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Emergency problem

  1. Nov 28, 2011 #1

    ShayanJ

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    Gold Member

    There is a rod rotating around the z axis with angular velocity [itex] \omega [/itex].The angle between the rod and z axis is constant and equal to [itex] \alpha [/itex].A mass m is constrained to move on the rod.The gravitational force is in the negative direction of z axis and the friction between the mass and rod is negligible.At time t=0 , the mass is at distance [itex] r_{0} [/itex] from the origin and is stationary relative to rod.

    1-In a proper coordinate system,write the newton's second law for the mass in a inertial frame of reference and write the differential equations of motion.Highlight the expressions indicating the reaction of rod.

    2-Solve the differential equations and write the mass's distance from origin as a function of time.

    3-Calculate the reaction force of the rod(direction and magnitude)


    thanks
     
  2. jcsd
  3. Nov 28, 2011 #2
    Does the following get you started?
     

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  4. Nov 28, 2011 #3

    ShayanJ

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    Gold Member

    Thanks alot.My mistake was that I didn't write N but only the centripetal force.

    But it seems sth is wrong there.Just imagine such a thing not rotating.you know that the mass slides down.So it seems that [itex] N \sin {\alpha} [/itex] is not equal to mg.
    Another question.
    Last night I worked on it and got some things but there is a big puzzle in my mind.
    Textbooks say that the acceleration in spherical coordinates is as following:

    [itex] \textbf{a}=(\ddot{r} - r \dot{\phi}^{2} \sin ^ {2} \theta -r \dot{\theta}^{2}) \hat{e_{r}} + [/itex] [itex] ( r \ddot{\theta} + 2 \dot{r} \dot{\theta} - r \dot{\phi} ^ {2} \sin {\theta} \cos{\theta}) \hat{e_{\theta}} + [/itex] [itex] (r \ddot{\phi} \sin{\theta} + 2 \dot {r} \dot {\phi} \sin {\theta} + 2 r \dot{\theta} \dot{\phi} \cos{\theta}) \hat{e_{\phi}}
    [/itex]

    Shoud I use the r coordinate of the above equation for writing Newton's 2nd law or simply write [itex] m \ddot{r} [/itex] ?
    I tried both.When I use the equation above I get crazy things.But when I use [itex] m \ddot{r} [/itex] I get a cosine.
    I'm just wondering that the above equation is the most general and can't understand why it becomes wrong in this problem.

    The last question.
    The [itex] \theta [/itex] coordinate of the acceleration should be zero.But when I write the Newton's 2nd law in that coordinate and replace [itex] \ddot{\theta} [/itex] with zero,I get r=constant,which is clearly wrong.Could you write that?

    thanks
     
    Last edited: Nov 29, 2011
  5. Apr 5, 2012 #4
    Do you know what i said?
     

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