Determination of the natural frequency of a Hartnell governor

  • #1
2
0

Homework Statement



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Homework Equations




The Attempt at a Solution


I found this solution for the nature frequency
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but here it does not include the Ball weight and centrifugal force in the moment balance equation about the pivot (O), it is wrong answer...is not it?

I tried to solve the problem this way, moment balance about pivot (O)
$$(mb^2)θ'' = -1/2*k*(1/100+a sinθ)a conθ + mgbsinθ + mrw^2bcosθ$$

where :
$$ -1/2*k*(1/100+a sinθ)a conθ $$ : is the moment of the spring force
$$ mgbsinθ $$: is the moment of the weight of the ball.
$$mrw2bcosθ $$: is the moment of the centrifugal force.
$$a=12cm,b=20cm,k=104N/m,mg=25N$$

for small displacment $$sinθ=θ,cosθ=1$$. then
$$(mb^2)θ′′=−1/2k(1/100+aθ)a+mgbθ+mrw^2b$$

r is the distance between ball center and the center of rotation.
$$r=(16/100+bsinθ)=(16/100+bθ)$$

the equation becomes
$$(mb^2)θ′′=−1/2∗k(1/100+aθ)a+mgbθ+m(16/100+bθ)w^2b$$
rearrangement of the equation
$$(m∗b^2)θ′′+(1/2∗ka^2+−mgb−mb^2w^2)θ+1/2∗k∗1/100∗a−m∗16/100∗w^2b=0$$
from this equation we could say that the nature frequency is
$$(w_n)^2=(1/2∗ka^2−mgb−mb^2w^2)/(mb^2)$$
which shows that the nature frequency changes with rotation speed. Is this a right solution?
 

Answers and Replies

  • #2
Spinnor
Gold Member
2,176
381
Did you have to do part a of the problem? If so what did you get.
 
  • #3
Spinnor
Gold Member
2,176
381
I think the last two equations are correct. In addition if you set θ and θ'' to zero you can solve for ω the angular velocity at equilibrium. I would not worry about the answers dependence on angular velocity. You look like you got the right answer, good work! Ask your teacher about your question.

Edit, remember ω for your problem is fixed.
 
  • #4
2
0
Thank you for your reply and check of the solution. and when you set θ and θ'' to zero you would get the answer to part (a).
 

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