Recent content by simphys

  1. S

    Why is there are precession frequency here?

    so I need to use the eq.: omega_pr = ##mgr / L_s## with r = 6cm. But from how I imagine/see it, the rod pierces the disk through the middle so how does this create precession if the weight will not cause a torque then and thus a hortizontal change in angular momentum? Thanks in advance!
  2. S

    Why is momentum not conserved?

    should've examined it more properly basically.. the problem is not reallly difficult if I would've thought a bit simpler about it. i.e. from the basic principles up.
  3. S

    Why is momentum not conserved?

    okay right.. it's the support forces then, correct?
  4. S

    Moment of inertia of a thin, square plate

    well.. that is exactly what I don't understand.. I didn't understand why the solution has used two integrals in such a way.. I haven't used a volume integral or smtn. I did differently by using a mass element and summing over that with one definite integral.
  5. S

    Why is momentum not conserved?

    exactly that is what I didn't do.. should've seen that there still is gravity as well.
  6. S

    Why is momentum not conserved?

    It's gravity, isn't it? That the one that creates the torque. If I were to take the 'rotation axis' in f.e. disk2, the gravitational force of disk 1 would still exert an angular impulse
  7. S

    Why is momentum not conserved?

    well.. a torque? but internally isn't it?
  8. S

    Why is momentum not conserved?

    Well, friction will occur. But I can't see how this creates non-conservation of AM Or.. is it the torque/moment that the force of your hand will apply to the disks when smooshed together that creates an angular impulse?
  9. S

    Why is momentum not conserved?

    Here is question + drawing.
  10. S

    Moment of inertia of a thin, square plate

    this is the solution. not mine :)
  11. S

    Moment of inertia of a thin, square plate

    Yep, that is what I did I expressed in terms of y^2 and then went ahead with that. But I just didn't understand how this solution came about basically.
  12. S

    Moment of inertia of a thin, square plate

    I don't really understand what the 2 integrals (dx and dxdy) for I_x represent. Could I get some explanation here please? Thanks in advance.
  13. S

    Correct Use of the Parallel Axis Theorem for Moment of Inertia

    yeah apologies it should have been 2 times the symbolic equation, just forgot to type it out. Omg... totally forgot that... thanks a lot... I needed to use the moment of inertia about the center of mass basically..,.
  14. S

    Correct Use of the Parallel Axis Theorem for Moment of Inertia

    so calculated, the moment of inertia for a rod about an axis at the end of the rod is I = 1/3 * M * L^2 here for case 1: arms to the side I is calculated to be ##I = 0.224## for case 2: arms stretched ## I = 1 / 3 * M * L^2 + M * d^2 ## with L = 0.6 m (length of rod) and d = 0.2 (dinstance from...
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