Study GuidesExam QuestionsPast Papers
AP®PhysicsCollege BoardPhysics 1: Algebra-BasedStudy GuidesUnit 6: Energy & Momentum of Rotating SystemsExamples of Rotational SystemsRolling

Rolling (College Board AP® Physics 1: Algebra-Based)

Study Guide

Katie M

Written by: Katie M

Reviewed by: Caroline Carroll

Total kinetic energy

total kinetic energy = translational kinetic energy + rotational kinetic energy

K subscript t o t a l end subscript space equals space K subscript t r a n s end subscript space plus space K subscript r o t end subscript

K subscript t o t a l end subscript space equals space 1 half m v squared space plus space 1 half I omega squared

  • Where:

    • m = mass of the object, in kg

    • v = linear velocity, in straight m divided by straight s

    • I = rotational inertia, in kg times straight m squared

    • omega = angular velocity, in rad divided by straight s

  • When considering the mechanical energy of a system, the kinetic energy can be broken down into its translational and rotational components

E space equals space increment K space plus thin space increment U

E space equals space increment open parentheses K subscript t r a n s end subscript space plus space K subscript r o t end subscript close parentheses space plus thin space increment U

  • Two special cases involving rotational kinetic energy are

    • rolling motion without slipping

    • rolling motion while slipping

Rolling without slipping

  • Rolling motion is a combination of rotating and sliding (translational) motion

  • When a circular rigid body (e.g. a ball, wheel, or cylinder) of radius r rolls without slipping:

    • the translational motion of the system's center of mass is directly related to its rotational motion

    • the point of contact between the rotating body and the surface is at rest

  • The distance traveled by the system's center of mass is:

increment x subscript c m end subscript space equals space r increment theta

  • Where:

    • increment x subscript c m end subscript = translational distance moved by the object's center of mass, in straight m

    • increment theta = angular displacement of the rolling object, in rad

  • The linear velocity of the system's center of mass is:

v subscript c m end subscript space equals space r omega

  • Where:

    • v subscript c m end subscript = translational velocity of the object's center of mass, in straight m divided by straight s

    • omega = angular speed of the rolling object, in rad divided by straight s

  • The linear acceleration of the system's center of mass is:

a subscript c m end subscript space equals space r alpha

  • Where:

    • a subscript c m end subscript = translational acceleration of the object's center of mass, in straight m divided by straight s squared

    • alpha = angular acceleration of the rolling object, in rad divided by straight s squared

  • These equations appear to be identical to the equations for fixed-axis rotation, but there are some key differences

    • For fixed axis rotation: s, v, and a refer to the tangential quantities at a point on the object

    • For a rolling object: increment x subscript c m end subscript, v subscript c m end subscript, and a subscript c m end subscript refer to the translational quantities of the object's center of mass

Translational and rotational components of a rotating wheel

A rotating wheel with radius r and angular velocity ω. The distance traveled by the wheel is equal to the arc length that has been in contact with the ground. The relationships between velocity, acceleration and their equivalent angular variables are also shown.
The linear velocity and acceleration of the center of mass of a rotating wheel are proportional to its angular counterparts. As the wheel rolls on a surface, the arc length that has been in contact with the ground is equal to the distance traveled by the center of mass

The role of friction in rolling without slipping

  • When an object rolls without slipping, static friction causes the rotation about the center of mass

  • This happens when there is no relative motion between the surface and the point of contact

  • The direction of the frictional force always acts to oppose the potential slipping motion

Static friction causing rolling on a horizontal plane

Two diagrams: a ball rolling on a flat surface with speed lines and a free-body diagram showing forces (gravity, normal, friction, and applied force).
When a circular rigid body is pulled across a horizontal surface by a force F, the force of static friction is large enough to keep it from slipping.

Static friction causing rolling on an inclined plane

A ball rolling down an inclined plane with angle θ without slipping. The free-body diagram shows the normal force, the static friction force, and the weight.
When a circular rigid body rolls down an inclined plane, the force of static friction is large enough to make it roll instead of slip

  • In the case of rolling motion without slipping:

    • no kinetic friction is present

    • only static friction is present

  • Since no work is done by static friction, no energy is dissipated when a system rolls without slipping

Rolling on a horizontal surface without slipping

  • When a disk rotates on a horizontal surface:

    • all points on the disk have a constant angular speed

    • each point on the disc has a different linear velocity depending on its distance from the center open parentheses v space proportional to space r close parentheses

    • the linear speed is the same at all points on the circumference

  • When a disk slips or slides on a horizontal surface:

