Mech4Sols

# Mech4Sols - Physics 105B Problem Set 4 Jeff Schonert...

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Physics 105B Problem Set 4 April 29, 2008 Jef Schonert: schonert (at) physics.ucsb.edu Taylor 10.19 a) We will work out the x -component oF the cross product; the other components Follow identically. Given two vectors ±ω and ± r , their cross product in component notation is × ± r =( ω y z - ω x y, ω z x - ω x z, ω x y - ω y x ) (1) Now consider the x -component oF the cross product ± r × ( × ± r ): ( ± r × ( × ± r )) x = y ( ω x y - ω y x ) - z ( ω z x - ω x z )=( y 2 + z 2 ) ω x - xyω y - xzω z (2) This is the x -component oF equation 10.35; the calculation oF the y and z components is now straightForward. b) The BAC - CAB rule For the cross product in this problem is ± r × ( × ± r )= ( ± r · ± r ) - ± r ( · ± r ). Then the x -component is ( ± r × ( × ± r )) x = ω x ( x 2 + y 2 + z 2 ) - x ( ω x x + ω y y + ω z z y 2 + z 2 ) ω x - xyω y - xzω z (3) which is the same as in a). Again, the calculation oF the y and z components is identical. Taylor 10.24 a) Let the center oF mass be located at ± r 0 x 0 ,y 0 ,z 0 ). We are told that the point P is displaced by Δ ξ, η, ζ ) From this point. Now consider the displacement oF the mass From both oF these points. IF q q 1 ,q 2 3 ) is the displacement oF the mass M From P , and q ± q ± 1 ± 2 ± 3 ) is the displacement oF M From the center oF mass, then the two displacement vectors are related by q ± - q = Δ . This allows us to work out how the components oF q and q ± are related to each other: q 1 = q ± 1 - ξ q 2 = q ± 2 - η q 3 = q ± 3 - ζ So now we have written the displacement oF the mass From P in terms oF the known quantities (since we are assuming we already know the moment oF inertia about the center oF mass). Now plug these into the de±nitions oF the inertia tensor components: I xx = ± i m i ( q ± 2 2 - η ) 2 +( q ± 3 2 - ζ ) 2 = ± i m i ( q ± 2 2 + q ± 2 3 + η 2 + ζ 2 - 2 ηq ± 2 - 2 ζq ± 3 ) = ± i m i ( q ± 2 2 + q ± 2 3 )+ ± i m i ( η 2 + ζ 2 ) (4) 1

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The sums involving ηq ± 2 and ζq ± 3 were dropped because they are zero according to equation (10.7). Therefore, I xx is related to the moment of inertia about the center of mass I cm xx by I xx = I cm xx + M ( η 2 + ζ 2 ) (5) Similarly, I yy = ± i m i ( ( q ± 1 - ξ ) 2 +( q ± 3 - ζ ) 2 ) (6) = ± i m i ( q ± 2 1 - 2 q ± 1 ξ + ξ 2 + q ± 2 3 - 2 q ± 3 ζ + ζ 2 ) (7) = ± i m i ( ( q ± 2 1 + q ± 2 3 )+ ξ 2 + ζ 2 ) (8) = I cm + M ( ξ 2 + ζ 2 ) (9) The calculation of the other diagonal component follows identically: I zz = I cm + M ( η 2 + ξ 2 ) (10) Now turn to the products of inertia, which are calculated in the same way. For example, I xz = - ± i m i q 1 q 3 (11) = - ± i m i ( q ± 1 q ± 3 - q ± 1 ζ - q ± 3 ξ + ξζ ) (12) = I xz - Mξζ, (13) where again we used (10.7) to drop the q ± 1 ζ and q ± 3 ξ terms. Since the inertia tensor is
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Mech4Sols - Physics 105B Problem Set 4 Jeff Schonert...

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