Rolling without slipping

The top of a rolling wheel can move faster than the axle while the bottom is instantaneously at rest relative to the road. No part has detached. Translation and rotation simply combine differently at different points.

After developing rolling dynamics, we introduce angular momentum about a stated origin. It provides a new conservation method for changing mass distributions, central forces and impacts where linear momentum is not conserved.

What you will learn to do

Add centre translation and rotation to obtain point velocities.
Use the no-slip condition only when the contact can sustain it.
Combine rotational and translational energy for rolling.
Determine static-friction direction from force and torque conditions.
Define angular momentum about a stated origin and identify zero external torque.
Handle changing inertia and angular impacts without assuming kinetic-energy conservation.

Useful before this chapterTorque, rotational inertia and fixed-axis dynamics →Impulse, momentum and collisions →Static friction: find what is needed →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Rolling motion and angular momentum
Learning goals:
- Add centre translation and rotation to obtain point velocities.
- Use the no-slip condition only when the contact can sustain it.
- Combine rotational and translational energy for rolling.
- Determine static-friction direction from force and torque conditions.
- Define angular momentum about a stated origin and identify zero external torque.
- Handle changing inertia and angular impacts without assuming kinetic-energy conservation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

BEFORE YOU READ

Bring back one idea

Try a sentence or a sketch before opening the comparison. If this is new, use the explanation to get started.

What does “no slipping” constrain at the point touching a stationary road?

Compare your reasoning

The contacting point has zero instantaneous velocity relative to the road. For a circular wheel this gives v of the centre = ωR. It does not make every point on the wheel stationary.

Revisit the explanation →

01 / Rolling motion and angular momentum

Rolling without slipping is a contact-velocity condition

A wheel can translate, rotate or do both. Rolling without slipping on stationary ground is the special motion in which the material point touching the ground has zero instantaneous velocity relative to it. This condition links the centre speed to angular speed: vcom = Rω for the corresponding magnitudes.

The velocity of any point is the centre-of-mass velocity plus the velocity produced by rotation about the centre. For rightward rolling, the wheel rotates clockwise. At the bottom, the rotational velocity points left and cancels the centre’s rightward velocity. At the top, the rotational velocity points right and adds, giving speed 2vcom.

The point at contact is only instantaneously at rest. A moment later another material point is in contact, while the former one rises. Zero instantaneous velocity does not imply zero acceleration. For constant-speed rolling, the bottom material point has an upward centripetal acceleration relative to the centre even though its instantaneous ground velocity is zero.

\mathbf v_P=\mathbf v_{\mathrm{com}}+\boldsymbol\omega\times\mathbf r_{P/\mathrm{com}},\qquad v_{\mathrm{com}}=R\omega\quad\text{for no slip on fixed ground}

The vector relation applies to every point of a rigid body. The scalar no-slip relation assumes the stated ground and rolling direction.

INTERACTIVE LAB

Add translation and rotation at a marked point

Start with the point at the top. Move it to 90°, 180° and 270°. Predict each velocity direction before reading the components. Then change the centre speed and check the top and bottom again.

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WORKED EXAMPLE

The side of a wheel moves diagonally

A wheel rolls right without slipping at centre speed 3 m/s on stationary ground. Find the velocity of the rightmost rim point at an instant. Use +x right and +y up.

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STOP & THINK

A material point on the rim is momentarily at rest at the ground contact. What follows?

It remains fixed to the ground permanently
Its instantaneous velocity is zero, but it can have nonzero acceleration
The entire wheel is at rest

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Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Rolling motion and angular momentum
Focus: Rolling without slipping is a contact-velocity condition
Learning goals:
- Add centre translation and rotation to obtain point velocities.
- Use the no-slip condition only when the contact can sustain it.
- Combine rotational and translational energy for rolling.
- Determine static-friction direction from force and torque conditions.
- Define angular momentum about a stated origin and identify zero external torque.
- Handle changing inertia and angular impacts without assuming kinetic-energy conservation.
Equations in this section (LaTeX):
\mathbf v_P=\mathbf v_{\mathrm{com}}+\boldsymbol\omega\times\mathbf r_{P/\mathrm{com}},\qquad v_{\mathrm{com}}=R\omega\quad\text{for no slip on fixed ground}
Meaning: The vector relation applies to every point of a rigid body. The scalar no-slip relation assumes the stated ground and rolling direction.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

02 / Rolling motion and angular momentum

Split rigid-body kinetic energy about the centre of mass

For planar rigid-body motion, total kinetic energy is the translational energy of the centre of mass plus rotational energy about the centre of mass. The cross terms in the particle sum cancel because mass-weighted displacements from the centre sum to zero.

