Sect. 10.4: Rotational Kinetic Energy

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

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 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. 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.

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: 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:  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

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:

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: