11.1 Rolling Motion

Example

Rolling Down an Inclined Plane

A solid cylinder rolls down an inclined plane without slipping, starting from rest. It has mass m and radius r. (a) What is its acceleration? (b) What condition must the coefficient of static friction [latex]{\mu }_{\text{S}}[/latex] satisfy so the cylinder does not slip?

Strategy

Draw a sketch and free-body diagram, and choose a coordinate system. We put x in the direction down the plane and y upward perpendicular to the plane. Identify the forces involved. These are the normal force, the force of gravity, and the force due to friction. Write down Newton’s laws in the x– and y-directions, and Newton’s law for rotation, and then solve for the acceleration and force due to friction.

Solution

1. The free-body diagram and sketch are shown in Figure, including the normal force, components of the weight, and the static friction force. There is barely enough friction to keep the cylinder rolling without slipping. Since there is no slipping, the magnitude of the friction force is less than or equal to [latex]{\mu }_{S}N[/latex]. Writing down Newton’s laws in the x– and y-directions, we have
   [latex]\sum {F}_{x}=m{a}_{x};\enspace\sum {F}_{y}=m{a}_{y}.[/latex]

   

   Substituting in from the free-body diagram,

   [latex]\begin{array}{ccc}\hfill mg\,\text{sin}\,\theta -{f}_{\text{S}}& =\hfill & m{({a}_{\text{CM}})}_{x},\hfill \\ \hfill N-mg\,\text{cos}\,\theta & =\hfill & 0,\hfill \\ \hfill {f}_{\text{S}}& \le \hfill & {\mu }_{\text{S}}N,\hfill \end{array}[/latex]

   we can then solve for the linear acceleration of the center of mass from these equations:

   [latex]{({a}_{\text{CM}})}_{x}=g(\text{sin}\,\theta -{\mu }_{S}\text{cos}\,\theta ).[/latex]

   However, it is useful to express the linear acceleration in terms of the moment of inertia. For this, we write down Newton’s second law for rotation,

   [latex]\sum {\tau }_{\text{CM}}={I}_{\text{CM}}\alpha .[/latex]

   The torques are calculated about the axis through the center of mass of the cylinder. The only nonzero torque is provided by the friction force. We have

   [latex]{f}_{\text{S}}r={I}_{\text{CM}}\alpha .[/latex]

   Finally, the linear acceleration is related to the angular acceleration by

   [latex]{({a}_{\text{CM}})}_{x}=r\alpha .[/latex]

   These equations can be used to solve for [latex]{a}_{\text{CM}},\alpha ,\,\text{and}\,{f}_{\text{S}}[/latex] in terms of the moment of inertia, where we have dropped the x-subscript. We write [latex]{a}_{\text{CM}}[/latex] in terms of the vertical component of gravity and the friction force, and make the following substitutions.

   [latex]{a}_{\text{CM}}=g\text{sin}\,\theta -\frac{{f}_{\text{S}}}{m}[/latex]
   [latex]{f}_{\text{S}}=\frac{{I}_{\text{CM}}\alpha }{r}=\frac{{I}_{\text{CM}}{a}_{\text{CM}}}{{r}^{2}}[/latex]

   From this we obtain

   [latex]\begin{array}{cc}\hfill {a}_{\text{CM}}& =g\,\text{sin}\,\theta -\frac{{I}_{\text{CM}}{a}_{\text{CM}}}{m{r}^{2}},\hfill \\ & =\frac{mg\,\text{sin}\,\theta }{m+({I}_{\text{CM}}\text{/}{r}^{2})}.\hfill \end{array}[/latex]

   Note that this result is independent of the coefficient of static friction, [latex]{\mu }_{\text{S}}[/latex].

   Since we have a solid cylinder, from Figure, we have [latex]{I}_{\text{CM}}=m{r}^{2}\text{/}2[/latex] and

   [latex]{a}_{\text{CM}}=\frac{mg\,\text{sin}\,\theta }{m+(m{r}^{2}\text{/}2{r}^{2})}=\frac{2}{3}g\,\text{sin}\,\theta .[/latex]

