HW9_prob4(a)_GPS3_G7 - d L dt = N(9 Next take the dot...

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Problem GPS 5.3 (2 nd ed. 5-7) We have a collection of mass points rotating about a fi xed point, and the positions r i and velocities v i of these points are measured in a space fi xed, inertial frame whose origin is at the fi xed point. Start with the de fi nition of T T = 1 2 X i m i v i · v i , (1) and take a total time derivative (in the inertial, space fi xed frame, naturally) to fi nd dT dt = X i m i d v i dt · v i . (2) Next, we fi rst note from Newton’s Second Law that m i d v i dt = F i , (3) where F i is the force on mass point i , and we also express the velocity of point i in terms of the angular velocity vector ω as v i = d r i dt = ω × r i , (4) because in the rotating frame ( d r i /dt ) b = 0 for all points of a rigid body. After substituting Eqs.(3) and (4) into (2) we fi nd dT dt = X i F i · ( ω × r i ) , (5) which we cyclically permute to obtain dT dt = X i ω · ( r i × F i ) . (6) Recognizing that the total torque on the rigid body is N = X i ( r i × F i ) , (7) we fi nally obtain the desired result dT dt = ω · N . (8) 1
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Alternate derivation: Start with the fundamental equation governing the rate of change of angular momentum from Chapter 1 or Chapter 5
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Unformatted text preview: d L dt = N . (9) Next, take the dot product of Eq.(9) with ω to get ω · d L dt = ω · N . (10) Now use the rate of change of a vector formula to rewrite the left dot product of Eq.(10) as ω · d L dt = ω · μ d L dt ¶ b + ω · ( ω × L ) , (11) and recognize that ω · ( ω × L ) = . Then, express ω and L as vectors in the principal axis body frame (with constant values of the principal moments I 1 , I 2 , and I 3 ) to expand the dot product as ω · μ d L dt ¶ b = ( ω 1 I 1 dω 1 dt + ω 2 I 2 dω 2 dt + ω 3 I 3 dω 3 dt ) = 1 2 d dt ( I 1 ω 2 1 + I 2 ω 2 2 + I 3 ω 2 3 ) = dT dt , (12) and obtain the desired result, dT dt = ω · N . (13) The distinction between a space frame and body frame derivative in the f nal equation is super F uous because T is a scalar. 2...
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