Kinetic Energy of Rotation Moment of Inertia

Kinetic Energy of Rotation KEi = ½mivi2 = ½ mi(riω)2 = ½ miri2ω2 KEtotal = Σ½ miri2ω2 = ½ (Σ miri2) ω2

Moment of inertia :

I = Σ miri2

mi o

ri

Moment of inertia = rotational mass m depends only on the quantity of matter in an object I depends on both the quantity of matter and its distribution in the rigid body

KErotational = ½ I ω2 KEtranslational = ½ mv2

Moment of Inertia Example 1: mass of the connecting rod is neglected. (very light rod) m1= 1 kg

L = 0.50 m

m2= 2 kg

L = 0.50 m

I = Σ miri2 = m1L2 + m2L2 I = (1 kg)(0.50 m)2 + (2 kg)(0.50 m)2 = 0.75 kgm2 Example 2 :(example 8.8) Example 3: (example 8.9)

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Moments of Inertia for various Rigid Objects of Uniform Composition Composition

Relationship between Torque and Angular Acceleration Newton’s Second Law: ΣF = Fnet = ma Fnet = ma = mrα Fnet r = mr2α

Στ = τnet = I α Newton’ Newton’s second Law for rotation Problem 36 Examples 8.12, Problem 39

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Angular Momentum Στ = I α

Στ = I

ωf − ωi Iωf − Iωi = ∆t ∆t

Angular momentum: L = I ω Units: kgm2rad/s Linear momentum: p = m v

Angular Momentum Στ =

Iωf − Iωi ∆L = ∆t ∆t F=

∆p ∆t

If Στ = 0, ∆L = 0 ∆L = Li - Lf = 0 angular momentum is conserved

Li = Lf

Iiωi = Ifωf Example 8.13, Problem 52

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