Motion Diagram Car Speeds Up Stops Then Speeds Up Again

xi Angular Momentum

11.1 Rolling Motion

Learning Objectives

Past the terminate of this section, you volition be able to:

Describe the physics of rolling motion without slipping
Explain how linear variables are related to angular variables for the case of rolling move without slipping
Observe the linear and angular accelerations in rolling motion with and without slipping
Summate the static friction forcefulness associated with rolling motion without slipping
Use free energy conservation to analyze rolling motion

Rolling motion is that common combination of rotational and translational motion that we see everywhere, every day. Think about the different situations of wheels moving on a auto along a highway, or wheels on a plane landing on a runway, or wheels on a robotic explorer on another planet. Understanding the forces and torques involved in rolling motion is a crucial cistron in many different types of situations.

For analyzing rolling motion in this affiliate, refer to (Effigy) in Fixed-Centrality Rotation to detect moments of inertia of some common geometrical objects. You may likewise observe it useful in other calculations involving rotation.

Rolling Movement without Slipping

People have observed rolling motion without slipping ever since the invention of the wheel. For example, we tin can await at the interaction of a auto's tires and the surface of the route. If the driver depresses the accelerator to the flooring, such that the tires spin without the automobile moving forward, at that place must be kinetic friction between the wheels and the surface of the route. If the driver depresses the accelerator slowly, causing the car to move forrard, and then the tires roll without slipping. It is surprising to nearly people that, in fact, the bottom of the bike is at rest with respect to the ground, indicating there must be static friction between the tires and the road surface. In (Figure), the bicycle is in motion with the rider staying upright. The tires have contact with the road surface, and, even though they are rolling, the bottoms of the tires deform slightly, do not slip, and are at rest with respect to the road surface for a measurable amount of time. At that place must be static friction between the tire and the route surface for this to exist and so.

Figure a is a photograph of a person riding a bicycle. The camera followed the bike, so the image of the bike and rider is sharp, the background is blurred due to bike's motion. Figure b is a photograph of a bicycle wheel rolling on the ground, with the camera stationary relative to the ground. The wheel and spokes are blurred at the top but clear at the bottom. — **Effigy xi.2** (a) The bicycle moves forward, and its tires do not sideslip. The bottom of the slightly deformed tire is at rest with respect to the road surface for a measurable amount of time. (b) This epitome shows that the top of a rolling wheel appears blurred by its move, simply the bottom of the wheel is instantaneously at rest. (credit a: modification of work by Nelson Lourenço; credit b: modification of work by Colin Rose)

To analyze rolling without slipping, nosotros first derive the linear variables of velocity and dispatch of the center of mass of the bike in terms of the angular variables that depict the wheel's motion. The situation is shown in (Effigy).

Figure a shows a free body diagram of a wheel, including the location where the forces act. Four forces are shown: M g is a downward force acting on the center of the wheel. N is an upward force acting on the bottom of the wheel. F is a force to the right, acting on the center of the wheel, and f sub s is a force to the left acting on the bottom of the wheel. The force f sub s is smaller or equal to mu sub s times N. Figure b is an illustration of a wheel rolling without slipping on a horizontal surface. Point P is the contact point between the bottom of the wheel and the surface. The wheel has a clockwise rotation, an acceleration to the right of a sub C M and a velocity to the right of v sub V M. The relations omega equals v sub C M over R and alpha equals a sub C M over R are given. A coordinate system with positive x to the right and positive y up is shown. Figure c shows wheel in the center of mass frame. Point P has velocity vector in the negative direction with respect to the center of mass of the wheel. That vector is shown on the diagram and labeled as minus R omega i hat. It is tangent to the wheel at the bottom, and pointing to the left. Additional vectors at various locations on the rim of the wheel are shown, all tangent to the wheel and pointing clockwise. — **Figure eleven.3** (a) A wheel is pulled across a horizontal surface by a force
$\overset{\to }{F}$

. The force of static friction

${\overset{\to }{f}}_{\text{S}},|{\overset{\to }{f}}_{\text{S}}|\le {\mu }_{\text{S}}N$

is large enough to go on information technology from slipping. (b) The linear velocity and acceleration vectors of the center of mass and the relevant expressions for

$\omega \,\text{and}\,\alpha$

. Indicate P is at rest relative to the surface. (c) Relative to the center of mass (CM) frame, point P has linear velocity

$\text{−}R\omega \hat{i}$

.

