kin_gear_prob1_sol

kin_gear_prob1_sol - Given 24 2 = N 18 3 = N 30 4 = N 36 6...

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Unformatted text preview: Given: 24 2 = N , 18 3 = N , 30 4 = N , 36 6 = N , 150 7 = ω RPM CW, 250 2 = ω RPM CCW Find: 5 ω Solution: We first need to find 7 N . Since 6 7 4 3 2 2 r r r r r + = + + ( 1 ) and since “all gears have the same module” we can replace r with N in equation (1) and obtain 6 7 4 3 2 2 N N N N N + = + + , ( 2 ) i.e. ( )( ) 54 36 30 18 2 24 2 6 4 3 2 7 = − + + = − + + = N N N N N . (3) We assume that arm 5 rotates CW and obtain the velocity profile for the gear system shown below (separate explanation for the velocities in gear 3 is given later). We therefore have ( ) 3 2 2 3 2 5 3 r r r r ω ω ω + + = , ( 4 ) ( ) ( ) ( ) ( ) ( ) ( ) 4 2 2 7 6 3 2 5 4 7 6 5 2 2 3 2 5 3 2 5 4 7 6 5 3 3 3 2 5 4 2 2 r r r r r r r r r r r r r r r r r r r r ω ω ω ω ω ω ω ω ω ω + − − + = + − + + + + = + − + + = (5) by virtue of (4), and ( ) 6 7 6 5 7 7 6 r r r r + − = ω ω ω . ( 6 ) 2 3 4 6 7 7 ω 2 ω 2 2 r ω ( ) 3 2 5 r r + ω 5 5 ( ) 3 3 3 2 5 r r r ω ω + + 7 7 r ω ( ) 7 6...
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This note was uploaded on 12/02/2011 for the course ME 4133 taught by Professor Ram during the Fall '06 term at LSU.

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kin_gear_prob1_sol - Given 24 2 = N 18 3 = N 30 4 = N 36 6...

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