Circular Motion Kinematics

  • Thread starter Nikstykal
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  • #1
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Homework Statement


The rod OA rotates clockwise with a constant angular velocity of 6 rad/s. Two pin-connected slider blocks located at B move freely on OA and the curved rod whose shape is a limacon described by the equation r = 200*(2-cosθ) mm. Determine the magnitude of the acceleration of the slider blocks at the instant θ = 135°.
QcsaAag.jpg

Homework Equations


αr=r''-rθ'2
αθ=rθ''+2r'θ'
α = √(αr2θ2)

The Attempt at a Solution


I am trying to solve for the magnitude using the normal and tangential acceleration components.
r = 200*(2-cosθ) --> r(135) = 541.42 mm θ' = 6 rad/s
r' = 200 sinθ * θ' --> r'(135) = 848.53 mm/s θ'' = 0
r'' = 200cosθ * θ'2= -5091.17 mm/s2

When I plug all those values in I keep getting the wrong answer. Can someone tell me where I am going wrong?
 

Answers and Replies

  • #2
TSny
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Careful with the signs. Note that the rod is rotating clockwise.
 
  • #3
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Careful with the signs. Note that the rod is rotating clockwise.
Can you explain that more? Is this because there is a negative change in θ over time?
 
  • #4
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I ended up getting the right answer, must have just been a miscalculation.
 
  • #5
TSny
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Can you explain that more? Is this because there is a negative change in θ over time?
Yes. In the figure, ##\theta## increases in the counterclockwise direction. Therefore, if the rod rotates clockwise, ##\dot{\theta}## is a negative number. You should find that ##\dot{r}## is negative when ##\theta = 135^o##.
 
  • #6
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Yes. In the figure, ##\theta## increases in the counterclockwise direction. Therefore, if the rod rotates clockwise, ##\dot{\theta}## is a negative number. You should find that ##\dot{r}## is negative when ##\theta = 135^o##.
Okay I understand that. Thank you for explaining. I just go lucky in my calculations because they ended up cancelling the negatives out.
 

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