From (Figure)(a), we encounter the force vectors involved in preventing the cycle from slipping. In (b), bespeak P that touches the surface is at residue relative to the surface. Relative to the heart of mass, point P has velocity

$\text{−}R\omega \hat{i}$

, where R is the radius of the wheel and

$\omega$

is the bicycle'southward angular velocity about its axis. Since the wheel is rolling, the velocity of P with respect to the surface is its velocity with respect to the center of mass plus the velocity of the center of mass with respect to the surface:

${\overset{\to }{v}}_{P}=\text{−}R\omega \hat{i}+{v}_{\text{CM}}\hat{i}.$

Since the velocity of P relative to the surface is zero,

${v}_{P}=0$

, this says that

${v}_{\text{CM}}=R\omega .$

Thus, the velocity of the bicycle's center of mass is its radius times the angular velocity about its axis. We show the correspondence of the linear variable on the left side of the equation with the athwart variable on the right side of the equation. This is done below for the linear dispatch.

If we differentiate (Figure) on the left side of the equation, we obtain an expression for the linear acceleration of the center of mass. On the right side of the equation, R is a abiding and since

$\alpha =\frac{d\omega }{dt},$

we have

${a}_{\text{CM}}=R\alpha .$

Furthermore, we can find the distance the bicycle travels in terms of angular variables by referring to (Figure). As the wheel rolls from point A to point B, its outer surface maps onto the basis by exactly the distance travelled, which is

${d}_{\text{CM}}.$

Nosotros see from (Figure) that the length of the outer surface that maps onto the footing is the arc length

$R\theta \text{}$

. Equating the two distances, we obtain

${d}_{\text{CM}}=R\theta .$

A wheel, radius R, rolling on a horizontal surface and moving to the right at v sub C M is drawn in two positions. In the first position, point A on the wheel is at the bottom, in contact with the surface, and point B is at the top. The arc length from A to B along the rim of the wheel is highlighted and labeled as being R theta. In the second position, point B on the wheel is at the bottom, in contact with the surface, and point A is at the top. The horizontal distance between the wheel's point of contact with the surface in the two illustrated positions is d sub C M. The arc length A B is now on the other side of the wheel. — **Figure 11.4** As the wheel rolls on the surface, the arc length
$R\theta$

from A to B maps onto the surface, corresponding to the distance

${d}_{\text{CM}}$

that the heart of mass has moved.

Case

Rolling Down an Inclined Plane

A solid cylinder rolls down an inclined aeroplane without slipping, starting from remainder. It has mass m and radius r. (a) What is its acceleration? (b) What condition must the coefficient of static friction

${\mu }_{\text{S}}$

satisfy so the cylinder does not slip?

Strategy

Draw a sketch and costless-body diagram, and cull a coordinate organisation. Nosotros put 10 in the management downward the airplane and y upward perpendicular to the aeroplane. Place the forces involved. These are the normal force, the force of gravity, and the force due to friction. Write downward Newton's laws in the x– and y-directions, and Newton's law for rotation, and then solve for the dispatch and strength due to friction.

Solution

The free-body diagram and sketch are shown in (Figure), including the normal forcefulness, components of the weight, and the static friction forcefulness. In that location is barely enough friction to continue the cylinder rolling without slipping. Since there is no slipping, the magnitude of the friction force is less than or equal to
${\mu }_{S}N$

. Writing downwardly Newton's laws in the x– and y-directions, we have

$\sum {F}_{x}=m{a}_{x};\enspace\sum {F}_{y}=m{a}_{y}.$

Effigy xi.five A solid cylinder rolls down an inclined plane without slipping from rest. The coordinate system has x in the direction downwardly the inclined airplane and y perpendicular to the plane. The free-body diagram is shown with the normal force, the static friction strength, and the components of the weight
$m\overset{\to }{g}$

. Friction makes the cylinder roll downwards the plane rather than skid.

Substituting in from the free-body diagram,

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

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

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

All the same, information technology is useful to express the linear acceleration in terms of the moment of inertia. For this, we write down Newton'southward second law for rotation,

$\sum {\tau }_{\text{CM}}={I}_{\text{CM}}\alpha .$

The torques are calculated near the axis through the centre of mass of the cylinder. The only nonzero torque is provided by the friction forcefulness. We accept

${f}_{\text{S}}r={I}_{\text{CM}}\alpha .$

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

${({a}_{\text{CM}})}_{x}=r\alpha .$

These equations can be used to solve for

${a}_{\text{CM}},\alpha ,\,\text{and}\,{f}_{\text{S}}$

in terms of the moment of inertia, where nosotros have dropped the x-subscript. We write

${a}_{\text{CM}}$

in terms of the vertical component of gravity and the friction force, and make the following substitutions.

