Sect. 10.4: Rotational Kinetic Energy

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Sect. 10.4: Rotational Kinetic Energy

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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)mv 2 ? CONNECTIONS - PowerPoint PPT Presentation

Transcript of Sect. 10.4: Rotational Kinetic Energy

Page 1: Sect. 10.4: Rotational Kinetic Energy

Sect. 10.4: Rotational Kinetic Energy

Page 2: 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

Page 3: Sect. 10.4: Rotational Kinetic Energy

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)

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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

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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.

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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)

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• 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

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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) =

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Moment of Inertia of a Uniform Thin Hoop

• This is a thin hoop, so all mass elements are the same distance from the center

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• 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

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• 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

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Moments of Inertia of Some Rigid Objects

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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

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• 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