Ch11 - Chapter 11: Rotational Vectors and Angular Momentum...

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Chapter 11: Rotational Vectors and Angular Momentum Vector (cross) products ± Definition of vector product and its properties Axis of rotation a r b r b a r r × θ sin ˆ b a n b a r r r r = × n ˆ unit vector normal to the plane defined by a r and b r a b b a r r r r × = × 0 r r r r r = × = × a a a a
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Vector (cross) products (cont’d) ± Properties of vector product k j i ˆ ˆ ˆ = × x y z i ˆ j ˆ k ˆ k j i ˆ , ˆ , ˆ i k j ˆ ˆ ˆ = × j i k ˆ ˆ ˆ = × Consider three vectors: ) , , ( z y x a a a a : unit vector in x,y,z direction = r ) , , ( z y x b b b b = r ) , , ( z y x c c c c = r k b a b a j b a b a i b a b a k b j b i b k a j a i a b a c x y y x z x x z y z z y z y x z y x ˆ ) ( ˆ ) ( ˆ ) ( ) ˆ ˆ ˆ ( ) ˆ ˆ ˆ ( + + = + + × + + = × = r r r Then
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Vector (cross) products (cont’d) ± Properties of vector product (cont’d) z j ˆ k ˆ y z z y x b a b a c = z x x z y b a b a c = y x y y x z b a b a c = x i ˆ z y x z y x b b b a a a k j i ˆ ˆ ˆ = b a c r r r × =
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Torque Torque is a quantitative measure of the tendency of a force to cause or change the rotational motion. ± Case for 1 point-like object of mass m with 1 force x y r r θ m F r r ˆ ˆ massless rigid rod φ • r is constant only tangential component of causes rotation F r ˆ = F F t r α ω mr r r r m a m F = = = ˆ ) ˆ ˆ ( ˆ ˆ 2 r r 0 τ I mr r F t = = 2 t F moment of inertia torque unit Nm
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Torque ± Case for 1 point-like object of mass m with 1 force x y r r F r r ˆ φ sin r r t = lever arm F r Fr Fr r F t t r r × = = = = τ sin F r t F Define r v r × = r is aligned with the rotation axis
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Torque (cont’d) ± Case for 2 point-like objects with 2 forces 2 1 θ = = 1 r r 2 r r t F 2 t F 1 2 2 1 1 r F r F t t net + = τ > 0 < 0 x y increases decreases Example: A see-saw in balance Mm r R Mg 0 2 2 1 1 = = + = mgr MgR r F r F t t net r R M m / / = mg
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Work & energy (I) ± A massive body on massless rigid rod y r r θ m t F r Work done by the force: ∫∫ ∫ = = = τ d rd F ds F W t t Also ∫∫ = = ω α d dt d I d I W ) / ( = d dt d d d I ) / )( / ( 1 0 1 0 | ) 2 / 1 ( 2 = = I d I K I I = = 2 0 2 1 ) 2 / 1 ( ) 2 / 1 ( x W= K
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Work & energy (I) (cont’d) ± Power in rotational motion Work done by the force: ∫∫ ∫ = = = θ τ d rd F ds F W t t d dW = dt d dt dW / / = τω = P Power :
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Correspondence between linear & angular quantities linear angular displacement velocity acceleration mass force Newton’s law kinetic energy work x θ dt dx v / = dt d / ω= dt dv a / = dt d / ω α= m = 2 i i r m I F r α τ I = ma F = F r r r r × = 2 ) 2 / 1 ( mv K = 2 ) 2 / 1 ( I K = = Fdx W = d W
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Work & energy (II) ± A massive body in rotational & translational motion Kinetic energy: = i i i v v m K r r ) 2 / 1 ( V v v i i r r r + = ' where ' i v r Now is the velocity with respect to V r the center of mass (COM) and ∑∑ + = + + = + + = + + = + + = = 2 2 ' ' 2 2 ' ' 2 2 ' 2 ' 2 ' ' ' ) 2 / 1 ( ) 2 / 1 ( / ) ( ) 2 / 1 ( ) 2 / 1 ( / ) 2 / 1 ( ) 2 / 1 ( ) 2 / 1 ( ) 2 / 1 ( ) ( ) ( ) 2 / 1 ( ) 2 / 1 ( MV v m dt r m d V MV v m dt r d m V m V v m V m v V m v m V v V v m v v m K i i i i i i i i i i i i i i i i i i i i i i r r r r r r r r r r the velocity of COM. Then r r 0 com ' i v r V r
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Work & energy (II) cont’d ±
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This note was uploaded on 05/28/2011 for the course PHY 126 taught by Professor Staff during the Summer '08 term at SUNY Stony Brook.

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Ch11 - Chapter 11: Rotational Vectors and Angular Momentum...

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