# 11.1 Rolling motion  (Page 3/6)

 Page 3 / 6

## Significance

1. The linear acceleration is linearly proportional to $\text{sin}\phantom{\rule{0.2em}{0ex}}\theta .$ Thus, the greater the angle of the incline, the greater the linear acceleration, as would be expected. The angular acceleration, however, is linearly proportional to $\text{sin}\phantom{\rule{0.2em}{0ex}}\theta$ and inversely proportional to the radius of the cylinder. Thus, the larger the radius, the smaller the angular acceleration.
2. For no slipping to occur, the coefficient of static friction must be greater than or equal to $\left(1\text{/}3\right)\text{tan}\phantom{\rule{0.2em}{0ex}}\theta$ . Thus, the greater the angle of incline, the greater the coefficient of static friction must be to prevent the cylinder from slipping.

Check Your Understanding A hollow cylinder is on an incline at an angle of $60\text{°}.$ 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. ${\mu }_{\text{S}}\ge \frac{\text{tan}\phantom{\rule{0.2em}{0ex}}\theta }{1+\left(m{r}^{2}\text{/}{I}_{\text{CM}}\right)}$ ; inserting the angle and noting that for a hollow cylinder ${I}_{\text{CM}}=m{r}^{2},$ we have ${\mu }_{\text{S}}\ge \frac{\text{tan}\phantom{\rule{0.2em}{0ex}}60\text{°}}{1+\left(m{r}^{2}\text{/}m{r}^{2}\right)}=\frac{1}{2}\text{tan}\phantom{\rule{0.2em}{0ex}}60\text{°}=0.87;$ 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 ${\mu }_{\text{S}}\ge \frac{1}{3}\text{tan}\phantom{\rule{0.2em}{0ex}}\theta =\frac{1}{3}\text{tan}\phantom{\rule{0.2em}{0ex}}60\text{°}=0.58.$ 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:

${a}_{\text{CM}}=\frac{mg\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta }{m+\left({I}_{\text{CM}}\text{/}{r}^{2}\right)}.$

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.

## Rolling motion with slipping

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, $\omega \ne \frac{{v}_{\text{CM}}}{R},\alpha \ne \frac{{a}_{\text{CM}}}{R}$ .

## Rolling down an inclined plane with slipping

A solid cylinder rolls down an inclined plane from rest and undergoes slipping ( [link] ). It has mass m and radius r . (a) What is its linear acceleration? (b) What is its angular acceleration about an axis through the center of mass?

## Strategy

Draw a sketch and free-body diagram showing the forces involved. The free-body diagram is similar to the no-slipping case except for the friction force, which is kinetic instead of static. Use Newton’s second law to solve for the acceleration in the x -direction. Use Newton’s second law of rotation to solve for the angular acceleration.

## Solution

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

$\sum {F}_{x}=m{a}_{x},$
$mg\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta -{\mu }_{\text{k}}mg\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta =m{\left({a}_{\text{CM}}\right)}_{x},$

or

${\left({a}_{\text{CM}}\right)}_{x}=g\left(\text{sin}\phantom{\rule{0.2em}{0ex}}\theta -{\mu }_{\text{K}}\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta \right).$

The friction force provides the only torque about the axis through the center of mass, so Newton’s second law of rotation becomes

$\sum {\tau }_{\text{CM}}={I}_{\text{CM}}\alpha ,$
${f}_{\text{k}}r={I}_{\text{CM}}\alpha =\frac{1}{2}m{r}^{2}\alpha .$

Solving for $\alpha$ , we have

$\alpha =\frac{2{f}_{\text{k}}}{mr}=\frac{2{\mu }_{\text{k}}g\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta }{r}.$

## Significance

We write the linear and angular accelerations in terms of the coefficient of kinetic friction. The linear acceleration is the same as that found for an object sliding down an inclined plane with kinetic friction. The angular acceleration about the axis of rotation is linearly proportional to the normal force, which depends on the cosine of the angle of inclination. As $\theta \to 90\text{°}$ , this force goes to zero, and, thus, the angular acceleration goes to zero.

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