EAS208_HW09_112910_Solutions

EAS208_HW09_112910_Solutions - EAS 208(Fall 2010 Homework#9...

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EAS 208 (Fall 2010) Homework #9 – Solutions Problem 18–50 The slender bar and 0. cylindrical disk are released from rest with the bar horizontal. The disk rolls on the curved surface. What is the bar’s angular acceleration at the instant it is released? 0.1 kg 2 kg Answer: The F.B.D for the bar and the disk, at the instant shown, are shown in the figure below: 1
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Disk The equation of motion in the y -direction is: , , disk y disk disk y y disk disk disk y F m a f B W m a Σ = = , (1) The equation of rotational motion (take moments about the center of mass): disk disk disk M I α Σ = disk disk Rf I α = (2) The disk rolls, therefore the kinematic condition for its center is: , disk y disk a R α = − (3) Bar The equation of motion in the -direction is: y , , bar y bar bar y y y bar bar bar y F m a A B W m a Σ = + = , (4) The equation of rotational motion (take moments about the center of mass): bar bar bar M I α Σ = 2 2 bar bar y y bar bar l l A B I α + = (5) The kinematic conditions for the bar is found from the acceleration of the center of the mass of the bar relative to point A : , , bar y A bar bar y bar bar bar y A A r r α ω ω = + × + × × a a r r r r r , ( ) ( ) ( ) , 0 0 0 2 2 bar bar bar y bar l l α = + × + × × a k i k k r i , 2 bar bar y bar l α = a j or , 2 bar bar y bar l a α = (6) 2
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From Eq. (1), Eq. (2) and Eq. (3): , 2 disk y disk disk disk y I B W m a R = + (7) From Eq. (4), Eq. (5) and Eq. (6): 2 bar bar y y bar bar m l A B W α + = (8) 2 2 bar bar y y bar bar l l A B I α + = (9) The kinematic condition for the disk and the bar is found from the acceleration of the center of the mass of the disk relative to point : A , , disk y A bar disk y bar bar disk y A A r r α ω ω = + × + × × a a r r r r r , ( ) ( ) ( ) ( ) ( ) ( ) , 0 0 0 disk y bar bar bar l l α = + × + × × a k i k k r i , disk y bar bar l α = a j or , disk y bar bar a l α = (10) From Eq. (7) and Eq. (10): 2 disk y disk disk bar bar I B W m l R α = + (11) We have the set of three equations Eq. (8), Eq. (9) and Eq. (11) and we have three unknowns y A , y B , and bar α : 2 2 2 2 2 2 2 2 12 2 bar bar bar bar y y bar bar y y bar bar bar bar bar bar bar bar y y bar bar y y bar disk disk y disk disk bar bar y disk disk bar bar m l m l A B W A B W l l l l m l A B I A B I m B W m l B W m l R α α α α α α + = + = + = + = = + = + 3
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( ) 0.981 0.006 0.981 0.006 0.06 0.06 0.00012 0.06 0.06 0.00012 1.962 0.2 0.1 0.12 1.962 0.036 y y bar y y bar y y bar y y bar y bar y bar A B A B A B A B B B α α α α α α + = + = + = + = = + = − (1) (2) 61.3125 0.042 2.943 0.08 4.905 0.038 1.962 0.038 1.962 0.038 1.962 1.962 0.036 1.962 0.036 1.962 0.036 bar y bar bar y bar y bar y bar y bar y bar y b rad A s A A A B B B α α α α α α α α = − = = + = − + = − + = − = − = − = − ar α 61.3125 0.367875 0.24525 bar y y rad s A N B N α = − = =
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