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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.

$mrw^2bcosθ$ : is the moment of the centrifugal force.

$$a = 12 cm, b = 20 cm, k = 10^4 N/m, mg = 25N$$

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 natural frequency is

$$ (w_n)^2 = (1/2*ka^2 + - mgb - mb^2w^2)/(mb^2) $$

which shows that the natural frequency changes with rotation speed. Is this a right solution?

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  • $\begingroup$ Thank you for showing your work. Are you asking us to check your answer? Why do you think it might be wrong? Is it the same as the answer you found? $\endgroup$
    – sammy gerbil
    Commented Jun 28, 2016 at 0:53
  • $\begingroup$ Yes, I am asking if my solution is right, and it is not as the solution from the solution manual. $\endgroup$
    – Ahmed
    Commented Jun 28, 2016 at 1:40
  • $\begingroup$ Sorry, I did not see your comment immediately below the manual solution. I think your solution is correct. Does the solution to part (a) suggest any approximations? $\endgroup$
    – sammy gerbil
    Commented Jun 28, 2016 at 16:29
  • $\begingroup$ Think you for the check of the answer...... In part (a) the solution does not use any approximations. you get the speed of the governor at the vertical arms when you set θ and θ'' to zero in the equation before the last one. $\endgroup$
    – Ahmed
    Commented Jun 28, 2016 at 21:46

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Yes I agree, the solution in the manual misses out some terms in the equation of motion, as well as the factor $\frac12$ in front of $k$. Probably some approximations have been made without explanation. Your equation seems correct to me.

When the ball arm is vertical, and there is no rotational motion in the vertical plane, then balancing moments about O gives
$b(mc\Omega^2) = a(\frac12 kx_0)$
where $c=16cm$ is the distance of the ball from the axis and $x_0=1cm$ is the initial compression of the spring. $\Omega$ is the angular velocity of rotation of the ball in a horizontal plane when the ball arm is vertical.

The above relation can be substituted into your eqn of motion to eliminate the constant terms.

Without doing so explicitly, the manual solution seems to have ignored the effect of gravity (the $mgb$ term). That might be a reasonable approximation because the spring is very stiff and dominates the motion : the compression force in the spring is $1000N$ which is $40\times$ the weight of each ball ($25N$).

I don't see any reason why the $\Omega^2$ term has been ignored.

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