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Mechanics tough question

  1. Sep 13, 2010 #1
    A particle moves on the surface of right circular cone of semi cone angle alpha, coaxial with z axis along a helical path of fixed pitch L starting from some r(0) radius.
    The angular velocity of the particle omega varies linearly with time, omega=kt.
    Determine velocity and acceleration of the particle vectorially in a cylindrical polar coordinate system.

    I am unable to relate its z axis velocity and time.
    At least its sure that time period of rotation is not constant.…

    plzzzz help me out
     
  2. jcsd
  3. Sep 14, 2010 #2

    berkeman

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    Staff: Mentor

    Can you please show a bit more work to get this problem started? Can you define the particle's initial starting point, and what direction you want to have it move in?

    Also, is there any more information given? Like, are you supposed to include the effects of gravity at all, or assume that there are no other forces other than from centripital acceleration and from the acceleration that is causing the speed increase?
     
  4. Sep 15, 2010 #3
    We have to neglect gravity and any other force like friction, air drag are to be neglected..
    The particle starts at t=0 with r(0) radius of cone.
    And it moves in a spiral path such that radius of spiral keeps increasing or it moves towards open side of the cone..

    What i am most confused about is whether time period of one rotation will be constant or not.
    If not then how can we relate its velocity along cone's axis with the angle theta rotated by it on inside the cone on its walls.

    As i don't have computer currently so i use mobile due to which i am unable to write equations properly here.

    So please bear with me.
    And sorry sir,for posting it in wrong forum
     
  5. Sep 17, 2010 #4
    Anyone pls reply
     
  6. Sep 17, 2010 #5

    berkeman

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    Staff: Mentor

    The key to the question is that the pitch of the helical path is constant. Use that fact, and the angular velocity equation that you are given, and show us your equations so far...
     
  7. Sep 28, 2010 #6
    Finally, now I have got the opportunity to sit on a PC.Earlier, i only add access through mobile.

    If vz is the velocity along the vertical axis of the cone,
    [tex]\omega[/tex] be the angular velocity = kt
    z be the z coordinate w.r.t. the tip of the cone.
    r be the radius of its path in cone

    So,
    r=ztan[tex]\alpha[/tex]

    and
    the velocity vector in polar coordinates is
    v = [tex]\dot{r}[/tex][tex]\hat{r}[/tex] + r[tex]\dot{\theta}[/tex][tex]\hat{\theta}[/tex] + [tex]\dot{z}[/tex][tex]\hat{z}[/tex]

    Hence,using previous relation, we get,
    v = tan[tex]\alpha[/tex] ([tex]\dot{z}[/tex][tex]\hat{r}[/tex] + zkt [tex]\hat{\theta}[/tex]) + [tex]\dot{z}[/tex][tex]\hat{z}[/tex]

    So, what is left is to get a relation between z and [tex]\omega[/tex]

    That is where i am stuck.I'm unable to get that relation.
    My friend says its [tex]\omega[/tex] = (2[tex]\pi[/tex]/L) vz


    But, i am not able to figure out the reason for it.As, [tex]\omega[/tex] or vz are not constant so we can't write such a linear relation,i think..

    Please clarify.....
     
  8. Oct 2, 2010 #7
    Pls reply
     
  9. Oct 3, 2010 #8
    A spiral (elic) drawn on a cone with fixed pitch L in cylindrical coordinates is

    [tex] z= L\ cos \alpha\ \frac{t'}{2\pi} [/tex]

    [tex] r= L\ sen \alpha\ \frac{t'}{2\pi} [/tex]

    [tex] \theta= \ t' [/tex]

    After each turn of [tex] \theta [/tex] which takes a time [tex] 2\pi [/tex], the distance between the 2 points will be [tex] \sqrt((L\ cos \alpha)^2+(L\ sen \alpha)^2) = L [/tex]

    They told you [tex] \omega = kt [/tex] so since [tex] \omega =\frac{d\theta}{dt} [/tex], we integrate [tex] \omega [/tex] to find [tex] \theta= \frac{kt^2}{2}[/tex]

    So we need to replace [tex] t' = \frac{kt^2}{2}[/tex] in the expressions above.

    Now we got the parametric expessions of [tex] \theta, z, r[/tex], we can find the speed and acceleration.

    To find speed and acceleration we use the versors

    [tex]{u_{\theta}} = -sin\theta\ \vec{i}+cos\theta\ \vec{j} [/tex]
    [tex]{u_{r} = cos\theta\ \vec{i}+sin\theta\ \vec{j} [/tex]
    [tex]{u_{z} = \vec{k} [/tex]

    To find acceleration and speed you should use this expressions:

    [tex]\vec{v} = \dot{r}\vec{u_r} + \dot{\theta}r\ \vec{u_\theta} + \dot{z}\ \vec{u_z}[/tex]
    [tex]\vec{a} = (\ddot{r}-r\dot{\theta}^2)\ \vec{u_r} + (\ddot{\theta}r+2\dot{r}\dot{\theta})\ \ \vec{u_\theta} + \ddot{z}\ \vec{u_z}[/tex]

    I made it short.
    If you don't understand some passages, ask.
     
    Last edited: Oct 3, 2010
  10. Oct 4, 2010 #9
    i couldn't understand only the 1st 3 expressions of z , r, and theta that u have wrtten, in them, they can be written only if there is a linear relation between z and t', r and t'.so how can we say there is linear relation between them?

    Sorry Sir,
    it may be a silly question but only that is what is confusing me..
    thank you...
     
  11. Oct 4, 2010 #10
    Don't call me sir, please. :)

    Just to be sure, we are talking about this object (the left one).

    spiral22.gif

    Basically it's a radious moving at a constant speed, so e.g.
    [tex]\theta =kt, \ \ \ \omega = k[/tex]

    while the radious sweeps, (thinks of a radar on the screen of a boat control room), the lenght of the radious increases linearly, so [tex]r = k_1 t[/tex].
    While it sweeps and increase its lenght, it rises linearly over the xy plane, so [tex]z = k_2 t[/tex].
    If you're convinced of this, then you can try to find the pitch between two near arcs of the spiral. You can take all the point which are on the same angle [tex]\theta[/tex] then find their distance or pitch.
     
  12. Oct 5, 2010 #11
    Thanks Sir(sorry).
    Just a bit more confusion

    But [tex]\omega=kt[/tex] which is not constant as u took in previous post..,i.e., it has an angular acceleration
    Then how to explain it??

    Thank you once again...
     
  13. Oct 5, 2010 #12
    Ok, so susbstitute [tex]t[/tex] with an appropriate expression.
    If you look in my first message there is already an answer to this, if not the complete solution.
     
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