Lecture20

# Lecture20 - Torque and Angular momentum vectors Torque...

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i.e. only y component contribute Fcos φ Fsin φ F r Torque and Angular momentum vectors τ is perpendicular to r and F τ has magnitude τ = r·F·sin Φ Torque τ = Ι α τ ext = = r Fsin φ τ ext = r x F ext Vector form: Lever Arm Another way of looking at τ = Fr sin φ is by defining the lever arm, d r F φ φ d d is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force τ = d F r F tan F tan =

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Torque and Angular momentum vectors Angular Momentum L = r x p L is perpendicular to r and p L has magnitude L = r·p·sin Φ L = Ι ω Vector form: τ is perpendicular to r and F τ has magnitude τ = r·F·sin Φ Torque τ = Ι α τ ext = r F tan = r Fsin φ τ ext = r x F ext Vector form: τ = d F
Angular momentum ( l ) of a particle l = r x p Definition l = r x p = r p sin φ l = m r 2 ω v= ω r = Ι ω I l = Ι ω = mvr The value of l depends on the choice of the axis of rotation. If we choose the center of the circle r perpendicular to p Relate angular momentum to torque Σ τ = d l dt Valid

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