${a}_{\text{CM}}=g\text{sin}\,\theta -\frac{{f}_{\text{S}}}{m}$

${f}_{\text{S}}=\frac{{I}_{\text{CM}}\alpha }{r}=\frac{{I}_{\text{CM}}{a}_{\text{CM}}}{{r}^{2}}$

From this we obtain

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

Note that this event is independent of the coefficient of static friction,

${\mu }_{\text{S}}$

.

Since we have a solid cylinder, from (Figure), we have

${I}_{\text{CM}}=m{r}^{2}\text{/}2$

and

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

Therefore, nosotros have

$\alpha =\frac{{a}_{\text{CM}}}{r}=\frac{2}{3r}g\,\text{sin}\,\theta .$
Considering slipping does not occur,
${f}_{\text{S}}\le {\mu }_{\text{S}}N$

. Solving for the friction force,

${f}_{\text{S}}={I}_{\text{CM}}\frac{\alpha }{r}={I}_{\text{CM}}\frac{({a}_{\text{CM}})}{{r}^{2}}=\frac{{I}_{\text{CM}}}{{r}^{2}}(\frac{mg\,\text{sin}\,\theta }{m+({I}_{\text{CM}}\text{/}{r}^{2})})=\frac{mg{I}_{\text{CM}}\,\text{sin}\,\theta }{m{r}^{2}+{I}_{\text{CM}}}.$

Substituting this expression into the condition for no slipping, and noting that

$N=mg\,\text{cos}\,\theta$

, we take

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

or

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

For the solid cylinder, this becomes

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

Significance

The linear acceleration is linearly proportional to
$\text{sin}\,\theta .$

Thus, the greater the angle of the incline, the greater the linear dispatch, as would be expected. The angular acceleration, however, is linearly proportional to

$\text{sin}\,\theta$

and inversely proportional to the radius of the cylinder. Thus, the larger the radius, the smaller the angular acceleration.
For no slipping to occur, the coefficient of static friction must be greater than or equal to
$(1\text{/}3)\text{tan}\,\theta$

. Thus, the greater the angle of incline, the greater the coefficient of static friction must be to prevent the cylinder from slipping.

Check Your Agreement

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

[reveal-answer q="275472″]Show Answer[/reveal-respond]
[hidden-answer a="275472″]a.

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

; inserting the angle and noting that for a hollow cylinder

${I}_{\text{CM}}=m{r}^{2},$

we have

${\mu }_{\text{S}}\ge \frac{\text{tan}\,60\text{°}}{1+(m{r}^{2}\text{/}m{r}^{2})}=\frac{1}{2}\text{tan}\,60\text{°}=0.87;$

we are given a value of 0.6 for the coefficient of static friction, which is less than 0.87, then the status isn't satisfied and the hollow cylinder volition slip; b. The solid cylinder obeys the condition

${\mu }_{\text{S}}\ge \frac{1}{3}\text{tan}\,\theta =\frac{1}{3}\text{tan}\,60\text{°}=0.58.$

The value of 0.6 for

${\mu }_{\text{S}}$

satisfies this status, and so the solid cylinder volition non slip.[/hidden-answer]

It is worthwhile to repeat the equation derived in this case for the acceleration of an object rolling without slipping:

${a}_{\text{CM}}=\frac{mg\,\text{sin}\,\theta }{m+({I}_{\text{CM}}\text{/}{r}^{2})}.$

This is a very useful equation for solving bug involving rolling without slipping. Note that the acceleration is less than that for an object sliding down a frictionless plane with no rotation. The dispatch volition besides be different for two rotating cylinders with different rotational inertias.

Rolling Motion with Slipping

In the case of rolling motility with slipping, we must use the coefficient of kinetic friction, which gives rise to the kinetic friction forcefulness since static friction is not present. The situation is shown in (Figure). In the example of slipping,

${v}_{\text{CM}}-R\omega \ne 0$

, because point P on the bike is not at rest on the surface, and

${v}_{P}\ne 0$

. Thus,

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

Case

Rolling Downwards an Inclined Plane with Slipping

A solid cylinder rolls down an inclined airplane 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 heart of mass?

