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Dynamics radial acceleration problem: Stone is placed in cup at end of rotating arm

  1. Oct 2, 2011 #1
    1. The stone of mass m=100 kg is placed inside the cup at the end of a rotating arm of length l=3 m. If arm is released from rest at theta=0 and its angular acceleration is given as theta (double dot) = 5 rad/s2 cos t (a) find the acceleration of the stone with theta=pi/4 rad; (b) what is the force acting on the stone from the cup at this instant (include gravity); (c) at what time does the stone fly off the arm?



    2. I'm thinking the equation to start with is a(p)=(r(double dot)-r*theta(dot)2)e(r) + (r*theta(double dot)+2r(dot)theta(dot))e(theta)



    3. Using the above equation, with the assumption that r(dot) and r(double-dot) are zero, I found that the acceleration at pi/4 was -14.8e(r) + 2.34 e(theta). But the answer given is -21.71e(r) + 12.64e(theta). I cannot figure out what I did wrong. Can anyone help me?
     
  2. jcsd
  3. Oct 3, 2011 #2

    Andrew Mason

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    Re: Dynamics radial acceleration problem: Stone is placed in cup at end of rotating a

    I am having difficulty following our notation. What does e(r) mean? or e(theta)? What is t? Is that the angle or time?

    To do this problem, it is easier to first determine [itex]\dot{\theta}, \text{ and } \ddot{\theta}[/itex] as functions t, whatever t is.

    Then work out the acceleration using: [itex]a = \ddot{\theta}r[/itex]

    AM
     
    Last edited: Oct 3, 2011
  4. Oct 3, 2011 #3
    Re: Dynamics radial acceleration problem: Stone is placed in cup at end of rotating a

    Thanks, Andrew. With your help, I've figured out that my problem was in the integration, forgetting to factor in the cosine of zero being 1.

    The e(r) and e(theta) are the unit vectors, as my professor uses them in this class. I don't know how to do the fancy equation editor stuff that you just did, but perhaps I should learn. :)
     
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