For rolling without slipping, substitute ω = v/R to express both contributions in terms of centre speed. Write Icom = βMR², where β captures the mass distribution about the symmetry axis. Then K = ½Mv²(1 + β). A thin hoop has β = 1, a uniform solid disc or cylinder has β = 1/2, and a uniform solid sphere has β = 2/5.

For a gravity-driven descent without dissipation, the same potential-energy loss is shared between these two kinetic-energy stores. A larger β leaves less energy in centre translation at the same total energy, so the centre speed is smaller. This comparison assumes the bodies actually satisfy the no-slip condition and have the stated shapes and axes.

K=\frac12Mv_{\mathrm{com}}^2+\frac12I_{\mathrm{com}}\omega^2=\frac12Mv_{\mathrm{com}}^2(1+\beta)

Use the centre-of-mass inertia in this split. Combining an arbitrary-axis inertia with a separate centre-translation term can double-count energy.

WORKED EXAMPLE

A rolling disc shares its energy between two motions

A uniform solid disc rolls from rest down a stationary rough track without slipping or dissipative losses, descending vertically by 1.5 m. Use g = 10 m/s². Find its centre speed.

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Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Rolling motion and angular momentum
Focus: Split rigid-body kinetic energy about the centre of mass
Learning goals:
- Add centre translation and rotation to obtain point velocities.
- Use the no-slip condition only when the contact can sustain it.
- Combine rotational and translational energy for rolling.
- Determine static-friction direction from force and torque conditions.
- Define angular momentum about a stated origin and identify zero external torque.
- Handle changing inertia and angular impacts without assuming kinetic-energy conservation.
Equations in this section (LaTeX):
K=\frac12Mv_{\mathrm{com}}^2+\frac12I_{\mathrm{com}}\omega^2=\frac12Mv_{\mathrm{com}}^2(1+\beta)
Meaning: Use the centre-of-mass inertia in this split. Combining an arbitrary-axis inertia with a separate centre-translation term can double-count energy.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

03 / Rolling motion and angular momentum

Static friction can supply the torque needed for rolling

Consider an unpowered circular body rolling down a fixed incline under gravity, with no axle torque. Gravity acts through the centre and has no torque about it. The normal also passes through the centre for the ideal circular contact. Static friction supplies the torque that increases the clockwise rolling rate, so in this situation it points uphill.

With downhill positive, translation gives Mg sin θ − f = Ma. Rotation about the centre gives fR = Icom α in compatible rotational sense. The no-slip constraint supplies a = Rα. Combining them gives a = g sin θ/(1 + β) and f = βMa.

Static friction is not automatically at its limiting value. The required f must be no greater than μsN, with N = Mg cos θ for the stated straight incline. This gives μs ≥ [β/(1 + β)]tan θ. If the contact cannot supply that value, sliding begins and the rolling constraint used in the derivation is invalid.

WORKED EXAMPLE

Verify grip after solving the no-slip equations

A uniform solid disc rolls from rest down an incline with sin θ = 0.6 and cos θ = 0.8. Take g = 10 m/s². Find its acceleration and the minimum static-friction coefficient needed for this no-slip motion.

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STOP & THINK

A thin hoop rolls without slipping down an incline with sin θ = 0.4. With g = 10 m/s² and β = 1, what is its centre acceleration?

Answer in m/s².

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Rolling motion and angular momentum
Focus: Static friction can supply the torque needed for rolling
Learning goals:
- Add centre translation and rotation to obtain point velocities.
- Use the no-slip condition only when the contact can sustain it.
- Combine rotational and translational energy for rolling.
- Determine static-friction direction from force and torque conditions.
- Define angular momentum about a stated origin and identify zero external torque.
- Handle changing inertia and angular impacts without assuming kinetic-energy conservation.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

04 / Rolling motion and angular momentum

Check the contact point before deciding what work friction does

For ideal rolling without slipping on a stationary rigid surface, the contact point has zero instantaneous velocity. The power of the static contact force on the rigid body is therefore zero at that point. Static friction can still change the division between translational and rotational motion without dissipating mechanical energy in this ideal model.

This statement has conditions. A moving surface can have a nonzero contact velocity in the ground frame, so static friction can transfer energy to or from the body. Real rolling can also dissipate energy through deformation, hysteresis or bearing losses even without visible sliding. The ideal rigid no-loss model excludes those effects.

The direction of static friction is likewise not universal. For the unpowered downhill body it points uphill; for a driven wheel an applied axle torque can make it point forward. Determine the tendency of relative slip from the forces and torques, then solve for the required static friction. Do not infer its direction solely from the centre’s velocity.

STOP & THINK

Does static friction always point opposite the centre-of-mass velocity of a rolling wheel?

Yes
No; its direction depends on the tendency to slip at contact
It never acts during rolling

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Rolling motion and angular momentum
Focus: Check the contact point before deciding what work friction does
Learning goals:
- Add centre translation and rotation to obtain point velocities.
- Use the no-slip condition only when the contact can sustain it.
- Combine rotational and translational energy for rolling.
- Determine static-friction direction from force and torque conditions.
- Define angular momentum about a stated origin and identify zero external torque.
- Handle changing inertia and angular impacts without assuming kinetic-energy conservation.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

05 / Rolling motion and angular momentum

Angular momentum measures motion about a stated origin

For a particle, angular momentum about an origin O is L = r × p, where r points from O to the particle. Its magnitude is momentum times the perpendicular distance from the origin to the line of motion. A particle can have angular momentum even while travelling in a straight line; it need not orbit the origin.

Changing the origin generally changes angular momentum. If the line of motion passes through the origin, r and p are parallel and L is zero. If the same straight line misses the origin, the cross product can be nonzero and constant for force-free motion.

Differentiate r × p about a fixed origin in an inertial frame. The term v × mv vanishes, leaving dL/dt = r × F = τ. For a system, the external torque changes total angular momentum under the usual classical interaction assumptions. This is the rotational counterpart of external force changing total linear momentum.

\mathbf L_O=\mathbf r\times\mathbf p,\qquad\frac{d\mathbf L_O}{dt}=\boldsymbol\tau_{\mathrm{ext},O}

The reference origin is fixed in an inertial frame for this form. Angular momentum has units kg·m²/s.

WORKED EXAMPLE

Straight motion can carry angular momentum

A 2 kg particle moves at 3 m/s along +x on the line y = 4 m. Find its angular momentum about the origin at an arbitrary position x. Use a right-handed frame with +z out of the page.

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Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Rolling motion and angular momentum
Focus: Angular momentum measures motion about a stated origin
Learning goals:
- Add centre translation and rotation to obtain point velocities.
- Use the no-slip condition only when the contact can sustain it.
- Combine rotational and translational energy for rolling.
- Determine static-friction direction from force and torque conditions.
- Define angular momentum about a stated origin and identify zero external torque.
- Handle changing inertia and angular impacts without assuming kinetic-energy conservation.
Equations in this section (LaTeX):
\mathbf L_O=\mathbf r\times\mathbf p,\qquad\frac{d\mathbf L_O}{dt}=\boldsymbol\tau_{\mathrm{ext},O}
Meaning: The reference origin is fixed in an inertial frame for this form. Angular momentum has units kg·m²/s.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

06 / Rolling motion and angular momentum

Conserve the angular-momentum component with zero external torque

If net external torque about a fixed origin is zero, total angular momentum about it is constant. If only one torque component is zero, the corresponding angular-momentum component is conserved. This distinction matters for an axle that can exert support torques in other directions.

For rotation about a fixed axis, the angular-momentum component along that axis is Iω. If a person or mechanism changes the mass distribution while net external torque about the axis is negligible, Iω remains constant but ω can change. Internal forces can do work, so rotational kinetic energy need not remain constant.

Writing K = L²/(2I) for this axis motion shows the energy consequence. At fixed L, reducing I increases K. The increased kinetic energy comes from internal work or another energy store, not from the angular-momentum conservation law itself.

WORKED EXAMPLE

Conserved angular momentum can accompany an energy increase

A rotating system changes its axis moment of inertia from 4 kg·m² to 2 kg·m² while external torque about that axis is negligible. Its initial angular speed is 3 rad/s. Find its final angular speed and the change in rotational kinetic energy.

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STOP & THINK

A system with I = 6 kg·m² rotates at 2 rad/s. With negligible external axis torque, its inertia changes to 3 kg·m². What is its final angular speed?

Answer in rad/s.

