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What Is the Maximum Speed the Cylinders Can Have Without the Top Cylinder Sliding Off?

Written By Wednesday, February 16, 2022 Add Comment Edit

Learning Objectives

By the end of this department, you lot will exist able to:

  • Describe the physics of rolling move without slipping
  • Explain how linear variables are related to athwart variables for the case of rolling motion without slipping
  • Find the linear and angular accelerations in rolling move with and without slipping
  • Calculate the static friction force associated with rolling movement 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 24-hour interval. Remember about the dissimilar situations of wheels moving on a car forth a highway, or wheels on a plane landing on a rails, or wheels on a robotic explorer on another planet. Understanding the forces and torques involved in rolling motion is a crucial factor in many different types of situations.

For analyzing rolling motion in this chapter, refer to (Figure) in Fixed-Axis Rotation to observe moments of inertia of some common geometrical objects. You lot may too find it useful in other calculations involving rotation.

Rolling Motion without Slipping

People have observed rolling motion without slipping ever since the invention of the wheel. For example, we tin can look at the interaction of a car's tires and the surface of the road. If the driver depresses the accelerator to the flooring, such that the tires spin without the auto moving forward, there must be kinetic friction between the wheels and the surface of the road. If the driver depresses the accelerator slowly, causing the car to movement forward, and so the tires roll without slipping. It is surprising to most people that, in fact, the bottom of the bicycle is at residue with respect to the ground, indicating there must be static friction between the tires and the route surface. In (Figure), the bicycle is in motion with the rider staying upright. The tires accept contact with the route surface, and, even though they are rolling, the bottoms of the tires deform slightly, practise not sideslip, and are at rest with respect to the road surface for a measurable amount of time. There must be static friction betwixt the tire and the road surface for this to be 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.ii (a) The bicycle moves frontwards, and its tires practice not skid. The bottom of the slightly deformed tire is at rest with respect to the road surface for a measurable corporeality of time. (b) This image shows that the top of a rolling cycle appears blurred past its motility, merely the bottom of the bicycle is instantaneously at rest. (credit a: modification of work past Nelson Lourenço; credit b: modification of piece of work by Colin Rose)

To analyze rolling without slipping, we starting time derive the linear variables of velocity and dispatch of the centre of mass of the wheel in terms of the athwart variables that describe the wheel'due south 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 11.3 (a) A wheel is pulled beyond a horizontal surface by a strength [latex] \overset{\to }{F} [/latex]. The force of static friction [latex] {\overset{\to }{f}}_{\text{S}},|{\overset{\to }{f}}_{\text{South}}|\le {\mu }_{\text{S}}Northward [/latex] is large plenty to continue it from slipping. (b) The linear velocity and acceleration vectors of the center of mass and the relevant expressions for [latex] \omega \,\text{and}\,\alpha [/latex]. Point P is at rest relative to the surface. (c) Relative to the centre of mass (CM) frame, betoken P has linear velocity [latex] \text{−}R\omega \hat{i} [/latex].

From (Figure)(a), we run into the force vectors involved in preventing the wheel from slipping. In (b), point P that touches the surface is at residual relative to the surface. Relative to the center of mass, point P has velocity [latex] \text{−}R\omega \hat{i} [/latex], where R is the radius of the wheel and [latex] \omega [/latex] is the bicycle's angular velocity nigh 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:

[latex] {\overset{\to }{five}}_{P}=\text{−}R\omega \hat{i}+{v}_{\text{CM}}\hat{i}. [/latex]

Since the velocity of P relative to the surface is zero, [latex] {v}_{P}=0 [/latex], this says that

[latex] {v}_{\text{CM}}=R\omega . [/latex]

Thus, the velocity of the wheel'south centre of mass is its radius times the athwart velocity about its centrality. 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 washed beneath for the linear acceleration.

If nosotros differentiate (Effigy) on the left side of the equation, we obtain an expression for the linear acceleration of the eye of mass. On the right side of the equation, R is a abiding and since [latex] \alpha =\frac{d\omega }{dt}, [/latex] we take

[latex] {a}_{\text{CM}}=R\alpha . [/latex]

Furthermore, we tin observe the distance the wheel travels in terms of angular variables by referring to (Effigy). As the wheel rolls from point A to point B, its outer surface maps onto the basis by exactly the distance travelled, which is [latex] {d}_{\text{CM}}. [/latex] We see from (Figure) that the length of the outer surface that maps onto the ground is the arc length [latex] R\theta \text{​} [/latex]. Equating the ii distances, we obtain

[latex] {d}_{\text{CM}}=R\theta . [/latex]

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 [latex] R\theta [/latex] from A to B maps onto the surface, corresponding to the distance [latex] {d}_{\text{CM}} [/latex] that the center of mass has moved.

