# p09fl21 - Lecture 21: Rotations (21 Oct 09) review take...

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Unformatted text preview: Lecture 21: Rotations (21 Oct 09) review take home exam grades: high 98, low 72; range: 90-100 ’ A; 70-90 ’ B. A. Rotation–moment of inertia 1. FW Sec. 26 2. build on material of Secs. 6,7, L6,7. 3. The relation between inertial and body frame time derivatives of a vector s is d s dt | inertial = d s dt | body + ~ω × s 4. In particular for the position r , the relation between inertial and body frame time derivatives becomes v = d r dt | inertial = d r dt | body + ~ω × r 5. Then if the vector r is fixed in the body frame coordinates, the total (rotational) kinetic energy is T = 1 2 X p m p v 2 p = 1 2 X p m p ( ~ω × r p ) · ( ~ω × r p ) 6. Use vector cross product identities [ a · ( b × c ) = ( a × b ) · c ] ( ~ω × r ) · ( ~ω × r ) = [( ~ω × r ) × ~ω ] · r =- [ ~ω × ( ~ω × r )] · r = ω 2 r 2- ( ~ω · r ) 2 = ω 2 r 2 ⊥ 7. Define the moment of inertia tensor (Cartesian components) I ij ≡ X p m p ( r 2 p δ i,j- x pi x pj ) 1 8. The rotational kinetic energy is8....
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## This note was uploaded on 12/14/2009 for the course PHYS 711 taught by Professor Bruch during the Fall '09 term at University of Wisconsin.

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p09fl21 - Lecture 21: Rotations (21 Oct 09) review take...

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