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Help finding the components of acceleration of a block on a wedge on a turntable.

  1. Apr 22, 2012 #1
    A wedge with face inclined at an angle theta to the horizontal is fixed on a rotating turntable. A block of mass m rests on the inclined plane and the coefficient of static friction between the block and the wedge is μ. The block is to remain at position R from the center of the turntable, as shown in the attachment. Find the components of the block's acceleration parallel and vertical to the inclined plane.

    a=-Rω^(2)[itex]\hat{r}[/itex]

    I have been struggling to get to grips with this question for a while now. I don't understand how the acceleration is pointing in the opposite direction of the position vector r, which has a z axis component although there is no displacement in that direction. Have I interpreted the polar coordinates wrongly? Any insight would be greatly appreciated. Thanks.
     

    Attached Files:

  2. jcsd
  3. Apr 22, 2012 #2
    yeah [itex] \hat{r}[/itex] is the unit vector in radial direction. So your problem is to find how much this given acceleration points(the projection of) parallel and vertical to the wedge i.e. change basis
    Also you have to express omega as something you know a general wedge problem.
    Edit:
    The r coordinate points from the axis of revolution(z) and out to. Thats why it is negative, cause the block is accelerated in the opposite direction
    And in fact there is an acceleration in z direction. Try to draw the situation where the polar coordinates(They look cylindrical to me) are defined by the axis in the middle.
     
    Last edited: Apr 22, 2012
  4. Apr 22, 2012 #3
    also omega is [itex] \omega = \dot{\theta}[/itex] the azimutal angle differentiated in time.
     
  5. Apr 22, 2012 #4
    Thanks for the reply. I still cant understand how there is an acceleration in the z direction. Surely its z coordinate does not change as its position on the wedge is locked?
     
  6. Apr 23, 2012 #5
    No but try to be strict when defining your coordinate system. This pre-defined is seen from the bottom and close to the vertical axis. When you define yours you should try to get a usual centripetal acceleration. So the z-component is maybe zero? Which would make sense considering that the object dosen't accelerate vertical.
    So my point is: Their definition looks like it is seen from the axis in the middle and your job is to use newtons laws to find the components of a in 2 directions.
     
  7. Apr 23, 2012 #6
    thanks i'm nearly there!
    when they say vertical component of the inclined plane, do they mean perpendicular to it or parallel to the z-axis? sorry if this seems obvious but it's pretty vital to answering the question!
     
  8. Apr 25, 2012 #7
    Sorry for the late answer, but vertical is parallel to their z-axis(probably also yours). Maybe it is 0 and maybe not.
     
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