Example

Rolling Down an Inclined Plane

A solid cylinder rolls down an inclined airplane without slipping, starting from remainder. Information technology has mass one thousand and radius r. (a) What is its dispatch? (b) What condition must the coefficient of static friction [latex] {\mu }_{\text{S}} [/latex] satisfy then the cylinder does not slip?

Strategy

Draw a sketch and gratis-body diagram, and choose a coordinate organization. We put 10 in the direction downward the plane and y upward perpendicular to the plane. Place the forces involved. These are the normal strength, the force of gravity, and the forcefulness due to friction. Write down Newton's laws in the x– and y-directions, and Newton'south police for rotation, and so solve for the dispatch and force due to friction.

Solution

  1. The free-trunk diagram and sketch are shown in (Figure), including the normal forcefulness, components of the weight, and the static friction strength. In that location 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'due south laws in the 10– and y-directions, nosotros have

    [latex] \sum {F}_{ten}=m{a}_{x};\enspace\sum {F}_{y}=m{a}_{y}. [/latex]

    A diagram of a cylinder rolling without 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 s i hat acts at the point of contact and points in the negative x direction.

    Figure 11.five A solid cylinder rolls down an inclined airplane without slipping from rest. The coordinate system has x in the management down the inclined airplane and y perpendicular to the airplane. The costless-body diagram is shown with the normal force, the static friction strength, and the components of the weight [latex] yard\overset{\to }{k} [/latex]. Friction makes the cylinder roll down the airplane rather than slip.

    Substituting in from the free-body diagram,

    [latex] \begin{array}{ccc}\hfill mg\,\text{sin}\,\theta -{f}_{\text{S}}& =\hfill & one thousand{({a}_{\text{CM}})}_{ten},\hfill \\ \hfill N-mg\,\text{cos}\,\theta & =\hfill & 0,\hfill \\ \hfill {f}_{\text{S}}& \le \hfill & {\mu }_{\text{S}}Northward,\hfill \cease{array} [/latex]

    nosotros tin can and so 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]

    Yet, 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 centre of mass of the cylinder. The only nonzero torque is provided by the friction strength. We take

    [latex] {f}_{\text{Due south}}r={I}_{\text{CM}}\blastoff . [/latex]

    Finally, the linear acceleration is related to the athwart acceleration past

    [latex] {({a}_{\text{CM}})}_{10}=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 nosotros have dropped the x-subscript. Nosotros 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{Due south}}}{m} [/latex]

    [latex] {f}_{\text{South}}=\frac{{I}_{\text{CM}}\blastoff }{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}^{two})}.\hfill \finish{array} [/latex]

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

    Since we have a solid cylinder, from (Effigy), we accept [latex] {I}_{\text{CM}}=one thousand{r}^{2}\text{/}two [/latex] and

    [latex] {a}_{\text{CM}}=\frac{mg\,\text{sin}\,\theta }{m+(thou{r}^{2}\text{/}two{r}^{2})}=\frac{2}{3}1000\,\text{sin}\,\theta . [/latex]

    Therefore, nosotros have

    [latex] \blastoff =\frac{{a}_{\text{CM}}}{r}=\frac{2}{3r}g\,\text{sin}\,\theta . [/latex]

  2. Because slipping does non occur, [latex] {f}_{\text{Due south}}\le {\mu }_{\text{S}}Northward [/latex]. Solving for the friction force,

    [latex] {f}_{\text{Southward}}={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 }{one thousand+({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] North=mg\,\text{cos}\,\theta [/latex], nosotros have

    [latex] \frac{mg{I}_{\text{CM}}\text{sin}\,\theta }{one thousand{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}^{two}\text{/}{I}_{\text{CM}})}. [/latex]

    For the solid cylinder, this becomes

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

Significance

  1. The linear dispatch is linearly proportional to [latex] \text{sin}\,\theta . [/latex] Thus, the greater the angle of the incline, the greater the linear dispatch, as would be expected. The angular dispatch, even so, 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 athwart dispatch.
  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.