    • there is not enough static friction present to allow the object to roll

    • all points on the disk have the same linear velocity

    • the angular velocity is zero

  • So, when a disk rolls without slipping:

    • there is enough static friction present to initiate rotational motion allowing the object to roll

    • the point in contact with the surface has a velocity of v space equals space 0

    • the center of mass has a velocity of v subscript c m end subscript space equals space omega r

    • the top point has a velocity of v space equals space 2 omega r, or 2 v

A rolling disc with rotational plus translational motion. The velocity vectors cancel and add together to demonstrate the net effect on a rolling object. At the top: v = 2ωr. At the center of mass: v = ωr. At the bottom: v = 0.
Rolling motion is a combination of rotational and translational motion. The resultant velocity at the bottom is zero and the resultant velocity at the top is 2v

Rolling on an inclined plane without slipping

  • Another common scenario is a circular rigid body rolling on an inclined plane without slipping

  • Since the object rolls without slipping, no work is done by frictional forces

  • Therefore, the total mechanical energy of the system is conserved

    • At the top of the slope, the object has gravitational potential energy only (assuming it is initially at rest)

    • As the object rolls down the slope, the gravitational potential energy is transferred to kinetic energy

    • At the bottom of the slope, the object has both translational and rotational kinetic energy

  • The linear or angular velocity can then be determined by:

    • applying energy conservation laws

    • using the equation for the rotational inertia of the object

    • using the relationship between linear and angular velocity v subscript c m end subscript space equals space omega r

The total kinetic energy acquired by a ball rolling without slipping down a slope is equal to the initial gravitational potential energy at the top of the slope. As it rolls down the slope it gains both translational and rotational kinetic energy
The gravitational potential energy store of the ball is transferred to the translational and rotational kinetic energy store as it rolls down the slope

Worked Example

A solid sphere of mass M, radius R, and rotational inertia I space equals space 2 over 5 M R squared is released from rest at the top of a ramp of height h which makes an angle of theta with the horizontal. Throughout its motion, the sphere rolls without slipping.

(A) Derive an expression for the linear velocity of the center of mass of the sphere at the bottom of the inclined plane in terms of M, R, theta, h and physical constants as appropriate.

(B) Derive an expression for the linear acceleration of the center of mass of the sphere in terms of M, R, theta, h and physical constants as appropriate.

Answer:

Part (A)

Step 1: Analyse the scenario

  • When an object rolls down the ramp without slipping:

    • no work is done by kinetic friction

    • the total mechanical energy of the system is conserved

  • Therefore, the sphere's gravitational potential energy U subscript g at the top of the ramp is completely transferred into kinetic energy K subscript t o t a l end subscript at the bottom

  • The rolling motion indicates the total kinetic energy of the sphere consists of both translational K subscript t r a n s end subscript and rotational K subscript r o t end subscript components

Step 2: Apply energy conservation laws to the scenario

  • The total mechanical energy of the system is equal to:

K subscript i space plus thin space U subscript i space equals space K subscript f space plus thin space U subscript f

  • Since K subscript i space equals space U subscript f space equals space 0, U subscript i space equals space increment U subscript g, and K subscript f space equals space increment K subscript t o t a l end subscript:

increment U subscript g space equals space increment K subscript t o t a l end subscript

Step 3: Write expressions for increment U subscript g and increment K subscript t o t a l end subscript

  • At the top of the ramp, the gravitational potential energy of the stationary sphere is equal to:

increment U subscript g space equals space m g increment y space equals space M g h

  • At the bottom of the ramp, the total kinetic energy of the sphere is equal to:

increment K subscript t o t a l end subscript space equals space K subscript t r a n s end subscript space plus space K subscript r o t end subscript space equals space 1 half M v subscript c m end subscript squared space plus space 1 half I omega squared

  • Equate the expressions for increment U subscript g and increment K subscript t o t a l end subscript:

M g h space equals space 1 half M v subscript c m end subscript squared space plus space 1 half I omega squared

  • Substitute the expressions for rotational inertia I space equals space 2 over 5 M R squared and angular velocity omega space equals space v subscript c m end subscript over R into the equation:

M g h space equals space 1 half M v subscript c m end subscript squared space plus space 1 half open parentheses 2 over 5 M R squared close parentheses open parentheses v subscript c m end subscript over R close parentheses squared

Step 4: Write the final expression for v subscript c m end subscript

M g h space equals space 1 half M v subscript c m end subscript squared space plus space 1 fifth M v subscript c m end subscript squared

g h space equals space 7 over 10 v subscript c m end subscript squared

v subscript c m end subscript space equals space square root of fraction numerator 10 g h over denominator 7 end fraction end root

Part (B)

Step 1: Analyze the scenario

  • The forces acting on the sphere during its motion are:

    • the force of gravity F subscript g acting directly downwards

    • the normal force F subscript N acting up and perpendicular to the ramp

    • the force of static friction F subscript f acting up and parallel to the ramp

Diagram showing a sphere on an inclined plane with forces labeled: normal force (FN), friction force (Ff), gravitational force (Fg=Mg), and components Mg cosθ and Mg sinθ.