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Rolling motion and angular momentum
Focus: Conserve the angular-momentum component with zero external torque
Learning goals:
- Add centre translation and rotation to obtain point velocities.
- Use the no-slip condition only when the contact can sustain it.
- Combine rotational and translational energy for rolling.
- Determine static-friction direction from force and torque conditions.
- Define angular momentum about a stated origin and identify zero external torque.
- Handle changing inertia and angular impacts without assuming kinetic-energy conservation.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

07 / Rolling motion and angular momentum

A radial force has zero torque about its centre

If a force always acts along the position vector from a fixed centre, its torque about that centre is zero. The particle’s angular momentum about the centre is therefore conserved when no other torque acts. This is true even if the force magnitude changes with radius.

In planar motion, angular momentum magnitude is mrvθ, where vθ is the tangential velocity component. A radial velocity contributes no angular momentum because its cross product with r vanishes. Thus a decreasing radius generally increases the tangential component when angular momentum is conserved, but it does not imply that the path is a sequence of circles.

The areal sweep rate is ½rvθ = L/(2m), so it is constant for a central-force orbit. Mechanical energy conservation is a separate question depending on whether the central force is conservative and whether its source changes with time. Zero torque alone does not imply zero work: a radial force can do work during radial motion.

WORKED EXAMPLE

A radial pull changes tangential speed without torque

A 1 kg particle moves on a smooth horizontal plane while a string pulls through a fixed small hole. Its radius decreases from 2 m to 1 m. Initially its tangential speed is 3 m/s. The string force is radial, and other horizontal forces are negligible. Find the final tangential speed.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Rolling motion and angular momentum
Focus: A radial force has zero torque about its centre
Learning goals:
- Add centre translation and rotation to obtain point velocities.
- Use the no-slip condition only when the contact can sustain it.
- Combine rotational and translational energy for rolling.
- Determine static-friction direction from force and torque conditions.
- Define angular momentum about a stated origin and identify zero external torque.
- Handle changing inertia and angular impacts without assuming kinetic-energy conservation.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

08 / Rolling motion and angular momentum

An axle can preserve angular momentum while changing linear momentum

During a short impact on a wheel mounted on a fixed axle, the axle can exert a large external impulse. Linear momentum of the wheel-plus-projectile system therefore need not be conserved. But if the axle force acts through the axis and axle friction torque is negligible, its torque about that axis is zero, so the axis angular momentum can be conserved during impact.

A projectile that sticks changes the final moment of inertia. Include its mass at its actual perpendicular distance from the axis. Using only the wheel’s original I would overestimate the final angular speed. As with a sticking linear collision, kinetic energy generally decreases during the impact.

Choose the conservation law from the external force or torque conditions, not from a vague statement that the event is quick. Specify the axis, check which impulses have zero moment about it and calculate the incoming angular momentum using the perpendicular impact parameter.

WORKED EXAMPLE

A sticking projectile changes both momentum and inertia

A fixed-axis wheel initially at rest has I = 0.45 kg·m². A 0.20 kg small clay pellet moving at 5 m/s strikes tangentially at radius 0.50 m and sticks. Axle friction torque and other external axis impulses are negligible during impact. Find the angular speed afterwards.

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Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Rolling motion and angular momentum
Focus: An axle can preserve angular momentum while changing linear momentum
Learning goals:
- Add centre translation and rotation to obtain point velocities.
- Use the no-slip condition only when the contact can sustain it.
- Combine rotational and translational energy for rolling.
- Determine static-friction direction from force and torque conditions.
- Define angular momentum about a stated origin and identify zero external torque.
- Handle changing inertia and angular impacts without assuming kinetic-energy conservation.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

09 / Rolling motion and angular momentum

Treat rolling as a proposed motion that friction must support

Writing a=αR proposes a no-slip constraint. Static friction must be able to supply the force that makes it true. Let I_C=βmR². For gravity-driven rolling down a fixed incline, translation gives ma=mg sinθ−f and rotation gives βmR²(a/R)=fR. Here f is the uphill friction magnitude. Eliminating f gives a=g sinθ/(1+β), then f=βmg sinθ/(1+β).

The normal reaction is mg cosθ in this simple model. Therefore no slip requires f≤μ_s mg cosθ. If the requirement fails, the rolling equations are not a slightly inaccurate answer; they describe an unavailable motion. Use kinetic friction and separate translational and angular equations after slip begins. Static friction equals its maximum only at impending slip.

\mu_s\geq\frac{\beta}{1+\beta}\tan\theta

The inequality tests feasibility, not a new acceleration formula. It assumes a rigid circular body on a stationary plane, no other applied forces and no rolling resistance.