Strategy

Describe a sketch and free-trunk 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'southward second law to solve for the dispatch in the 10-direction. Utilise Newton's second police of rotation to solve for the angular acceleration.

Solution

A diagram of a cylinder rolling and slipping down an inclined plane and a free body diagram of the cylinder. On the left is an illustration showing the inclined plane, which makes an angle of theta with the horizontal. The cylinder is shown to be at rest at the top, then moving along the incline when it is lower. On the right is a free body diagram. The x y coordinate system is tilted so that the positive x direction is parallel to the inclined plane and points toward its bottom, and the positive y direction is outward, perpendicular to the plane. Four forces are shown. N j hat acts at the center of the cylinder and points in the positive y direction. m g sine theta i hat acts at the center of the cylinder and points in the positive x direction. Minus m g cosine theta j hat acts at the center of the cylinder and points in the negative y direction. Minus f sub k i hat acts at the point of contact and points in the negative x direction. — **Figure 11.7** A solid cylinder rolls downwards an inclined plane from rest and undergoes slipping. The coordinate system has ten in the direction down the inclined airplane and y upward perpendicular to the airplane. The free-torso diagram shows the normal strength, kinetic friction force, and the components of the weight
$m\overset{\to }{g}.$

The sum of the forces in the y-direction is aught, and so the friction force is now

${f}_{\text{k}}={\mu }_{\text{k}}N={\mu }_{\text{k}}mg\text{cos}\,\theta .$

Newton'southward second law in the x-direction becomes

$\sum {F}_{x}=m{a}_{x},$

$mg\,\text{sin}\,\theta -{\mu }_{\text{k}}mg\,\text{cos}\,\theta =m{({a}_{\text{CM}})}_{x},$

${({a}_{\text{CM}})}_{x}=g(\text{sin}\,\theta -{\mu }_{\text{K}}\,\text{cos}\,\theta ).$

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 take

$\alpha =\frac{2{f}_{\text{k}}}{mr}=\frac{2{\mu }_{\text{k}}g\,\text{cos}\,\theta }{r}.$

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 downwardly an inclined airplane with kinetic friction. The angular acceleration nigh 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.

Conservation of Mechanical Free energy in Rolling Motility

In the preceding chapter, nosotros introduced rotational kinetic free energy. Any rolling object carries rotational kinetic energy, as well as translational kinetic energy and potential energy if the organization requires. Including the gravitational potential free energy, the total mechanical free energy of an object rolling is

${E}_{\text{T}}=\frac{1}{2}m{v}_{\text{CM}}^{2}+\frac{1}{2}{I}_{\text{CM}}{\omega }^{2}+mgh.$

In the absence of whatsoever nonconservative forces that would have free energy out of the system in the form of rut, the total free energy of a rolling object without slipping is conserved and is abiding throughout the motion. Examples where energy is non conserved are a rolling object that is slipping, product of heat as a result of kinetic friction, and a rolling object encountering air resistance.

You may inquire why a rolling object that is not slipping conserves energy, since the static friction force is nonconservative. The answer can be found by referring back to (Figure). Betoken P in contact with the surface is at rest with respect to the surface. Therefore, its minute displacement

$d\overset{\to }{r}$

with respect to the surface is zero, and the incremental piece of work done by the static friction forcefulness is zero. Nosotros can apply energy conservation to our study of rolling motion to bring out some interesting results.

Instance

Curiosity Rover

The Curiosity rover, shown in (Figure), was deployed on Mars on August vi, 2012. The wheels of the rover have a radius of 25 cm. Suppose astronauts make it on Mars in the yr 2050 and find the at present-inoperative Curiosity on the side of a bowl. While they are dismantling the rover, an astronaut accidentally loses a grip on ane of the wheels, which rolls without slipping downward 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 bowl?

A photograph of the NASA rover Curiosity during testing at the Jet Propulsion Laboratory. — **Effigy 11.8** The NASA Mars Science Laboratory rover Curiosity during testing on June three, 2011. The location is inside the Spacecraft Assembly Facility at NASA's Jet Propulsion Laboratory in Pasadena, California. (credit: NASA/JPL-Caltech)

Strategy

Nosotros apply mechanical energy conservation to analyze the problem. At the pinnacle 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 exist equal to the initial potential free energy by free energy conservation. Since the bike is rolling without slipping, we utilize the relation

${v}_{\text{CM}}=r\omega$

to chronicle the translational variables to the rotational variables in the energy conservation equation. Nosotros then solve for the velocity. From (Figure), we see that a hollow cylinder is a good approximation for the wheel, so we tin can employ this moment of inertia to simplify the adding.

