Minimum force required to make an object slipping

Question

Let's say we have a wheel, and I exert some force on the centre of mass of the wheel. The wheel starts rolling without slipping. At the contact point the static friction force is opposite to the force I am exerting. To find the the minimum force required to make the wheel slip, I have to find the torque of the maximum static friction, then find the angular acceleration produced by that torque, and then, since it will slip, the translational acceleration must be greater than that angular acceleration*radius of wheel and solve the equation. Is this correct?

Answer 1

Yes, it is correct. The maximum static friction is $F_A^{max}=\mu_S Mg$, being $M$ the mass of the wheel, and its torque is $\tau_A^{max}=R\mu_S Mg$, so the maximum angular acceleration without slipping is $$ \alpha^{max}=\frac{\tau_A^{max}}{I}=\frac{R\mu_S Mg}{I} $$ Let $F$ be the exerted force, which produces an acceleration $$ a_{CM}=\frac{F}{M} $$ so $$ \frac{F}{M}=a_{CM}=R\alpha^{max}=\frac{R^2\mu_S Mg}{I}\quad\Longrightarrow\quad F^{min}=\frac{R^2\mu_S M^2g}{I} $$

Answer 2

For questions like this, start with a free-body diagram. The radius of the wheel is $r$, the mass is $M$, and the coefficient of friction is $\mu$.

_{Using right hand angle conventions, + is CCW.}

Since the body never leaves the horizontal plane, the floor reaction always equals the weight $M g$.

As a result of an arbitrary force $F_A$ being applied, a reaction $F_R$ will appear that will try to resist the relative motion of the wheel to the ground at the contact point.

When dealing with finite friction, start with the problem as if the contact is rough and sticking (infinite coefficient of friction) to resolve the forces, and then afterward apply the frictional force limit due to dry friction.

The equations of motion of the body about the center of mass are

$$\begin{aligned} F_A - F_R & = M a \\ -r \; F_R & = I \alpha \\ \end{aligned}$$

and in the scenario that the contact is sticking, you set to link the rotational and translational acceleration with $a + r \alpha =0$. Rotational acceleration is eliminated with $\alpha =\mbox{-} \frac{a}{r}$ from the equations and the system above can be solved for $a$ and $F_R$.

$$\begin{aligned} a & =\left( \frac{ r^2 } { I + M r^2 } \right) F_A \\ F_R & =\left( \frac{ I } { I + M r^2 } \right) F_A \\ \end{aligned}$$

For example, with a uniform thin disk for a wheel $I = \tfrac{M}{2} r^2$ and the above become $F_R = \tfrac{1}{3} F_A$ and $a = \tfrac{2}{3}\,\tfrac{M}{a}$

And now to your question which is for what value of $F_A$ the disk will not purely roll, but it will start slipping.

The limit for friction of $F_R \le \mu M g$ so use the following equation to find when the transition happens

$$ \left( \frac{ I } { I + M r^2 } \right) F_A = \mu M g $$

with solution

$$ F_A = \frac{ \mu M g}{\left( \frac{ I } { I + M r^2 } \right) } $$

and for the uniform disk for a wheel, the result is $F_A = 3 \left( \mu M g\right)$. So you need three times the friction limit applied at the center of mass to make a uniform disk to start slipping.