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Check Your Understanding A hollow cylinder is on an incline at an angle of $60\text{\xb0}.$ The coefficient of static friction on the surface is ${\mu}_{S}=0.6$ . (a) Does the cylinder roll without slipping? (b) Will a solid cylinder roll without slipping?
a. ${\mu}_{\text{S}}\ge \frac{\text{tan}\phantom{\rule{0.2em}{0ex}}\theta}{1+(m{r}^{2}\text{/}{I}_{\text{CM}})}$ ; inserting the angle and noting that for a hollow cylinder ${I}_{\text{CM}}=m{r}^{2},$ we have ${\mu}_{\text{S}}\ge \frac{\text{tan}\phantom{\rule{0.2em}{0ex}}60\text{\xb0}}{1+(m{r}^{2}\text{/}m{r}^{2})}=\frac{1}{2}\text{tan}\phantom{\rule{0.2em}{0ex}}60\text{\xb0}=0.87;$ we are given a value of 0.6 for the coefficient of static friction, which is less than 0.87, so the condition isn’t satisfied and the hollow cylinder will slip; b. The solid cylinder obeys the condition ${\mu}_{\text{S}}\ge \frac{1}{3}\text{tan}\phantom{\rule{0.2em}{0ex}}\theta =\frac{1}{3}\text{tan}\phantom{\rule{0.2em}{0ex}}60\text{\xb0}=0.58.$ The value of 0.6 for ${\mu}_{\text{S}}$ satisfies this condition, so the solid cylinder will not slip.
It is worthwhile to repeat the equation derived in this example for the acceleration of an object rolling without slipping:
This is a very useful equation for solving problems involving rolling without slipping. Note that the acceleration is less than that for an object sliding down a frictionless plane with no rotation. The acceleration will also be different for two rotating cylinders with different rotational inertias.
In the case of rolling motion with slipping, we must use the coefficient of kinetic friction, which gives rise to the kinetic friction force since static friction is not present. The situation is shown in [link] . In the case of slipping, ${v}_{\text{CM}}-R\omega \ne 0$ , because point P on the wheel is not at rest on the surface, and ${v}_{P}\ne 0$ . Thus, $\omega \ne \frac{{v}_{\text{CM}}}{R},\alpha \ne \frac{{a}_{\text{CM}}}{R}$ .
The sum of the forces in the y -direction is zero, so the friction force is now ${f}_{\text{k}}={\mu}_{\text{k}}N={\mu}_{\text{k}}mg\text{cos}\phantom{\rule{0.2em}{0ex}}\theta .$
Newton’s second law in the x -direction becomes
or
The friction force provides the only torque about the axis through the center of mass, so Newton’s second law of rotation becomes
Solving for $\alpha $ , we have
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