Check Your Understanding

A hollow cylinder is on an incline at an angle of [latex] lx\text{°}. [/latex] The coefficient of static friction on the surface is [latex] {\mu }_{Due south}=0.half dozen [/latex]. (a) Does the cylinder roll without slipping? (b) Volition a solid cylinder ringlet without slipping

a. [latex] {\mu }_{\text{S}}\ge \frac{\text{tan}\,\theta }{1+(yard{r}^{2}\text{/}{I}_{\text{CM}})} [/latex]; inserting the angle and noting that for a hollow cylinder [latex] {I}_{\text{CM}}=m{r}^{2}, [/latex] we have [latex] {\mu }_{\text{S}}\ge \frac{\text{tan}\,lx\text{°}}{1+(m{r}^{2}\text{/}m{r}^{2})}=\frac{ane}{2}\text{tan}\,60\text{°}=0.87; [/latex] we are given a value of 0.half dozen for the coefficient of static friction, which is less than 0.87, so the status isn't satisfied and the hollow cylinder will slip; b. The solid cylinder obeys the condition [latex] {\mu }_{\text{S}}\ge \frac{i}{iii}\text{tan}\,\theta =\frac{1}{3}\text{tan}\,60\text{°}=0.58. [/latex] The value of 0.6 for [latex] {\mu }_{\text{Due south}} [/latex] satisfies this condition, then the solid cylinder will non slip.

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

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

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 downward a frictionless plane with no rotation. The acceleration will as well be different for two rotating cylinders with unlike rotational inertias.

Rolling Movement with Slipping

In the case of rolling motility with slipping, nosotros must apply the coefficient of kinetic friction, which gives ascent to the kinetic friction forcefulness since static friction is not present. The situation is shown in (Figure). In the case of slipping, [latex] {v}_{\text{CM}}-R\omega \ne 0 [/latex], because point P on the wheel is not at rest on the surface, and [latex] {five}_{P}\ne 0 [/latex]. Thus, [latex] \omega \ne \frac{{five}_{\text{CM}}}{R},\blastoff \ne \frac{{a}_{\text{CM}}}{R} [/latex].

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 k is a force to the left acting on the bottom of the wheel. The force f sub k is equal to mu sub k times N. Figure b is an illustration of the wheel rolling and slipping on a horizontal 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. omega does not equal v sub C M over R and alpha does not equal a sub C M over R. A coordinate system with positive x to the right and positive y up is shown.

Figure eleven.six (a) Kinetic friction arises betwixt the bike and the surface because the wheel is slipping. (b) The simple relationships between the linear and angular variables are no longer valid.

Example

Rolling Downwards an Inclined Plane with Slipping

A solid cylinder rolls down an inclined plane from rest and undergoes slipping ((Figure)). Information technology has mass thou and radius r. (a) What is its linear acceleration? (b) What is its athwart dispatch nigh an centrality through the centre of mass?

Strategy

Draw a sketch and costless-torso diagram showing the forces involved. The free-torso diagram is similar to the no-slipping instance except for the friction force, which is kinetic instead of static. Use Newton's 2nd law to solve for the acceleration in the x-direction. Utilise Newton'due south second law 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.

Effigy eleven.7 A solid cylinder rolls down an inclined plane from residue and undergoes slipping. The coordinate system has 10 in the direction downwardly the inclined plane and y upward perpendicular to the plane. The free-torso diagram shows the normal strength, kinetic friction force, and the components of the weight [latex] m\overset{\to }{g}. [/latex]

The sum of the forces in the y-direction is nix, then the friction force is now [latex] {f}_{\text{m}}={\mu }_{\text{k}}N={\mu }_{\text{k}}mg\text{cos}\,\theta . [/latex]

Newton's second law in the 10-direction becomes

[latex] \sum {F}_{x}=m{a}_{x}, [/latex]

[latex] mg\,\text{sin}\,\theta -{\mu }_{\text{k}}mg\,\text{cos}\,\theta =1000{({a}_{\text{CM}})}_{10}, [/latex]

or

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

The friction force provides the just torque about the axis through the centre of mass, and so Newton's 2nd law of rotation becomes

[latex] \sum {\tau }_{\text{CM}}={I}_{\text{CM}}\blastoff , [/latex]

[latex] {f}_{\text{k}}r={I}_{\text{CM}}\alpha =\frac{1}{two}m{r}^{2}\blastoff . [/latex]

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

[latex] \blastoff =\frac{two{f}_{\text{grand}}}{mr}=\frac{2{\mu }_{\text{thousand}}k\,\text{cos}\,\theta }{r}. [/latex]

Significance

We write the linear and athwart accelerations in terms of the coefficient of kinetic friction. The linear acceleration is the same equally that found for an object sliding down an inclined plane with kinetic friction. The athwart dispatch well-nigh the centrality of rotation is linearly proportional to the normal force, which depends on the cosine of the angle of inclination. As [latex] \theta \to 90\text{°} [/latex], this strength goes to nada, and, thus, the angular acceleration goes to cipher.

