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undamped free vibrations ( I have some of the work 9 easy k pt)

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Scholar
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Date Posted: 7/24/2008 2:01:17 PM  Status: Live
undamped free vibrations ( I have some of the work 9 easy k pt)
Course Textbook Chapter Problem
N/A N/A N/A N/A
Question Details:
 

  NOTE: I only need help with the first part! And my work is below

  1. The mechanism shown is a simplified version of what is used in an Impact Test. The radius of gyration about the point O is kO = 0.95 m. However, the radius to the center of gravity of the entire object from the point O is rG = 0.9 m. First derive the equation of motion using Newton’s Laws – do not plug in numbers for the variables yet….just keep them as letters (m, kO, rG, etc.). Then, if the hammer is rotated to -0.4 rad and given an initial velocity of 0.7rad/s, find:
    1. Natural period. Does this vary with initial conditions? (2.01 s)
    2. Phase angle. Does this vary with initial conditions? (-1.06 rad)
    3. Position θ as a function of time using BOTH solutions to the differential equation of motion. (θ = 0.224sin(3.13t) – 0.4cos(3.13t) OR                      θ = 0.458sin(3.13t – 1.06))
    4. Maximum displacement and first time which it occurs. (0.458 rad, 0.841 s)
Maximum velocity and first time which it occurs. (1.43 rad/s, 0.339 s)
 
 
given:
Ko = 0.95 m
Rg = 0.9 m
 
-Wrsinθ = Iα + mr^2α    =  -mgrθ = Ko*r*g^2 + mr^2α
 0 = (Ko +m)*r^2*α +mgrθ  = α + m*g*r/(Ko +m)*r^2
since τ natural period = √m*g*r/(Ko +m)*r^2 = gives me the wrong answer
 
 
 
 
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Oracle
Karma Points: 15,012
Date Posted: 7/25/2008 1:54:00 PM  Status: Live
Asker's Rating: Helpful   
Response:
-mgrGsinθ = Iα
where I = mko2
mko2α + mgrGθ = 0
ko2α + grGθ = 0
α + (grG/ko2)θ = 0
so ω2 = grG/ko2
ω = √(grG) /ko
period T = 2π/ω = 2πko/√(grG) = 2.01 s

chiefdog's Comment:
thanks!




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