WORKED EXAMPLE

Check the grip before using the rolling acceleration

A solid sphere has β=2/5 on an incline with tanθ=0.70. Is μ_s=0.18 enough for no-slip rolling under gravity alone?

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Friction need not point opposite the centre’s velocity. Consider a horizontal force F applied at height h above a wheel’s centre. Take rightward translation and clockwise rotation as positive, and let f be the signed rightward friction. Translation gives ma=F+f; torque about the centre gives βmR²α=Fh−fR. Combining these with a=αR determines both acceleration and the direction of friction.

At h=βR, the applied force already produces the correct ratio of linear and angular acceleration, so no friction is required. Below that height friction points left; above it friction points right. This result is valid only if the surface can provide the calculated static friction.

a=\frac{F(1+h/R)}{m(1+\beta)},\qquad f=\frac{F(h/R-\beta)}{1+\beta}

A positive f means rightward friction in this convention. The horizontal applied force leaves N=mg; require |f|≤μ_s mg.

Explain this equation

Read it in words

Rightward friction equals the applied force times the difference between relative pull height and inertia factor, divided by one plus that factor.

Symbols and units

$f$: Signed rightward static frictionN
$F$: Magnitude of the horizontal rightward pullN
$h$: Pull height above the centrem
$R$: Rolling radiusm
$\beta$: Central inertia divided by mR²dimensionless
$a$: Rightward centre accelerationm/s²
$m$: Body masskg

Signs and operations

Positive f acts rightward. A negative numerator means leftward friction. Clockwise angular acceleration is positive for rightward rolling.

When it applies

Rigid circular body, horizontal stationary plane, no slip, horizontal pull, no rolling resistance; check |f|≤μ_smg.

Make it concrete

For a top pull on a solid cylinder, h/R=1 and β=1/2, so f=F/3 rightward.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Equation (LaTeX): a=\frac{F(1+h/R)}{m(1+\beta)},\qquad f=\frac{F(h/R-\beta)}{1+\beta}
Read it as: Rightward friction equals the applied force times the difference between relative pull height and inertia factor, divided by one plus that factor.
Symbols:
f: Signed rightward static friction; N
F: Magnitude of the horizontal rightward pull; N
h: Pull height above the centre; m
R: Rolling radius; m
\beta: Central inertia divided by mR²; dimensionless
a: Rightward centre acceleration; m/s²
m: Body mass; kg
Signs and operations: Positive f acts rightward. A negative numerator means leftward friction. Clockwise angular acceleration is positive for rightward rolling.
Conditions: Rigid circular body, horizontal stationary plane, no slip, horizontal pull, no rolling resistance; check |f|≤μ_smg.

Help me explain the equation in words, predict how changing a quantity affects the result, and apply it to a small numerical example.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

WORKED EXAMPLE

Friction can help the forward motion

Pull a 2 kg solid cylinder horizontally at its top with 6 N. Take β=1/2, g=10 m/s² and μ_s=0.20. Find a and f if the cylinder rolls.

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STOP & THINK

A hoop has β=1 and rolls under a horizontal force applied at its very top. In the ideal model, what fraction f/F of rightward friction is required?

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Rolling motion and angular momentum
Focus: Treat rolling as a proposed motion that friction must support
Learning goals:
- Add centre translation and rotation to obtain point velocities.
- Use the no-slip condition only when the contact can sustain it.
- Combine rotational and translational energy for rolling.
- Determine static-friction direction from force and torque conditions.
- Define angular momentum about a stated origin and identify zero external torque.
- Handle changing inertia and angular impacts without assuming kinetic-energy conservation.
Equations in this section (LaTeX):
\mu_s\geq\frac{\beta}{1+\beta}\tan\theta
Meaning: The inequality tests feasibility, not a new acceleration formula. It assumes a rigid circular body on a stationary plane, no other applied forces and no rolling resistance.
a=\frac{F(1+h/R)}{m(1+\beta)},\qquad f=\frac{F(h/R-\beta)}{1+\beta}
Meaning: A positive f means rightward friction in this convention. The horizontal applied force leaves N=mg; require |f|≤μ_s mg.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

10 / Rolling motion and angular momentum

Separate the brief collision from the motion that follows

A pivot can deliver a large impulse during a collision, so the combined projectile-and-target linear momentum is usually not conserved. Yet a force applied at an ideal fixed pivot has zero moment about that pivot. During a sufficiently brief collision, if other external angular impulses are negligible, angular momentum about the pivot is conserved. This is a statement about a chosen axis and a time interval, not a rule that angular momentum is always conserved.

