Sect. 10.4: Rotational Kinetic Energy
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Transcript of Sect. 10.4: Rotational Kinetic Energy
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 Objects4 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)
• 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.
Sect. 10.5: Moments of Inertia
2I r dV
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) =
Moment of Inertia of a Uniform Thin Hoop
• This is a thin hoop, so all mass elements are the same distance from the center
• 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 = ρASo, the moment of inertia is
Moment of Inertia of a Uniform Thin Rod
• Divide cylinder into concentric shells of radius r, thickness dr, length L.
Volume Mass Density:
ρ = (m/V) Then, I becomes:
Moment of Inertia of a Uniform Solid Cylinder
Moments of Inertia of Some Rigid Objects
2cmI I Md
NOTE! I depends on the rotation axis! Suppose we need I for rotation about the off-center axis in the figure:
If the axis of interest is a distance d from aparallel axis through the center of mass, the moment of inertia has the form:
The Parallel Axis Theorem
d
• 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:
Moment of Inertia of a Rod Rotating Around an End