Recent content by qweazy

Yea I got it. Thanks a lot for helping me. I appreciate it.

15(9.81)(.405)= 59.6

No, for I I got 2.91 and solved it like I how I did before and got 3.60rad/s

ok so it would be (1/2)9(.075)^2+9(.525)^2 for the disk?

I'm guessing its the pin?

I think its just saying that its rotating about that axis. Izz is rotating about the z axis and Ixx is rotating about the x-axis and so on

No, I'm assuming it equals mgh-1/2ks^2

Ixx=Iyy=(1/4)mr^2 Izz=(1/2)mr^2 Iz'z'=(3/2)mr^2 This is what the book gives me

I think its (3/2)mr^2

Dont have a camera but I drew a picture :) and the numbers are in mm

Homework Statement When Θ=0, the assembly is held at rest, and the torsional spring is untwisted. If the assembly is released and falls downward, determine its angular velocity at the instant Θ=90°. Rod AB has a mass of 6kg, and disk has a mass of 9kg. Rod is 450mm and disk has a radius of...

Oh ok I see. I just thought that Asin(kx) applied for everything. It makes sense now, thanks for your help!

Homework Statement A particle is confined to a two-dimensional box defined by the following boundary conditions: U(x, y) = 0 for \frac{-L}{2} ≤ x ≤ \frac{L}{2} and \frac{-3L}{2} ≤ y ≤ \frac{3L}{2}, and U(x, y) = ∞ outside these ranges. Determine the energies of the three lowest energy states...