10_1425_web_Lec_21_AngularMomentum

# 10_1425_web_Lec_21_AngularMomentum - Angular Momentum...

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Angular Momentum Physics 1425 Lecture 21 Michael Fowler, UVa

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A New Look for τ = I α We’ve seen how τ = I α works for a body rotating about a fixed axis . τ = I α i s not true in general if the axis of rotation is itself accelerating BUT it IS true if the axis is through the CM, and isn’t changing direction! This is quite tricky to prove—it’s in the book And τ CM = I CM α CM is often useful, as we’ll see.
Forces on Hoop Rolling Down Ramp Take no slipping, so v = R ω , a = R α Translational accn F = ma : mg sin θ - F fr = ma Rotational accn τ CM = I CM α CM : F fr R = mR 2 α = mRa so F fr = ma and mg sin = 2 ma, a = ( g sin )/2: the acceleration is one-half that of a sliding frictionless block—and independent of mass or radius . x mg sin F fr The only force having torque about the center of the hoop (its CM) is the frictional force : the total gravitational force and the normal force both act through the center.

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Yet Another Look at That Hoop… Take no slipping, so v = R ω , a = R α Since there’s no slipping , the point on the hoop in contact with the ramp is momentarily at rest , and the hoop is rotating about that point .
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10_1425_web_Lec_21_AngularMomentum - Angular Momentum...

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