1. Sect. 10.4: Rotational Kinetic Energy

2. Translation-Rotation Analogues & Connections
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

3. 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)

4. Translation-Rotation Analogues & Connections
Displacement x θ
Velocity v ω
Acceleration a α
Force (Torque) F τ
Mass (moment of inertia) m I
Kinetic Energy (K) (½)mv (½)Iω2
CONNECTIONS
s = rθ, v = rω, at= rα, ac = (v2/r) = ω2r

5. 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.
Figure 10.8: (Example 10.3) Four spheres form an unusual baton. (a) The baton is rotated about the y axis. (b) The baton is rotated about the z axis.

6. 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)

7. 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:
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.

8. Notes on Densities ρ Volume Mass Density
Solid mass m in volume V:
 r = mass per unit volume: r = (m/V)
Surface Mass Density
Mass m on a thin surface of thickness t:
 σ = mass per unit thickness: σ = rt
Linear Mass Density
Mass m in a rod of length L & cross sectional area A:
 λ = mass per unit length: λ = (m/L) = rA

9. Moment of Inertia of a Uniform Thin Hoop
This is a thin hoop, so all mass elements are the same distance from the center

10. 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

11. 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:

12. Moments of Inertia of Some Rigid Objects

13. 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:

14. 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: