If we consider the following equation:
squareroot ("K"/"M") where K=mgd and M=Io+md^2 P.A.T.
We know that Io=((1/12)mL^2 for a bar and we can symplify the following equation:
squareroot((mgd) / ((1/12)mL^2 + md^2)) which is also:
squareroot((mgd) / (m((1/12)L^2+d^2)))
and so we see that the mass cancel out of the equation and we get:
squareroot(gd / ((1/12)L^2 +d^2))
This work also if Ic=Io .
So we can conclude that the mass does not affect the equation for the frequency and consequently; we estimate that the frequency will stay the same for whatever mass we used at G.
5 comments:
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BACKGROUND (for those not in Division 3):
A compound pendulum is shown in the figure in the original post. The natural frequency for small oscillations of the pendulum was shown in lecture to be:
omega_n = sqrt(m*g*d/(I_O + m*d^2))
where O is the cm of the pendulum.
QUESTION POSED IN LECTURE:
If a point mass is added to the pendulum at O, does the natural frequency increase, decrease or remain the same? A prize of a Purdue Engineering t-shirt is to be awarded to the first person who posts the correct answer, with explanation, on the blog.
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Sebastien has an excellent solution showing that the natural frequency of the rigid body modeled as a thin bar is independent of the mass of the body. That was not exactly the question that was asked (we were asked to add a POINT mass at the cm). However, I actually like his question better than the one I posed. I say that he wins the prize!
I have another Purdue Engineering prize that can be given out for this problem. The first person who posts the solution (after noon Saturday) for the original problem (of adding a POINT mass to O) wins the prize. (I am asking you to wait until Saturday to post to allow more people a chance to see the new contest.)
HINT: The final answer to this question is different than the one that Sebastien posted. Here the natural frequency INCREASES.
You may treat the body as a thin rod if you want. Or better, write the mass moment of inertia of the body in terms of its radius of gyration for a general proof.
Maybe this will be a more suitable answer:
Assume the bar has some moment about its center of mass 'I', some mass 'M', and the center of mass distance 'd' away from the pin.
To find the omega, using this equation, under the assumption sin(theta)= theta
I(alpha)=-Mg(theta)(d)
yielding:
w=sqrt(Mgd/(I +Md^2)
Now assuming you add a mass 'm' a distance 'r' away from the pin.
Now:
I = I_o + mr^2
and the equation changes to
I(alpha)=-Mg(theta)(d)-mg(theta)(r)
and
w_new=sqrt(g(Md+mr)/(I_o+mr^2+Md^2)
In order to estimate whether this is greater or less than the previous w:
remove sqrt(g) as it is a constant.
r=kd; whether k is an arbitrary constant
and in the interest of simplification, assume I_o is dependent on the M and d^2, therefore:
I_o = c*Md^2
w_new=sqrt(d(M+km)/[d^2(M+cM+k^2m))
once again removing the constant sqrt(1/d)
w_new=sqrt((M+km)/(M+cM+k^2*m))
Applying the same standards to the old equation:
w_old=sqrt((M/(M+cM))
So which is bigger? Set them equal and k= c+1 is the break point.(note, this is independent of M or m)
This means that if k < c+1, than w_new > w_old, if k= c+1 than w_new = w_old and if k>c+1 w_new < w_old.
Lets test it to make sure it make sense. For a bar d = 1/2*L and I_o= 1/12 m L^2 so:
c = 1/3
this means, that k must be less than c+1, or 4/3. remember that
r = kd, so if r < 2/3*L, than the frequency increases.
I don't need anymore frisbees, so I'll do my best to stay anonymous.
We have another winner! Floyd showed that the frequency went up with the placement of a particle at the cm. He went further and showed that placing the particle at the center of percussion of a thin bar does not change the natural frequency regardless of the size of the particle.
Good job, Floyd.
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