- #1
Starwatcher16
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Imagine we have a Equilateral triangle in the xy plane with one vertex at the origin. The triangle will be rotated about the z axis. The depth of the triangle, d, will be constant.[tex]B_r=B(r)[/tex]
From the picture I see that [tex] B_r=\frac{r}{h}B , V=\frac{B_rhd}{2}=\frac{Bdr}{2}[/tex]
First, I need to find I.
[tex]
I=\sum M_i*r_i^2=\int r^2 dm. p=M/V, so, dp=p dv[/tex]
[tex]
I=p\int r^2 dv[/tex] Substituting using[tex] dp=p dv and dv=\frac{Bd}{2}dr[/tex], we get:
[tex]
I=\frac{pBd}{2}*\int r^2 dr=\frac{pBd}{2}*\frac{r^3}{3}.[/tex] Using p=M/V:
[tex]
I=\frac{Mr^3}{3h}=\frac{Mr^2}{3}.[/tex] R_total/h=1.
Now, let's say I want to find I for the same shape, but now there is a distance d from the vertex to the z axis.
Everything is the same as the above, except, [tex]B_r[/tex] is now a piecewise function, where B(r)=0 for r<d and B(r)=[tex] B_r=\frac{r-d}{h}B for r>d. [/tex]
Reworking it with the new B_r, I can get the new I. After that I just solve for x in [tex] I_2=I_1+x[/tex] to get the parallel axis theorem.
The thing is, I don't know how to generalize this one example to work for all shapes. Any help is appreciated.
From the picture I see that [tex] B_r=\frac{r}{h}B , V=\frac{B_rhd}{2}=\frac{Bdr}{2}[/tex]
First, I need to find I.
[tex]
I=\sum M_i*r_i^2=\int r^2 dm. p=M/V, so, dp=p dv[/tex]
[tex]
I=p\int r^2 dv[/tex] Substituting using[tex] dp=p dv and dv=\frac{Bd}{2}dr[/tex], we get:
[tex]
I=\frac{pBd}{2}*\int r^2 dr=\frac{pBd}{2}*\frac{r^3}{3}.[/tex] Using p=M/V:
[tex]
I=\frac{Mr^3}{3h}=\frac{Mr^2}{3}.[/tex] R_total/h=1.
Now, let's say I want to find I for the same shape, but now there is a distance d from the vertex to the z axis.
Everything is the same as the above, except, [tex]B_r[/tex] is now a piecewise function, where B(r)=0 for r<d and B(r)=[tex] B_r=\frac{r-d}{h}B for r>d. [/tex]
Reworking it with the new B_r, I can get the new I. After that I just solve for x in [tex] I_2=I_1+x[/tex] to get the parallel axis theorem.
The thing is, I don't know how to generalize this one example to work for all shapes. Any help is appreciated.