What is the principal moment of inertia tensor for a lamina?

LeoChan
Messages
5
Reaction score
1
Homework Statement
A thin uniform rectangular plate (lamina) is of mass m and dimensions 2a by a. Choose a coordinate system Oxyz such that the plate lies in the xy plane with origin at a corner, the long dimension being along the x-axis.

(a) The moment of inertia tensor about the origin
(b) The principal moments of inertia about the origin
Relevant Equations
Inertia tensor
The moment of inertia tensor I found out is
(1/3) (-1/2) 0
(-1/2) (4/3) 0
0 0 (5/3)

The principal moment of inertia tensor I found out is
(1/3-I) . (-1/2) . 0
(-1/2) . (4/3-I) . 0
0 . 0 . (5/3-I)

det of principal of moment inertia = 0
So (1/3-I) (4/3-I)(5/3-I)-(-1/2)(-1/2)(5/3-I)=0
I=5/3, (5+√18)/6, (5-√18)/6

The answer look so weird to me and differ from common lamina moment of inertia
 
Physics news on Phys.org
Looks OK to me, except that the mass ##m## and factors of ##a## should be included.
 
  • Like
Likes LeoChan
TSny said:
Looks OK to me, except that the mass ##m## and factors of ##a## should be included.
Thanks. But just didn’t expect the answer to be irrational for a lamina, it made me so confused.
 
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...

Similar threads

Back
Top