Solution

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

$mgh=\frac{1}{2}m{v}_{\text{CM}}^{2}+\frac{1}{2}{I}_{\text{CM}}{\omega }^{2}.$

The known quantities are

${I}_{\text{CM}}=m{r}^{2}\text{,}\,r=0.25\,\text{m,}\,\text{and}\,h=25.0\,\text{m}$

We rewrite the energy conservation equation eliminating

$\omega$

by using

$\omega =\frac{{v}_{\text{CM}}}{r}.$

We have

$mgh=\frac{1}{2}m{v}_{\text{CM}}^{2}+\frac{1}{2}m{r}^{2}\frac{{v}_{\text{CM}}^{2}}{{r}^{2}}$

$gh=\frac{1}{2}{v}_{\text{CM}}^{2}+\frac{1}{2}{v}_{\text{CM}}^{2}⇒{v}_{\text{CM}}=\sqrt{gh}.$

On Mars, the acceleration of gravity is

$3.71\,{\,\text{m/s}}^{2},$

which gives the magnitude of the velocity at the bottom of the basin equally

${v}_{\text{CM}}=\sqrt{(3.71\,\text{m}\text{/}{\text{s}}^{2})25.0\,\text{m}}=9.63\,\text{m}\text{/}\text{s}\text{.}$

Significance

This is a fairly authentic result considering that Mars has very piddling temper, and the loss of energy due to air resistance would be minimal. The result besides assumes that the terrain is smooth, such that the wheel wouldn't encounter rocks and bumps forth the way.

Also, in this example, the kinetic energy, or free energy of motion, is equally shared between linear and rotational motion. If we look at the moments of inertia in (Effigy), nosotros 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 past solid cylinders, for example, there would be more kinetic free energy in linear motion than in rotational motion. This would requite the wheel a larger linear velocity than the hollow cylinder approximation. Thus, the solid cylinder would attain the bottom of the basin faster than the hollow cylinder.

Summary

In rolling motion without slipping, a static friction force is present betwixt the rolling object and the surface. The relations
${v}_{\text{CM}}=R\omega ,{a}_{\text{CM}}=R\alpha ,\,\text{and}\,{d}_{\text{CM}}=R\theta$

all utilise, such that the linear velocity, acceleration, and distance of the center of mass are the angular variables multiplied past the radius of the object.
In rolling motion with slipping, a kinetic friction forcefulness arises between the rolling object and the surface. In this case,
${v}_{\text{CM}}\ne R\omega ,{a}_{\text{CM}}\ne R\alpha ,\,\text{and}\,{d}_{\text{CM}}\ne R\theta$

.
Free energy conservation can be used to clarify rolling motility. Energy is conserved in rolling move without slipping. Energy is not conserved in rolling motion with slipping due to the estrus generated by kinetic friction.

Conceptual Questions

Can a round object released from rest at the superlative of a frictionless incline undergo rolling motion?

[reveal-respond q="fs-id1165036812159″]Show Solution[/reveal-reply]

[hidden-answer a="fs-id1165036812159″]

No, the static friction forcefulness is nix.

[/hidden-answer]

A cylindrical can of radius R is rolling beyond a horizontal surface without slipping. (a) Afterward one complete revolution of the can, what is the distance that its center of mass has moved? (b) Would this altitude be greater or smaller if slipping occurred?

A bicycle is released from the top on an incline. Is the wheel virtually likely to slip if the incline is steep or gently sloped?

[reveal-reply q="fs-id1165037047596″]Show Solution[/reveal-respond]

[subconscious-respond a="fs-id1165037047596″]

The wheel is more likely to slip on a steep incline since the coefficient of static friction must increase with the angle to keep rolling motility without slipping.

[/hidden-answer]

Which rolls down an inclined plane faster, a hollow cylinder or a solid sphere? Both have the same mass and radius.

A hollow sphere and a hollow cylinder of the same radius and mass roll upwards an incline without slipping and have the same initial center of mass velocity. Which object reaches a greater tiptop earlier stopping?

[reveal-answer q="fs-id1165036990641″]Bear witness Solution[/reveal-answer]

[hidden-reply a="fs-id1165036990641″]

The cylinder reaches a greater height. By (Figure), its acceleration in the direction down the incline would be less.

