Analytical Mech Homework Solutions 118

Analytical Mech - 16 a a 2b α 2b a b After some algebra … the angular velocity of B is found to be … aα ωB = β −α φ = 2b For A we take

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Unformatted text preview: 16 a ( a + 2b ) α 2b ( a + b ) After some algebra … the angular velocity of B is found to be … aα ωB = β −α + φ = 2b For A , we take moments about O and for B we take moments about its center. Call TA and TB the components of the reaction forces tangent to A and B (the “upward-going” TA acts on disk B . The “downward-going” TA acts on disk A) so that φ = Thus ( a + 2b )α 2 (a + b) and β = K − TA a = I Aα (Torque on Disk A) ( α ) 2b ) 1 M B aα 2 −TAb − TB b = − I B β − α + φ = − I B a ( TA − TB = M B ( a + b ) α − φ = (Torque on Disk B) (Force on Disk B) Eliminate TA and TB α ( 4b2 I A + M B a 2b2 + a 2 I B ) 4b 2 Integrating this equation gives : K= α ( 4b2 I A + M B a 2b2 + a 2 I B ) 4b 2 Putting ω A = α at t = t gives Kt = 4b 2 Kt ωA = ( 4b2 I A + M B a 2b2 + a 2 I B ) Putting in values for I A and I B gives 4 Kt ωA = 3 a 2 2M A + M B 2 Since the angular velocity of B is ω B = β − α + φ = ωB = a ωA = 2b 2 Kt 3 ab 2M A + M B 2 16 aα , we have 2b ...
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This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.

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