- #1
Marcheline
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- 0
1. A rigid uniform slender beam weighing 8kg is hinged at A and is in dynamic motion. A spring of stiffness 4.5kN/m and a damper giving a viscous resistance of 35Ns/m is connected at B. The beam carries a flywheel of mass 28kg at a distance of 0.9m from A. The beam, AB is 1.2m long, Find the natural frequency of the system
2. [tex]\sum[/tex]M=J[tex]\ddot{}\theta[/tex]
3. I have no problems summing up all the moments. What I am unsure of is the moment of inertia, J in this case. For a beam pivoted at one end, J=(1/3)mL^2. How to I account for the added mass of the flywheel? Do i treat it as a concentrated mass and just add the J of the flywheen to the beam's moment of inertia?
-kL^2 (Theta) - cL^2 (Theta/dot) = {(1/3)ML^2 + m[(3/4)L]^2} (Theta/dotdot)
When I do this and solve the equation harmonically, I get the natural frequency as 15.63 rad/s. Is this right? Or has the mass of the flywheel already been taken into account in the moment balance?
I tried to use Latex but failed miserably, as can be seen from the top.
Thank you.
2. [tex]\sum[/tex]M=J[tex]\ddot{}\theta[/tex]
3. I have no problems summing up all the moments. What I am unsure of is the moment of inertia, J in this case. For a beam pivoted at one end, J=(1/3)mL^2. How to I account for the added mass of the flywheel? Do i treat it as a concentrated mass and just add the J of the flywheen to the beam's moment of inertia?
-kL^2 (Theta) - cL^2 (Theta/dot) = {(1/3)ML^2 + m[(3/4)L]^2} (Theta/dotdot)
When I do this and solve the equation harmonically, I get the natural frequency as 15.63 rad/s. Is this right? Or has the mass of the flywheel already been taken into account in the moment balance?
I tried to use Latex but failed miserably, as can be seen from the top.
Thank you.