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Recently I encountered a misunderstanding with applying Newton's second law to rotational dynamics. Imagine a yoyo of radius $R$ on a plane (with non-zero friction) being pulled towards the right.

YoYo

Then imagine that the force applied at some inner radius $b$ (red circle) will be precisely equal to that of the magnitude of the static friction $f$ on the yoyo at the contact point with the ground.

$f=Mg\mu$

$F=f$

Then, if we sum the torques on the yoyo about the contact point with the ground we get:

$\tau=-(R-b)F$

Clearly, there is some nonzero torque which will cause some sort of angular acceleration, and thus, a translational acceleration along the plane. Summing the torques about the circles center of mass yeilds the same result; a non-zero torque being applied onto the yoyo causing rotation.

But when we apply Newton's second law to the system, my understanding begins to fall apart. $F=ma$ applied on the yoyo's center of mass yields

$F_{net}=ma=F-f=0$

It sums to zero, since as we remember $F=f$. This suggests no translational acceleration whatsoever, in conflict with the result we derived using rotational dynamics. (Remember: all forces applied to a rigid body can be treated as being applied to its center of mass).

How is this conflict between $F\propto a$ and $\tau\propto\alpha$ then explained?

TLDR: I know I am making a mistake using $F=ma$ somewhere, but I'm not sure where. Could somebody point it out with an intuitive explanation and a resolution of the two equations that adds up?

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  • $\begingroup$ $f=Mg\mu$ is the limiting static friction force, I.e, where the yo yo loses traction $\endgroup$
    – Bob D
    Commented Feb 2 at 9:59
  • $\begingroup$ In such a case the condition $a=R\alpha$, that is, rotation without slipping, is not valid. $\endgroup$
    – Wolphram jonny
    Commented Feb 2 at 14:35

3 Answers 3

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Then imagine that the force applied at some inner radius $b$ (red circle) will be precisely equal to that of the magnitude of the static friction $f$ on the yoyo at the contact point with the ground.

Imagining that the applied force $F$ is precisely equal to the magnitude of the static friction force $f$ doesn't necessarily mean it is possible. In this case, the applied force $F$ will be greater than the static friction force $f$.

The conditions that must be met are as follows:

$$a_{com}=\frac{F-f}{M}\tag{1}$$

$$\alpha=\frac{\tau_{net}}{I_{com}}\tag{2}$$

$$\tau_{net}=Rf-bF\tag{3}$$

$$a_{com}=R\alpha\tag{4}$$

$$f_{max}=\mu_{s}Mg\tag{5}$$

where

$a_{com}=$ the translational acceleration of the center of mass (COM)

$\alpha=$ the angular acceleration about the COM and

$\tau_{net}=$ the net torque about the COM

$I_{com}=$the moment of inertia of the yoyo about its COM

The first two equations are from Newton's 2nd law applied to translational and rotational acceleration, respectively. Equation (4) is the required relationship between translational and rotational acceleration for rolling without slipping. Equation (5) gives the maximum possible static friction force where slipping is impending.

Combining equations (1) through (4) yields the following relationship between the applied force $F$ and the static friction forceh $f$

$$F=f\biggl (\frac{R+I_{com}/MR}{b+I_{com}/MR}\biggr )\tag{6}$$

Since $R\gt b$, the equation tells us that the applied force $F$ must be greater than the static friction force $f$.

Moreover, since the maximum possible static friction force $f$ is given by equation (5), the maximum possible applied force, $F_{max}$ becomes

$$F_{max}=\mu Mg\biggl (\frac{R+I_{com}/MR}{b+I_{com}/MR}\biggr )\tag{7}$$

Hope this helps.

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  • $\begingroup$ Oh, I think I understand what is meant. When F = Mg(\mu), in other words F is the maximum friction value, the ACTUAL friction force is not yet at its maximum value. Hence, a_com is nonzero and the circles COM translates. The friction reaches its maximum value when F_max (7) is applied, at which point it begins to slip. Is my understanding correct? $\endgroup$
    – Stefan Karwasiecki
    Commented Feb 2 at 22:19
  • $\begingroup$ You are absolutely right. $\mu Mg$ is the maximum possible value of the static friction force to have angular and translational acceleration without slipping at which point the friction becomes kinetic. It is not necessarily the actual static friction force. The actual static friction force adjusts on the basis of the applied force and the other constraints of the problem, in particular equation (4) of my answer. $\endgroup$
    – Bob D
    Commented Feb 2 at 22:26
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When you apply the Newton's second law to a rotational system you should use the form:

$$I\ddot \theta = \sum_i T_i$$

Where $I$ is the moment of inertia of the object, $\ddot \theta$ is the angular acceleration and $T$ is the torque applied to the body.

You can not use the form $ma=F_{net}$ as you did because that is the linear translational analogue.

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If $f = F$ the yoyo rotates in clockwise direction without translational movement ($a = 0$). (It is easy to see if $f$ is supposed to be another applied force instead of a friction force). But in this case there is slipping and instead of static friction, we have kinetic friction between the wheel and the ground, and that force has the same direction of $F$. So the equation should be:

$$F + K_f = 0$$

What is impossible unless both are zero.

The conclusion is that the static friction force can not be equal to the applied force.

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