Choose the projectile and target as the system. Before impact, a projectile contributes mu times the perpendicular distance from the pivot to its line of motion. After sticking, its contribution to the moment of inertia is mb², where b is its final radial distance from the pivot. Those two distances coincide for a tangential hit. For an oblique hit, the incoming angular momentum needs the perpendicular component but the final inertia still uses the full radius.

Sticking is inelastic. First find the common angular speed from angular momentum; then compute kinetic energy before and after to find the energy converted to deformation, vibration and heat. Do not impose kinetic-energy conservation during impact. If the assembly subsequently rises in gravity, a separate energy balance may describe that later swing.

WORKED EXAMPLE

A small projectile changes both momentum and inertia

A uniform 2 kg rod of length 1.2 m is initially at rest about an ideal fixed end pivot. A 0.50 kg projectile travelling at 4 m/s strikes the free end perpendicular to the rod and sticks. Neglect other angular impulses. Find the immediate angular speed and lost kinetic energy.

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STOP & THINK

For the fixed-pivot sticking impact, which balance is justified during the collision?

Conserve both linear momentum and kinetic energy.
Conserve angular momentum about the pivot; calculate kinetic-energy loss afterward.
Conserve kinetic energy but ignore angular momentum.

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Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Rolling motion and angular momentum
Focus: Separate the brief collision from the motion that follows
Learning goals:
- Add centre translation and rotation to obtain point velocities.
- Use the no-slip condition only when the contact can sustain it.
- Combine rotational and translational energy for rolling.
- Determine static-friction direction from force and torque conditions.
- Define angular momentum about a stated origin and identify zero external torque.
- Handle changing inertia and angular impacts without assuming kinetic-energy conservation.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

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BRING THE IDEAS TOGETHER

What should stay with you

No slip on fixed ground means zero instantaneous contact-point velocity.
Rolling energy splits into centre translation and rotation about the centre.
Static friction supplies the required contact constraint within its limit.
Angular momentum depends on the stated origin or axis.
Zero external axis torque conserves the corresponding angular-momentum component.
Changing inertia or sticking can change kinetic energy while angular momentum stays constant.

RETRIEVE, THEN EXPLAIN

Try it without the examples

Work these out before checking. If an answer surprises you, identify which assumption or step needs another look. A correct choice is most useful when you can explain why the alternatives fail.

SELF CHECK 1

A wheel rolls without slipping on stationary ground at centre speed 4 m/s. What is the instantaneous speed of its top rim point?

Answer in m/s.

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SELF CHECK 2

A rotating isolated-axis system reduces its moment of inertia while conserving angular momentum. What happens to its rotational kinetic energy?

It must stay constant
It increases, with an energy source associated with the rearrangement
It becomes zero

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SELF CHECK 3

A no-slip incline calculation requires friction above μsN. What follows?

Use a larger static friction anyway
The assumed no-slip motion is inconsistent and sliding must be considered
Set the angular acceleration to zero without changing other equations

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SELF CHECK 4

Why can angular momentum about a fixed axle be conserved during a short impact even when linear momentum is not?

The axle impulse can have zero moment about the axle
All impacts conserve every quantity
Torque and force are identical

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SELF CHECK 5

A rotor has constant inertia 2 kg·m² and initial angular velocity −1 rad/s. A positive torque pulse supplies angular impulse +6 N·m·s. Find the final angular velocity.

Answer in rad/s.

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APPLY THE CHAPTER

Take the ideas into a problem

Try a complete problem next. Commit to a method before revealing a hint, then compare your reasoning with the worked solution.

COME BACK TO THE IDEA

What can you explain later?

Try these from memory at your next study session, then again a few days later. Say your reasoning aloud or sketch it before looking. Adjust the gap to what you can explain.

Next study session

How can static friction supply torque to an ideally rolling wheel without doing work at its contact with a stationary rigid road?

Compare your reasoning

The point where the force acts has zero instantaneous velocity in that ground frame, so the contact power is zero. Friction can still redistribute the translational and rotational parts of the motion.

Revisit the explanation →

A few days later: change the situation

A rotating skater pulls their arms inward with negligible external torque. What stays fixed, and where can extra rotational kinetic energy come from?

Compare your reasoning

Angular momentum stays fixed while moment of inertia decreases and angular speed rises. The skater does internal work, so rotational kinetic energy need not remain constant.

Revisit the explanation →

Compare the physical reason, the assumptions and the conclusion. Your wording can differ. If one part is missing, revisit that section and try explaining it again after a break.

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