[/subconscious-answer]

Problems

What is the angular velocity of a 75.0-cm-bore tire on an automobile traveling at 90.0 km/h?

[reveal-answer q="719142″]Show Answer[/reveal-reply]
[hidden-answer a="719142″]

${v}_{\text{CM}}=R\omega \,⇒\omega =66.7\,\text{rad/s}$

[/hidden-answer]

A boy rides his bicycle 2.00 km. The wheels have radius thirty.0 cm. What is the total angle the tires rotate through during his trip?

If the male child on the cycle in the preceding trouble accelerates from rest to a speed of 10.0 grand/s in ten.0 s, what is the angular dispatch of the tires?

[reveal-answer q="fs-id1165037846364″]Bear witness Solution[/reveal-respond]

[hidden-reply a="fs-id1165037846364″]

$\alpha =3.3\,\text{rad}\text{/}{\text{s}}^{2}$

[/hidden-answer]

Formula Ane race cars take 66-cm-diameter tires. If a Formula Ane averages a speed of 300 km/h during a race, what is the angular displacement in revolutions of the wheels if the race machine maintains this speed for 1.5 hours?

A marble rolls down an incline at

$30\text{°}$

from rest. (a) What is its acceleration? (b) How far does it get in 3.0 south?

[reveal-answer q="fs-id1165038000616″]Show Solution[/reveal-respond]

[hidden-answer a="fs-id1165038000616″]

${I}_{\text{CM}}=\frac{2}{5}m{r}^{2},\,{a}_{\text{CM}}=3.5\,\text{m}\text{/}{\text{s}}^{2};\,x=15.75\,\text{m}$

[/hidden-answer]

Echo the preceding problem replacing the marble with a solid cylinder. Explain the new result.

A rigid body with a cylindrical cross-department is released from the top of a

$30\text{°}$

incline. It rolls ten.0 g to the bottom in 2.threescore southward. Detect the moment of inertia of the torso in terms of its mass thou and radius r.

[reveal-reply q="fs-id1165037169593″]Show Solution[/reveal-answer]

[hidden-answer a="fs-id1165037169593″]

positive is downwards the incline plane;

${a}_{\text{CM}}=\frac{mg\,\text{sin}\,\theta }{m+({I}_{\text{CM}}\text{/}{r}^{2})}⇒{I}_{\text{CM}}={r}^{2}[\frac{mg\,\text{sin}30}{{a}_{\text{CM}}}-m]$

$x-{x}_{0}={v}_{0}t-\frac{1}{2}{a}_{\text{CM}}{t}^{2}⇒{a}_{\text{CM}}=2.96\,{\text{m/s}}^{2},$

${I}_{\text{CM}}=0.66\,m{r}^{2}$

[/hidden-answer]

A yo-yo tin can be idea of a solid cylinder of mass m and radius r that has a light string wrapped effectually its circumference (encounter below). One end of the string is held fixed in infinite. If the cylinder falls as the string unwinds without slipping, what is the acceleration of the cylinder?

An illustration of a cylinder, radius r, and the forces on it. The force m g acts on the center of the cylinder and points down. The force T acts on the right hand edge and points up.

A solid cylinder of radius ten.0 cm rolls down an incline with slipping. The angle of the incline is

$30\text{°}.$

The coefficient of kinetic friction on the surface is 0.400. What is the angular acceleration of the solid cylinder? What is the linear acceleration?

[reveal-answer q="fs-id1165038304382″]Show Solution[/reveal-answer]

[hidden-answer a="fs-id1165038304382″]

$\alpha =67.9\,\text{rad}\text{/}{\text{s}}^{2}$

${({a}_{\text{CM}})}_{x}=1.5\,\text{m}\text{/}{\text{s}}^{2}$

[/hidden-answer]

A bowling ball rolls upward a ramp 0.5 m high without slipping to storage. It has an initial velocity of its center of mass of three.0 m/s. (a) What is its velocity at the top of the ramp? (b) If the ramp is i m high does it go far to the pinnacle?

A 40.0-kg solid cylinder is rolling across a horizontal surface at a speed of 6.0 m/s. How much work is required to stop it?

[reveal-answer q="fs-id1165037047359″]Prove Solution[/reveal-answer]

[subconscious-answer a="fs-id1165037047359″]

$W=-1080.0\,\text{J}$

[/hidden-answer]

A twoscore.0-kg solid sphere is rolling across a horizontal surface with a speed of 6.0 m/south. How much work is required to stop it? Compare results with the preceding problem.