Conservation of Mechanical Free energy in Rolling Motion

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

[latex] {Eastward}_{\text{T}}=\frac{1}{2}g{v}_{\text{CM}}^{2}+\frac{one}{ii}{I}_{\text{CM}}{\omega }^{2}+mgh. [/latex]

In the absence of any nonconservative forces that would accept energy out of the organisation in the class of oestrus, the total energy of a rolling object without slipping is conserved and is abiding throughout the movement. Examples where free energy is non conserved are a rolling object that is slipping, production of heat as a consequence of kinetic friction, and a rolling object encountering air resistance.

You may ask why a rolling object that is not slipping conserves energy, since the static friction forcefulness is nonconservative. The answer can exist found by referring back to (Figure). Point P in contact with the surface is at residuum with respect to the surface. Therefore, its minute displacement [latex] d\overset{\to }{r} [/latex] with respect to the surface is nada, and the incremental work done past the static friction strength is zero. Nosotros can utilize energy conservation to our study of rolling move to bring out some interesting results.

Example

Curiosity Rover

The Curiosity rover, shown in (Figure), was deployed on Mars on August 6, 2012. The wheels of the rover accept 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 downward into the bottom of the bowl 25 meters below. If the bike has a mass of 5 kg, what is its velocity at the bottom of the basin?

A photograph of the NASA rover Curiosity during testing at the Jet Propulsion Laboratory.

Figure 11.8 The NASA Mars Science Laboratory rover Marvel during testing on June three, 2011. The location is inside the Spacecraft Assembly Facility at NASA'south Jet Propulsion Laboratory in Pasadena, California. (credit: NASA/JPL-Caltech)

Strategy

We apply mechanical energy conservation to analyze the trouble. At the acme of the loma, the wheel is at rest and has only potential energy. At the bottom of the basin, the wheel has rotational and translational kinetic free energy, which must be equal to the initial potential energy past free energy conservation. Since the bike is rolling without slipping, nosotros use the relation [latex] {v}_{\text{CM}}=r\omega [/latex] to chronicle the translational variables to the rotational variables in the free energy conservation equation. Nosotros then solve for the velocity. From (Effigy), nosotros see that a hollow cylinder is a skillful approximation for the wheel, so we tin use this moment of inertia to simplify the adding.

Solution

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

[latex] mgh=\frac{1}{ii}m{5}_{\text{CM}}^{2}+\frac{1}{two}{I}_{\text{CM}}{\omega }^{two}. [/latex]

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

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

[latex] mgh=\frac{one}{2}m{5}_{\text{CM}}^{2}+\frac{one}{ii}1000{r}^{two}\frac{{v}_{\text{CM}}^{2}}{{r}^{2}} [/latex]

or

[latex] gh=\frac{1}{2}{v}_{\text{CM}}^{2}+\frac{i}{2}{five}_{\text{CM}}^{2}⇒{v}_{\text{CM}}=\sqrt{gh}. [/latex]

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

[latex] {v}_{\text{CM}}=\sqrt{(iii.71\,\text{m}\text{/}{\text{s}}^{2})25.0\,\text{grand}}=9.63\,\text{m}\text{/}\text{s}\text{.} [/latex]

Significance

This is a fairly accurate result because that Mars has very picayune temper, and the loss of free energy due to air resistance would exist minimal. The outcome also assumes that the terrain is shine, such that the wheel wouldn't encounter rocks and bumps forth the way.

Too, in this instance, the kinetic free energy, or free energy of motility, is equally shared between linear and rotational move. If nosotros await at the moments of inertia in (Figure), we run across 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 requite the wheel a larger linear velocity than the hollow cylinder approximation. Thus, the solid cylinder would reach the lesser of the bowl faster than the hollow cylinder.

