Resource Lesson
Rotational Dynamics: Rolling Spheres/Cylinders
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Rolling Spheres and Cylinders
Consider the following three diagrams. The first one shows the velocity vectors for an object experiencing only pure translation. The center diagram is for an object that is experiencing only rotation. The final diagram is a combination of the two - both translation and rotation.
Notice how the vectors add together. At the bottom of the object that is both rotating and translating, the contact point is instantaneously at rest. This is why static friction is used to calculate the torque that produces rotational motion. Also notice that the very top point on the wheel is moving with speed 2v
_{CM}
- faster than any other point on the wheel.
Rotational Dynamics
Now consider an object rolling down an incline plane. The first thing that we are going to do is draw a freebody diagram of the forces acting on this mass and then resolve those forces into their components which act parallel and perpendicular to the plane.
Let's consider the axis of rotation passing through the disk's center of mass, cm. Notice in this case that only the instantaneous static friction force will supply a torque since the lines of action of the other two forces (normal and weight) act through the center of mass and cannot produce a torque. As long as the mass
rolls without slipping
, we can use the relationships:
v = rω
and
a = rα
.
Rotationally,
net
τ
=
I
_{CM}
α
τ
= f
_{s}
r
f
_{s}
=
I
_{CM}
(α/r)
Remember linearly,
net F = ma
mg sinθ - f
_{s}
= mrα where a = rα
Simultaneously eliminating f
_{s}
and solving for α yields:
mg sinθ -
I
_{CM}
(α/r) = mrα
α = g sinθ /(r +
I
_{CM}
/mr)
The moment of inertia for a disk, or solid cylinder (see chart provided below), equals ½mr
^{2}
. Substituting in this value and simplifying gives us
α = 2g sinθ/(3r)
Since this angular acceleration is uniform, you would be free to use any of the
rotational kinematics equations
to solve for final angular velocity, time to travel down the incline, or the number of rotations it completes as it rolls along the incline's surface.
Moments of Inertia
Below you will find a chart of the three most popular "rolling objects." Notice that their rotational inertia increases from left to right as the mass distribution gets farther from the axis of rotation that passes through their center of mass.
solid sphere
I
= 2/5 MR
^{2}
solid cylinder
(about central axis)
I
= 1/2 MR
^{2}
thin-walled cylinder/hoop/ring
(about central axis)
I
= MR
^{2}
Based on this chart, which object in the following picture would you predict would reach the bottom of the incline first: the solid cylinder or the thin ring? To decide, consider the rotational inertia of each object and how inertia affects motion. Both objects have the same mass and equal diameters.
picture courtesy of Arbor Scientific
Using Energy Methods
As a solid sphere rolls without slipping down an incline, its initial gravitational potential energy is being converted into two types of kinetic energy: translational KE and rotational KE.
If the sphere were to both roll and slip, then conservation of energy could not be used to determine its velocity at the base of the incline. The slipping would result in kinetic friction doing work on the sphere and dissipating energy in the form of heat.
Remember that when an object rolls down an incline, it's linear (also called translational) kinetic energy is always less than what it would have been had it was only slid down a frictionless incline.
An Application of the Parallel-Axis Theorem
The principle that can unite a rolling object's rotational and translational kinetic energy into one expression of total KE is called the
Parallel Axis Theorem
.
I
_{P}
=
I
_{CM}
+ Mh
^{2}
where
I
_{P}
represents the object's moment of inertia from any location, P
I
_{CM}
represents the object's moment of inertia about its center of mass
h represents the perpendicular distance from P to the center of mass
For our purposes, let P represent the point of contact where the rolling thin ring, cylinder, or sphere touches the incline's surface.
Total KE = ½
I
_{P}
ω
^{2}
Total KE = ½(
I
_{CM}
+ mh
^{2}
)ω
^{2}
Total KE = ½
I
_{CM}
ω
^{2}
+ ½mh
^{2}
ω
^{2}
Total KE = ½
I
_{CM}
ω
^{2}
+ ½m(r
^{2}
ω
^{2}
)
Total KE = ½
I
_{CM}
ω
^{2}
+ ½mv
^{2}
Total KE = KE
_{rotational}
+ KE
_{translational}
This procedure can apply to any rolling object - just substitute in its correct moment of inertia.
Related Documents
Lab:
Labs -
A Physical Pendulum, The Parallel Axis Theorem and A Bit of Calculus
Labs -
Conservation of Momentum in Two-Dimensions
Labs -
Mass of a Paper Clip
Labs -
Moment of Inertia of a Bicycle Wheel
Labs -
Rotational Inertia
Resource Lesson:
RL -
A Chart of Common Moments of Inertia
RL -
A Further Look at Angular Momentum
RL -
Center of Mass
RL -
Centripetal Acceleration and Angular Motion
RL -
Discrete Masses: Center of Mass and Moment of Inertia
RL -
Hinged Board
RL -
Introduction to Angular Momentum
RL -
Rolling and Slipping
RL -
Rotary Motion
RL -
Rotational Dynamics: Pivoting Rods
RL -
Rotational Dynamics: Pulleys
RL -
Rotational Equilibrium
RL -
Rotational Kinematics
RL -
Rotational Kinetic Energy
RL -
Thin Rods: Center of Mass
RL -
Thin Rods: Moment of Inertia
RL -
Torque: An Introduction
Worksheet:
APP -
The Baton Twirler
APP -
The See-Saw Scene
CP -
Center of Gravity
CP -
Torque Beams
CP -
Torque: Cams and Spools
NT -
Center of Gravity
NT -
Center of Gravity vs Torque
NT -
Falling Sticks
NT -
Rolling Cans
NT -
Rolling Spool
WS -
Moment Arms
WS -
Moments of Inertia and Angular Momentum
WS -
Practice: Uniform Circular Motion
WS -
Rotational Kinetic Energy
WS -
Torque: Rotational Equilibrium Problems
TB -
Basic Torque Problems
TB -
Center of Mass (Discrete Collections)
TB -
Moment of Inertia (Discrete Collections)
TB -
Rotational Kinematics
TB -
Rotational Kinematics #2
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