   Therefore, we have

   [latex]\alpha =\frac{{a}_{\text{CM}}}{r}=\frac{2}{3r}g\,\text{sin}\,\theta .[/latex]
2. Because slipping does not occur, [latex]{f}_{\text{S}}\le {\mu }_{\text{S}}N[/latex]. Solving for the friction force,
   [latex]{f}_{\text{S}}={I}_{\text{CM}}\frac{\alpha }{r}={I}_{\text{CM}}\frac{({a}_{\text{CM}})}{{r}^{2}}=\frac{{I}_{\text{CM}}}{{r}^{2}}(\frac{mg\,\text{sin}\,\theta }{m+({I}_{\text{CM}}\text{/}{r}^{2})})=\frac{mg{I}_{\text{CM}}\,\text{sin}\,\theta }{m{r}^{2}+{I}_{\text{CM}}}.[/latex]

   Substituting this expression into the condition for no slipping, and noting that [latex]N=mg\,\text{cos}\,\theta[/latex], we have

   [latex]\frac{mg{I}_{\text{CM}}\text{sin}\,\theta }{m{r}^{2}+{I}_{\text{CM}}}\le {\mu }_{\text{S}}mg\,\text{cos}\,\theta[/latex]

   or

   [latex]{\mu }_{\text{S}}\ge \frac{\text{tan}\,\theta }{1+(m{r}^{2}\text{/}{I}_{\text{CM}})}.[/latex]

   For the solid cylinder, this becomes

   [latex]{\mu }_{\text{S}}\ge \frac{\text{tan}\,\theta }{1+(2m{r}^{2}\text{/}m{r}^{2})}=\frac{1}{3}\text{tan}\,\theta .[/latex]

Significance

1. The linear acceleration is linearly proportional to [latex]\text{sin}\,\theta .[/latex] 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 [latex]\text{sin}\,\theta[/latex] 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 [latex](1\text{/}3)\text{tan}\,\theta[/latex]. Thus, the greater the angle of incline, the greater the coefficient of static friction must be to prevent the cylinder from slipping.

Example

Rolling Down an Inclined Plane with Slipping

A solid cylinder rolls down an inclined plane from rest and undergoes slipping (Figure). 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 [latex]{f}_{\text{k}}={\mu }_{\text{k}}N={\mu }_{\text{k}}mg\text{cos}\,\theta .[/latex]

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 [latex]\alpha[/latex], we have

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 [latex]\theta \to 90^\circ[/latex], this force goes to zero, and, thus, the angular acceleration goes to zero.

Example

Curiosity Rover

The Curiosity rover, shown in Figure, was deployed on Mars on August 6, 2012. The wheels of the rover have a radius of 25 cm. Suppose astronauts arrive on Mars in the year 2050 and find the now-inoperative Curiosity on the side of a basin. While they are dismantling the rover, an astronaut accidentally loses a grip on one of the wheels, which rolls without slipping down into the bottom of the basin 25 meters below. If the wheel has a mass of 5 kg, what is its velocity at the bottom of the basin?

Strategy

We use mechanical energy conservation to analyze the problem. At the top of the hill, the wheel is at rest and has only potential energy. At the bottom of the basin, the wheel has rotational and translational kinetic energy, which must be equal to the initial potential energy by energy conservation. Since the wheel is rolling without slipping, we use the relation [latex]{v}_{\text{CM}}=r\omega[/latex] to relate the translational variables to the rotational variables in the energy conservation equation. We then solve for the velocity. From Figure, we see that a hollow cylinder is a good approximation for the wheel, so we can use this moment of inertia to simplify the calculation.

Solution

Energy at the top of the basin equals energy at the bottom:

The known quantities are [latex]{I}_{\text{CM}}=m{r}^{2}\text{,}\,r=0.25\,\text{m,}\,\text{and}\,h=25.0\,\text{m}[/latex].

We rewrite the energy conservation equation eliminating [latex]\omega[/latex] by using [latex]\omega =\frac{{v}_{\text{CM}}}{r}.[/latex] We have

or

On Mars, the acceleration of gravity is [latex]3.71\,{\,\text{m/s}}^{2},[/latex] which gives the magnitude of the velocity at the bottom of the basin as

Significance

This is a fairly accurate result considering that Mars has very little atmosphere, and the loss of energy due to air resistance would be minimal. The result also assumes that the terrain is smooth, such that the wheel wouldn’t encounter rocks and bumps along the way.

Also, in this example, the kinetic energy, or energy of motion, is equally shared between linear and rotational motion. If we look at the moments of inertia in Figure, we see that the hollow cylinder has the largest moment of inertia for a given radius and mass. If the wheels of the rover were solid and approximated by solid cylinders, for example, there would be more kinetic energy in linear motion than in rotational motion. This would give the wheel a larger linear velocity than the hollow cylinder approximation. Thus, the solid cylinder would reach the bottom of the basin faster than the hollow cylinder.