Summary

  • In rolling movement without slipping, a static friction force is present between the rolling object and the surface. The relations [latex] {five}_{\text{CM}}=R\omega ,{a}_{\text{CM}}=R\alpha ,\,\text{and}\,{d}_{\text{CM}}=R\theta [/latex] all employ, such that the linear velocity, dispatch, and distance of the centre of mass are the angular variables multiplied past the radius of the object.
  • In rolling motility with slipping, a kinetic friction forcefulness arises betwixt the rolling object and the surface. In this instance, [latex] {v}_{\text{CM}}\ne R\omega ,{a}_{\text{CM}}\ne R\blastoff ,\,\text{and}\,{d}_{\text{CM}}\ne R\theta [/latex].
  • Free energy conservation tin can be used to analyze rolling motility. Energy is conserved in rolling move without slipping. Energy is not conserved in rolling motion with slipping due to the heat generated past kinetic friction.

Conceptual Questions

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

No, the static friction forcefulness is zilch.

A cylindrical tin of radius R is rolling across a horizontal surface without slipping. (a) After ane complete revolution of the can, what is the distance that its eye of mass has moved? (b) Would this altitude exist greater or smaller if slipping occurred?

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

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

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

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

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

Problems

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

[latex] {v}_{\text{CM}}=R\omega \,⇒\omega =66.vii\,\text{rad/due south} [/latex]

A boy rides his wheel 2.00 km. The wheels have radius 30.0 cm. What is the total angle the tires rotate through during his trip?

If the boy on the bicycle in the preceding trouble accelerates from residue to a speed of 10.0 thousand/southward in 10.0 s, what is the angular acceleration of the tires?

[latex] \alpha =iii.iii\,\text{rad}\text{/}{\text{s}}^{two} [/latex]

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

A marble rolls downwards an incline at [latex] thirty\text{°} [/latex] from rest. (a) What is its dispatch? (b) How far does it go in 3.0 due south?

[latex] {I}_{\text{CM}}=\frac{two}{5}grand{r}^{ii},\,{a}_{\text{CM}}=3.5\,\text{m}\text{/}{\text{s}}^{ii};\,x=15.75\,\text{k} [/latex]

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

A rigid body with a cylindrical cross-section is released from the superlative of a [latex] xxx\text{°} [/latex] incline. It rolls x.0 m to the lesser in 2.60 s. Discover the moment of inertia of the body in terms of its mass m and radius r.

A yo-yo tin be thought of a solid cylinder of mass thou and radius r that has a light cord wrapped around its circumference (see below). 1 end of the string is held fixed in space. 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 10.0 cm rolls down an incline with slipping. The bending of the incline is [latex] xxx\text{°}. [/latex] 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?

A bowling ball rolls up a ramp 0.5 one thousand loftier without slipping to storage. It has an initial velocity of its eye of mass of three.0 yard/southward. (a) What is its velocity at the peak of the ramp? (b) If the ramp is i m high does information technology get in to the top?

A 40.0-kg solid cylinder is rolling across a horizontal surface at a speed of 6.0 grand/due south. How much work is required to terminate it?

[latex] W=-1080.0\,\text{J} [/latex]

A 40.0-kg solid sphere is rolling across a horizontal surface with a speed of half-dozen.0 yard/southward. How much piece of work is required to end it? Compare results with the preceding problem.

A solid cylinder rolls up an incline at an bending of [latex] twenty\text{°}. [/latex] If it starts at the bottom with a speed of x yard/southward, how far upward the incline does information technology travel?

A solid cylindrical wheel of mass G and radius R is pulled by a forcefulness [latex] \overset{\to }{F} [/latex] applied to the center of the bicycle at [latex] 37\text{°} [/latex] to the horizontal (see the post-obit effigy). If the wheel is to roll without slipping, what is the maximum value of [latex] |\overset{\to }{F}|? [/latex] The coefficients of static and kinetic friction are [latex] {\mu }_{\text{S}}=0.40\,\text{and}\,{\mu }_{\text{k}}=0.thirty. [/latex]

The forces on a wheel, radius R, on a horizontal surface are shown. The wheel is centered on an x y coordinate system that has positive x to the right and positive y up. Force F acts on the center of the wheel at an angle of 37 degrees above the positive x direction. Force M g acts on the center of the wheel and points down. Force N points up and acts at the contact point where the wheel touches the surface. Force f sub s points to the left and acts at the contact point where the wheel touches the surface.

A hollow cylinder is given a velocity of 5.0 m/s and rolls up an incline to a height of i.0 m. If a hollow sphere of the aforementioned mass and radius is given the same initial velocity, how high does it roll upwardly the incline?

Glossary

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

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Source: https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/11-1-rolling-motion/

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