The principle of a governor

1. Sep 8, 2006

fara0815

Hello there!
I am not sure how to get started with this one:

Two half shell rods are welded in a v-shape together and attached to a vertikal axis which rotates with 2 rounds per minute. The angle between the rods is 80 degress. In the rods, there a two balls on each side which are in a certain height (counting from the low end of the v) resting in a equilibrium. How great is the height?

So my guess is, that the centrifugal force has to be the same than the gravity force:
$$F_g=F_z = m*g = m*\omega^2*r \\ \omega=2*f\pi*f =12.57 \frac{rad}{sec}$$
So the radius would be :

$$r= \frac{g}{\omega^2} \\ \mbox{the height is then} h=\frac{r}{\tan40} = 0.074 m$$

However, it is supposed to be 0,088 m.

Thank you very much in advance!

Last edited: Sep 8, 2006
2. Sep 9, 2006

andrevdh

I currently get some ridiculous answer for the height which I find hard to accept, but maybe someone else will point out the error that I made. The problem is quite simple to solve. A ball is experiencing two forces: its weight and the normal force from the rail. Resolve the normal force into two perpendicular components. The vertical need to balance the weight of the ball and the horizontal component need to provide the centripetal force. Use trigonometry to get the radius in terms of the height.

Last edited: Sep 9, 2006
3. Sep 9, 2006

fara0815

mh, that sounds logical.
The normal force would be a component of the gravitational force:
$$F_n= m*g*\cos50$$
if I use that in the equation for the centripetal force equation and solve it for the radius, I get :

$$r= \frac{g*\cos50*\sin40}{\omega^2}$$ with $$\omega= 12, 57 \frac{rad}{sec}$$ I get then for r 0,026 m and the height would be $$h=\frac{r}{\tan40} = .031 m$$

Did you get the same ?

4. Sep 9, 2006

andrevdh

Even though the ball is stationary on the rail it is experiencing an acceleration since it is moving in a circle. This means that the forces do not balance out in general. So one cannot equate components (in general). What one can say is that the vertical forces need to balance out since the ball is not accelerating in the vertical direction. The resultant of the horizontal components will give the centripetal force. So construct the two separate sets of equations and see if you can relate them in some way. Should'nt the angular speed be

$$\frac{2 \times 2\pi}{60}\ rad/s ?$$

Last edited: Sep 9, 2006
5. Sep 10, 2006

fara0815

andre, really, thank you very much for taking the time to help me out!
I am sorry, I made a typo in the text. The angular velocity is $$\omega= 2\times 2\times \pi$$ since it does two revolutions a second, not a minute.
Is it always like that that one should only equate the components that are not accelerating in a problem with an unstable equilibrium?

However, I managed it with your help:

$$F_z= m\times\omega^2\times r$$

$$F_n= \frac{- F_z}{\cos40}$$

$$F_{ny}= F_n\times \sin40$$

Like you said I should only equate the vertikal components:

$$-F_g= F_{ny}$$
$$-m\times g=F_n\times\sin40 = m\times g =\frac{F_z}{cos40}\sin40$$
$$F_z= \frac{m\times g}{\tan40}$$
$$r= \frac{g}{\tan40\times \omega^2} =0,074 m$$

With that I get the height since $$h=\frac{r}{\tan40} = 0,088 m$$

It is soo great that there are guys like you who help others out!

Ps: what do I have to do to insert a new line with latex? \\ does not give me a new line

Last edited: Sep 10, 2006
6. Sep 11, 2006

andrevdh

Yes, if the object is not accelerating in a certain (any) direction then the sum of the force components in that particular direction should be zero.

I am glad you got the correct answer, because with the angular velocity in radians per minute I got 317 meters! That is why I remarked about the rediculous answer.

This analysis do bring one to a result that does not make sense, or contradicts intiution:

$$h = \frac{g}{\omega ^2 \tan ^2(\theta)}$$

this means that as the angular speed increases the object will go down and if g increases the object will go up, which cannot be right. So I am not so sure about the result of this analysis. I would like the opinion of someone who is proficient in mechanics to comment on this, maybe Doc Al or Andrew.

As for the lines in latex, I have not been able to solve the mystery myself. I just start a new latex construct for each new line with a normal line separation (Enter) between each in order to keep the the feet of the guys at the top out of the hair of the guys below.

Last edited: Sep 11, 2006