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Homework Help: Centrifugal and circular motion

  1. Mar 31, 2010 #1
    ell lets say theres a cabin and its rotating about its axisand we have a block of mass m
    placed (not at the the centre of the cabin)
    and theres some friction between the the block and the surface
    now lets say the speed of the cabin is increased a lil high so that friction can no more balance it
    to make the block rotate it in a circular motion
    so now if i see from the cabin frame then theres the centrifugal force which pushes the block to the wall
    so now my qs is
    firstly have i stated the trajectory correct that in the case friction cannot balance then the block will move outward in a straight line to towards the wall of the cabin

    2nd qs. in the cabin frame i used the centrifugal force to see the motion
    but how do i know the blocks motion in ground frame?
    in ground frame there is no centrifugal force so why now is the block moving outward?


    well i forgot that in the above qs there will be coroilis force acting
    so let me change the situation
    lets say i have a circular disk rotating about it axis and a grove(channel) from the centre to the outer end of the disk
    and i have placed a particle in the grove.

    now again the same qs how to find the force that makes the particle move outward from the ground frame
    Last edited: Mar 31, 2010
  2. jcsd
  3. Mar 31, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Once it starts moving there will also be a coriolis force to contend with, thus the trajectory won't be a straight radial line.

    Because it has a tangential velocity and insufficient force to keep it moving in a circle.
  4. Mar 31, 2010 #3
    ok tnx for that
    i forgot about the coriolis force
    anyways thats not in my sylabus
    so i am gonna edit the qs a bit
    plz refer to the first post
  5. Mar 31, 2010 #4
    One way to resolve the 'bead-in-a-groove' problem is to brute-force it with vector algebra. This approach is, however, less than satisfactory.

    Rewriting the vector identity for acceleration in terms of radial and polar coordinates:

    [tex]\vec a_{radial}=[\frac{d^2 r}{dt^2}-r(\frac{d\theta}{dt})^2]\hat r[/tex]

    [tex]\vec a_{tangential}=\frac{1}{r}[\frac{d}{dt}(r^2\frac{d\theta}{dt})]\hat \theta[/tex]

    These identities are far from trivial, and personally, I had to look them up, but they are valid in any frame of reference.

    Let's assume for the sake of simplicity that there's no friction, and that the normal force can only act perpendicular to the grooves. That means that the bead will only experience a normal force acting in the direction of the tangential velocity in the laboratory frame, while the groove rotates with constant angular velocity [tex]\omega[/tex]

    Using Newton's second law:
    [tex]a_{radial}=0[/tex] (No forces in the radial direction!)
    [tex]a_{tangential}=0[/tex] (The angular velocity is constant!)

    Plugging all we know into the two identities for the accelerations:
    [tex]\vec\frac{d^2 r}{dt^2} = \omega ^2 \vec r[/tex]
    [tex]\vec N=2m\omega\frac{dr}{dt}\hat \theta[/tex]

    These sorts of problems break down what we usually perceive as radial and tangential accelerations. For instance, we were just shown that there is 0 "radial" acceleration, but there is an acceleration in the radial direction!
    Things get wonky here.
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