$? $\endgroup$ – S. Rotos Apr 4 at 18:16
  • $\begingroup$ @S.Rotos It is some constant. For example for a cylinder $\gamma=1/2$ (i.e. $I=1/2mR^2$). For a hoop $\gamma=1$, etc. $\endgroup$ – Aaron Stevens Apr 4 at 18:17
  • $\begingroup$ Ah, of course! I asked too hastily. Thank you for the answer! $\endgroup$ – S. Rotos Apr 4 at 18:18
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No slipping is when the velocity of rolling object at the point of contact with surface is same as velocity of surface. For ground, the velocity of surface is zero, so we say the velocity of lowermost point of wheel is zero. The same is applicable to rolling on a moving surface, for a belt moving around a wheel, or two wheels/gears rotating with each other. In general , it is the friction with helps rolling(or there may be rolling motion provided initially) We check if the STATIC friction is enough to permit case of no slipping i.e., $v=r\omega$ in case friction is present and there is slipping initially. If friction is enough then there will be no slipping. Else there will be slipping and friction is dynamic. If there is no friction, then no change in velocities, if $v=r\omega$ then no slipping else slipping occurs.

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This is how I have approached these problems

  • Assume there is no slipping and calculate the required friction force to enforce this constraint.
  • Check that the friction force is less than the available traction. This is usually done in scalar form $\| \vec{F} \| \leq \mu \| \vec{N} \|$ where $\mu$ is whatever static coefficient of friction is appropriate for the materials.
  • If the friction exceeds traction, then the problem changes, and you de-couple the rotational and translational degrees of freedom, but friction force is no longer an unknown, but it is set to oppose the relative motion $ \vec{F} = - \mu \| \vec{N} \| \frac{\vec{v}}{\| \vec{v} \| }$. Re-solving is needed at this point.
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