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# ch8sol-to-class - 10 CHAPTER 8 MULTI DOF VIBRATION Problems...

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10 CHAPTER 8 MULTI DOF VIBRATION Figure 8.40: Double pendulum. Solution: We assume the pendula links to be inextensible strings. , we need a free-body diagram for each mass, where x is to the right and y is up. Take point O as the origin with x axis pointing to the right and y axis pointing up. Then, since the motion of each mass is planar, we will have two EOM for each mass, even though there are only 2 degrees-of-freedom, 1 and 2 : The additional equations are due to the unknown tensions in the strings. Therefore, if the tensions can be removed from the equations then result is two EOMs. Let 1 and 2 be the generalized coordinates. Consider mass 1. The positions are given by x = l 1 sin 1 y = l 1 cos 1 :

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11 The x and y components of velocities are: _ x = l 1 _ 1 cos 1 _ y = l 1 _ 1 sin 1 : Di/erentiate these expressions with respect to time t; and recalling the chain rule of di/erentiation, x and y directions, x = l 1 ( 1 cos 1 _ 2 1 sin 1 ) y = l 1 ( 1 sin 1 + _ 2 1 cos 1 ) : Applying Newton±s second law to each mass, we obtain: For mass 1: X F x = m 1 a x = ) T 1 sin 1 + T 2 sin 2 = m 1 l 1 1 cos 1 _ 2 1 sin 1 ± (1) X F y = m 1 a y = ) T 1 cos 1 T 2 cos 2 = m 1 g + m 1 l 1 1 sin 1 + _ 2 1 cos 1 ± : (2) Similarly for mass 2: x = l 1 sin 1 + l 2 sin 2 y = l 1 cos 1 l 2 cos 2 : Taking two derivatives of each expression results in the accelerations in the x and y directions, and therefore, X F x = m 1 a x = ) T 2 sin 2 = m 2 l 2 2 cos 2 _ 2 2 sin 2 ± + m 2 l 1 1 cos 1 _ 2 1 sin 1 ± (3) X F y = m 1 a y = ) T 2 cos 2 = m 2 g + m 2 l 2 2 sin 2 + _ 2 2 cos 2 ± + m 2 l 1 1 sin 1 + _ 2 1 cos 1 ± : (4) We can eliminate T 2 from Equations (3) and (4) by adding Equation (3) ± cos 2 and Equation (2) ± sin 2 ; so that (3) ± cos 2 + (2) ± sin 2 = 0 m 2 l 2 2 + m 2 l 1 h 1 cos ( 2 1 ) + _ 2 1 sin ( 2 1 ) i + m 2 g sin 2 = 0 : (EOM 1) This is one of the equations of motion. The second equation of motion is obtained by eliminating T 1 and T 2 from Equations (1) and (2). Equations (3) and (4) are substituted into Equations (1) and (2) to eliminate T 2 T 1 sin 1 = ( m 1 + m 2 ) l 1 1 cos 1 _ 2 1 sin 1 ± + m 2 l 2 2 cos 2 _ 2 2 sin 2 ± T 1 cos 1 = ( m 1 + m 2 ) g + ( m 1 + m 2 ) l 1 1 sin 1 + _ 2 1 cos 1 ± + m 2 l 2 2 sin 2 + _ 2 2 cos 2 ± :
12 CHAPTER 8 MULTI DOF VIBRATION cos 1 , multiplying the second by sin 1 ; and adding the two, we obtain ( m 1 + m 2 ) l 1 1 + m 2 l 2

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## This note was uploaded on 02/26/2012 for the course 650 443 taught by Professor Benaroya during the Fall '11 term at Rutgers.

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ch8sol-to-class - 10 CHAPTER 8 MULTI DOF VIBRATION Problems...

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