Incorrect derivation of tangential acceleration in polar coordinates

  • #1
yucheng
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I am trying to derive the tangential acceleration of a particle. We have tangential velocity, radius and angular velocity. $$v_{tangential}= \omega r$$ then by multiplication rule, $$\dot v_{tangential} = a_{tangential} = \dot \omega r + \omega \dot r$$ and $$a_{tangential} = \ddot \theta r + \dot \theta \dot r$$ However, we also have $$\vec{a} = (\ddot r - r \dot \theta^2)\hat{r} + (r \ddot \theta + 2 \dot r \dot \theta)\hat{\theta}$$, which implies $$a_{tangential} = \ddot \theta r + 2 \dot \theta \dot r$$

Now, what's wrong?
 

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  • #2
PeroK
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$$v_{tangential}= \omega r$$ then by multiplication rule, $$\dot v_{tangential} = a_{tangential} = \dot \omega r + \omega \dot r$$
You have essentially proved that ##a_T \ne \dot v_T##. The equation does not hold where the unit vectors change with position, hence time.
 
  • #3
yucheng
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You have essentially proved that ##a_T \ne \dot v_T##. The equation does not hold where the unit vectors change with position, hence time.
Hmmmm... Is there any way to make it hold when unit vectors change position? By the way, is there a reason why it does not hold?
 
  • #4
PeroK
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By the way, is there a reason why it does not hold?
You've proved it yourself. The mathematics in your original post is the reason.
 
  • #5
yucheng
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You've proved it yourself. The mathematics in your original post is the reason.
Oops... ahahaha
 

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