Sect. 10.4: Rotational Kinetic Energy
Translation-Rotation Analogues & Connections
Translation
Rotation
Displacement
x
θ
Velocity
v
ω
Acceleration
a
α
Mass
m
?
Kinetic Energy (K) (½)mv2
?
CONNECTIONS
s = rθ, v = rω, at= rα, ac = (v2/r) = ω2r
Sect. 10.4: Rotational Kinetic Energy
• Translational motion (Chs. 7 & 8): K = (½)mv2
• Rigid body rotation, angular velocity ω. Rigid means
 Every point has the same ω. Object is made of particles, masses mi.
• For each mi at a distance ri from the rotation axis: vi = riω.
So, each mass mi has kinetic energy Ki = (½)mi(vi)2. So,
The Rotational Kinetic Energy is:
KR = ∑[(½)mi(vi)2] = (½)∑mi(ri)2ω2 = (½)∑mi(ri)2ω2
ω2 goes outside the sum, since it’s the same everywhere in the body
– Define the moment of inertia of the object, I  ∑mi(ri)2
 KR = (½)Iω2 (Analogous to (½)mv2)
Translation-Rotation Analogues & Connections
Translation
Rotation
Displacement
x
θ
Velocity
v
ω
Acceleration
a
α
Force (Torque)
F
τ
Mass (moment of inertia) m
I
Kinetic Energy (K) (½)mv2
(½)Iω2
CONNECTIONS
s = rθ, v = rω, at= rα, ac = (v2/r) = ω2r
Example 10.3: Four Rotating Objects
4 tiny spheres on ends of 2
massless rods. Arranged in
the x-y plane as shown.
Sphere radii are very small.
 Masses are treated as
point masses. Calculate
moment of inertia I when
(A) The system is rotated
about the y axis, as in Fig. a.
(B) The system is rotated in
the x-y plane as in Fig. b.
Bottom line for rotational kinetic energy
• Analogy between the kinetic energies associated with
linear motion: K = (½)mv2, & the kinetic energy
associated with rotational motion: KR = (½)Iω2.
• NOTE: Rotational kinetic energy is not a new type of
energy. But the form of this kinetic energy is different
because it applies to a rotating object.
• Of course, the units of rotational kinetic energy are Joules (J)
Sect. 10.5: Moments of Inertia
• Definition of moment of inertia:
I  ∑mi(ri)2
• Dimensions of I are ML2 & its SI units are kg.m2
• Use calculus to calculate the moment of inertia of an object:
Assume it is divided into many small volume elements, each
of mass Δmi. Rewrite expression for I in terms of Δmi. & take
the limit as Δmi  0. So
Make a small volume segment assumption & the sum changes to an
integral:
2
I    r dV
ρ ≡ density of the object. If ρ is constant, the integral can be evaluated
with known geometry, otherwise its variation with position must be known.
With this assumption, I can be calculated for simple geometries.
Notes on Densities ρ
• Volume Mass Density
Solid mass m in volume V:
  = mass per unit volume:  = (m/V)
• Surface Mass Density
Mass m on a thin surface of thickness t:
 σ = mass per unit thickness: σ = t
• Linear Mass Density
Mass m in a rod of length L & cross sectional area A:
 λ = mass per unit length: λ = (m/L) = A
Moment of Inertia of a Uniform Thin Hoop
• This is a thin hoop, so all
mass elements are the same
distance from the center
Moment of Inertia of a Uniform Thin Rod
• The hatched area has mass
dm = λdx = (M/L)dx
• Linear Mass Density: Mass
per unit length of a rod of
uniform cross-sectional area
λ = M/L = ρA
So, the moment of inertia is
Moment of Inertia of a Uniform Solid Cylinder
• Divide cylinder into concentric shells
of radius r, thickness dr, length L.
Volume Mass Density:
ρ = (m/V)
Then, I becomes:
Moments of Inertia of Some Rigid Objects
The Parallel Axis Theorem
NOTE! I depends on the rotation axis! Suppose we need
I for rotation about the off-center axis in the figure:
d
If the axis of interest is a distance d from a
parallel axis through the center of mass,
the moment of inertia has the form:
I  I cm  Md
2
Moment of Inertia of a Rod Rotating Around an End
• We’ve seen: Moment of inertia
ICM of a rod of length L about the
center of mass is:
• What is the moment of inertia I
about one end of the rod?
Shift rotation axis by D = (½)L
& use Parallel Axis Theorem: