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In my physics course, we are doing an experiment rolling disks and spheres down an incline (assuming there is no slipping). In doing the derivations (assuming a moment of inertia of $\frac25mR^2$ for sphere and $\frac12mR^2$ for disk) I have derived that the final velocity should be $\sqrt{\frac{4gh}{3}}$ for the disk and its acceleration should be $\frac{2g\sin{\theta}}{3}$. I have derived similar equations for the sphere. My question relates to the coefficient of static friction.

Logically, I think that when the coefficient of static friction increases, it should increase the force that is applying torque to the rolling object and thus increase the final speed at which the object is rolling. This should result in a lower final velocity since more of the initial energy ($mgh$) is "allocated" to the rotational kinetic energy as opposed to the translational kinetic energy.

Yet, according to the derivations, the final velocity does not depend on the coefficient of static friction. Why is this? I believe my equations are correct so I know that the final velocity is entirely determined by the object's moment of inertia, and the angle of the incline, but I intuitively think that a higher coefficient of friction should change how much energy is "allocated" between translational and rotational kinetic energy.

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    $\begingroup$ To be strictly correct, problems like this ought to include a step where you determine if it will slip or not. But that is conceptualy difficult for students first meeting the problem and if it slipts the mathematical difficulty goes up so it is usually just assumed in the problem statement. $\endgroup$ – dmckee Oct 21 at 23:46
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It is a common mistake among introductory physics students to assume that the coefficient of static friction $\mu_s$ determines the strength of the static friction force $F_s$ as expressed through the equation $F_s=\mu_sN$, where $N$ is the normal force interaction between the two surfaces in question. This equation is actually false except for a very specific scenario.

An easy way to see that this is false is to imagine the scenario of a book sitting on a table with no other forces acting on it other than its own weight and the normal force from the table. Therefore, $\mu_sN>0$, as both $\mu_s$ and $N$ are greater than $0$. Yet is there a static friction force acting on the book? Of course not. If there was we would see the book accelerating in the direction of this friction force due to Newton's second law, and I typically don't see my books fly away from me after I put them on a table.

The correct understanding of the static friction force and $\mu_s$ is that $\mu_sN$ determines the maximum static friction force before the object starts sliding. In other words, it is always true that $$F_s\leq\mu_sN$$ We can only say $F_s=\mu_sN$ when the surfaces are just at the threshold of slipping. If we are not there, then the static friction force is whatever it needs to be to prevent slipping. In the case of the book, you can imaging pushing horizontally on it with a force $F$ that slowly increases from $0$. In this case, $F_s=F$ until $F=\mu_sN$, at which point the book will start sliding across the table. Notice then that $F_s=\mu_sN$ only at the point of slipping.

In your case of the rolling objects, the same idea holds. If your object is rolling without slipping, then all you can say is $F_s\leq\mu_sN$. You cannot say $F_s=\mu_sN$ unless you know your object is on the verge of slipping. This is why you don't see a dependence on $\mu_s$ in your derivations. It is incorrect to say $F_s=\mu_sN$. The force of friction would not change for a different $\mu_s$, only the point of slipping would change. For example, if you did experiments to see find the maximum angle the incline could be at before slipping occurred, you would find a dependence on $\mu_s$ (shown below). This is also why you don't get different "energy allocation": the static friction force does not depend on $\mu_s$ if there is no slipping.


How do you correctly do problems with static friction then? Well, just bring $F_s$ along for the ride without assuming any value for it. Then at the end, if it is valid, you can substitute in $F_s=\mu_sN$. I will show you how this is done through the example below for the object rolling down an incline (Similar analysis of an object rolling on a flat surface can be found here and here).

You can determine what the friction force needs to be so that slipping does not occur through using Newton's second law and imposing rolling without slipping. Along the incline at an angle $\theta$ with the horizontal we have friction acting up the incline and a component of gravity acting down the incline. Therefore, $$\sum F=ma=F_s-mg\sin\theta$$ where $a$ is the linear acceleration of the object, and $m$ is the mass of the object.

Additionally, friction exerts a torque about the center of the rolling object, so we also have $$\sum\tau=I\alpha=F_sR$$ where $I$ is the moment of inertia of the object and $\alpha$ is the angular acceleration of the object (both about the object's center), and $R$ is the radius of the object.

Imposing our no slipping condition $a=-\alpha R$ (the negative sign is needed due to the sign conventions in the above equations), we can determine $F_s$ needed for no slipping to occur: $$F_s=\frac{Img\sin\theta}{I+mR^2}$$

If we are working with nice objects such that $I=\gamma mR^2$, this reduces to $$F_s=\frac{mg\sin\theta}{1+1/\gamma}$$

If you see that your object is not slipping, then this is the value of the static friction force acting on your object. We do not have any reference to $\mu_s$ yet because we have not assumed we are at the verge of slipping. If you wanted to find that maximum angle though, then you would say $F_s=\mu_sN=\mu_smg\cos\theta_\text{max}$ and arrive at

$$\tan\theta_\text{max}=\left(1+\frac1\gamma\right)\mu_s$$

Note that for objects that cannot rotate ($\gamma\to\infty$), we arrive at the usual equations for the static friction force and maximum angle before slipping for a block on an incline.

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The coefficient of static friction predicts the maximum possible value of static friction. Static friction can have any value below that. If the required value exceeds the maximum, the the friction becomes kinetic (usually smaller).

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