  • The net torque exerted on the sphere's center of mass is due to the force of static friction only

    • This is because it is the only force which does not act directly through the sphere's center of mass

  • The net force exerted on the sphere's center of mass is proportional to its linear acceleration

Step 2: Write an expression for the frictional force acting on the sphere

  • Newton's second law in rotational form is given by:

sum tau space equals space I alpha

  • The torque due to the frictional force is:

tau space equals space F r space sin space theta space equals space F subscript f R space sin space 90 degree space equals space F subscript f R

  • Substitute the expressions for rotational inertia I space equals space 2 over 5 M R squared and angular acceleration alpha space equals space a subscript c m end subscript over R into the equation:

F subscript f R space equals space open parentheses 2 over 5 M R squared close parentheses open parentheses a subscript c m end subscript over R close parentheses

  • Therefore, the frictional force acting on the sphere is:

F subscript f space equals space 2 over 5 M a subscript c m end subscript

Step 3: Write an expression for the net force exerted on the sphere

  • The net force, and therefore, the linear acceleration of the sphere act down and parallel to the ramp

  • Newton's second law is given by:

sum F space equals space M a subscript c m end subscript

  • Two forces act parallel to the ramp, in opposite directions, these are:

    • a component of the gravitational force (weight): M g space sin space theta

    • the frictional force F subscript f

  • The net force on the sphere is therefore equal to:

M g space sin space theta space minus space F subscript f space equals space M a subscript c m end subscript

Step 4: Write the final expression for a subscript c m end subscript

M g space sin space theta space minus space 2 over 5 M a subscript c m end subscript space equals space M a subscript c m end subscript

M g space sin space theta space equals space 7 over 5 M a subscript c m end subscript

a subscript c m end subscript space equals space 5 over 7 g space sin space theta

Examiner Tips and Tricks

The equation for translational displacement straight capital delta x subscript c m end subscript space equals space r straight capital delta theta is given on your equation sheet, but the equations for v subscript c m end subscript and a subscript c m end subscript are not.

However, the equations for v subscript c m end subscript and a subscript c m end subscript have the same form as their tangential forms, and these are given on the equation sheet. Furthermore, both can easily be derived by dividing the straight capital delta x subscript c m end subscript equation by time. Similarly, dividing the equation for v subscript c m end subscript by time gives the equation for a subscript c m end subscript.

Rolling while slipping

  • When a rigid body rolls while slipping, the translational motion of the system's center of mass cannot be directly related to its rotational motion

increment x subscript c m end subscript space not equal to space r increment theta

v subscript c m end subscript space not equal to space r omega

a subscript c m end subscript space not equal to space r alpha

  • This is because the point of contact between the rotating body and the surface is not at rest

  • In the case of rolling motion while slipping:

    • there is not enough static friction to cause the rotation about the system's center of mass

    • only kinetic friction is present as there is relative motion between the surface and the point of contact

  • As with the no-slipping case, the direction of the frictional force acts to oppose the slipping motion

  • Since work is done by kinetic friction, energy is dissipated from the system when slipping occurs

  • The amount of energy dissipation (work done) depends on

    • the coefficient of kinetic friction

    • the distance over which slipping occurs

Examiner Tips and Tricks

You will not be expected to quantitively explain the relationships between linear and angular quantities for an object rolling while slipping, as these are beyond the scope of both AP Physics 1 and 2.

However, you may be expected to qualitatively explain the changes to linear and angular quantities for a rigid body rolling while slipping.

You may also come across questions where a rigid body is released from the top of a frictionless slope, be aware that this results in slipping, but neither static nor kinetic frictional forces are acting in this scenario, so this results in purely translational motion.

You may come across the term 'rolling friction' in your studies, but rest assured this is beyond the scope of AP Physics 1 and will not be tested.

Last updated:

You've read 0 of your 5 free study guides this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Test yourself

    Did this page help you?

    Previous:Angular Momentum of a SystemNext:Motion of Orbiting Satellites
    Author
    Reviewer
    Katie M

    Author: Katie M

    Expertise: Physics

    Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

    Caroline Carroll

    Author: Caroline Carroll

    Expertise: Physics Subject Lead

    Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.