EAS208_HW09_112910_Solutions

EAS208_HW09_112910_S - EAS 208(Fall 2010 Homework#9 Solutions Problem 1850 The 0.1kg slender bar and 0.2kg cylindrical disk are released from rest

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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 Fm a f B W m a Σ= = , (1) The equation of rotational motion (take moments about the center of mass): disk disk disk MI α Σ = disk disk Rf I = (2) The disk rolls, therefore the kinematic condition for its center is: , disk y disk aR = − (3) Bar The equation of motion in the -direction is: y bar y bar bar y y y bar bar bar y a A B W m a + = , (4) The equation of rotational motion (take moments about the center of mass): bar bar bar Σ = 22 bar bar y y bar bar ll AB I −+= (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 AA rr αω ω ⎛⎞ =+ × + × × ⎜⎟ ⎝⎠ aa r , () ()() , 00 0 bar bar bar y bar × + × × ak i k k r i , 2 bar bar y bar l = a j or , 2 bar bar y bar l a = (6) 2
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From Eq. (1), Eq. (2) and Eq. (3): , 2 disk y disk disk disk y I BW m a R ⎛⎞ −− = + ⎜⎟ ⎝⎠ (7) From Eq. (4), Eq. (5) and Eq. (6): 2 bar bar y y bar bar ml ABW α +− = (8) 22 bar bar y y bar bar ll AB I −+= (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 AA rr αω ω =+ × + × × aa r , () ( ) ( ) ( ) ( ) ( ) , 00 0 disk y bar bar bar × + × × ak i k k r i , disk y bar bar l = a j or , disk y bar bar al = (10) From Eq. (7) and Eq. (10): 2 disk y disk disk bar bar I m l R = + (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 : 2 2 2 2 1 2 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 I Im m l m l R αα ⎧⎧ = = ⎪⎪ ⎨⎨ = + = + ⎩⎩ 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 yy b a r b a r b a r b a r y bar y bar AB BB αα ⎧⎧ +− = = ⎪⎪ += ⎨⎨ −− = + = ⎩⎩ (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 yb a r b a r a r a r a r a r a r rad A s AA A B α =− −= = ⇒+ = + = = a r 61.3125 0.367875 0.24525 bar y y rad s A N BN ⇒= = At the instant it is released, the angular velocity of the bar is 61.3125 bar rad s . 4
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Problem 18–60 Let the total moment of inertia of the car’s two rear wheels and axle be R I , and let the total moment of inertia of the two front wheels be F I . The radius of the tires is R , and the total mass of the car, including the wheels, is . If the car’s engine exerts a torque (couple) T on the rear wheels and the wheels do not slip, show that the car’s acceleration is: m 2 RF RT a mR I I = + + Answer:
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This document was uploaded on 10/27/2011 for the course EAS 208 at SUNY Buffalo.

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EAS208_HW09_112910_S - EAS 208(Fall 2010 Homework#9 Solutions Problem 1850 The 0.1kg slender bar and 0.2kg cylindrical disk are released from rest

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