A solid cylinder rolls upward an incline at an bending of

$20\text{°}.$

If it starts at the bottom with a speed of 10 one thousand/south, how far up the incline does it travel?

[reveal-answer q="fs-id1165038133403″]Show Solution[/reveal-answer]

[hidden-reply a="fs-id1165038133403″]

Mechanical energy at the lesser equals mechanical energy at the peak;

$\frac{1}{2}m{v}_{0}^{2}+\frac{1}{2}(\frac{1}{2}m{r}^{2}){(\frac{{v}_{0}}{r})}^{2}=mgh⇒h=\frac{1}{g}(\frac{1}{2}+\frac{1}{4}){v}_{0}^{2}$

$h=7.7\,\text{m,}$

so the distance upwards the incline is

$22.5\,\text{m}$

[/hidden-answer]

A solid cylindrical wheel of mass M and radius R is pulled by a force

$\overset{\to }{F}$

applied to the center of the wheel at

$37\text{°}$

to the horizontal (see the post-obit figure). If the wheel is to roll without slipping, what is the maximum value of

$|\overset{\to }{F}|?$

The coefficients of static and kinetic friction are

${\mu }_{\text{S}}=0.40\,\text{and}\,{\mu }_{\text{k}}=0.30.$

A hollow cylinder is given a velocity of v.0 thousand/south and rolls up an incline to a height of 1.0 chiliad. If a hollow sphere of the same mass and radius is given the same initial velocity, how high does information technology roll upwards the incline?

[reveal-answer q="fs-id1165038369522″]Prove Solution[/reveal-answer]

[hidden-answer a="fs-id1165038369522″]

Use energy conservation

$\frac{1}{2}m{v}_{0}^{2}+\frac{1}{2}{I}_{\text{Cyl}}{\omega }_{0}^{2}=mg{h}_{\text{Cyl}}$

$\frac{1}{2}m{v}_{0}^{2}+\frac{1}{2}{I}_{\text{Sph}}{\omega }_{0}^{2}=mg{h}_{\text{Sph}}$

Subtracting the two equations, eliminating the initial translational free energy, nosotros have

$\frac{1}{2}{I}_{\text{Cyl}}{\omega }_{0}^{2}-\frac{1}{2}{I}_{\text{Sph}}{\omega }_{0}^{2}=mg({h}_{\text{Cyl}}-{h}_{\text{Sph}})$

$\frac{1}{2}m{r}^{2}{(\frac{{v}_{0}}{r})}^{2}-\frac{1}{2}\frac{2}{3}m{r}^{2}{(\frac{{v}_{0}}{r})}^{2}=mg({h}_{\text{Cyl}}-{h}_{\text{Sph}})$

$\frac{1}{2}{v}_{0}^{2}-\frac{1}{2}\frac{2}{3}{v}_{0}^{2}=g({h}_{\text{Cyl}}-{h}_{\text{Sph}})$

${h}_{\text{Cyl}}-{h}_{\text{Sph}}=\frac{1}{g}(\frac{1}{2}-\frac{1}{3}){v}_{0}^{2}=\frac{1}{9.8\,\text{m}\text{/}{\text{s}}^{2}}(\frac{1}{6})(5.0\,\text{m}\text{/}{\text{s)}}^{2}=0.43\,\text{m}$

Thus, the hollow sphere, with the smaller moment of inertia, rolls up to a lower height of

$1.0-0.43=0.57\,\text{m}\text{.}$

[/subconscious-answer]

Glossary

rolling move: combination of rotational and translational motion with or without slipping

riveradiseld.blogspot.com

Source: https://opentextbc.ca/universityphysicsv1openstax/chapter/11-1-rolling-motion/

Motion Diagram Car Speeds Up Stops Then Speeds Up Again

11.1 Rolling Motion

Learning Objectives

Rolling Movement without Slipping

Case

Rolling Down an Inclined Plane

Strategy

Solution

Check Your Agreement

Rolling Motion with Slipping

Case

Rolling Downwards an Inclined Plane with Slipping

Strategy

Solution

Significance

Conservation of Mechanical Free energy in Rolling Motility

Instance

Curiosity Rover

Strategy

Solution

Significance

Summary

Conceptual Questions

Problems

Glossary

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