# Incorrect derivation of tangential acceleration in polar coordinates

• yucheng
In summary, the conversation discusses the tangential acceleration of a particle and the equations involved in deriving it. However, it is pointed out that the equation ##a_T = \dot v_T## does not hold when unit vectors change position, and the reason for this is due to the mathematics in the original post.

#### yucheng

Homework Statement
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Relevant Equations
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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?

yucheng said:
$$v_{tangential}= \omega r$$ 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.

PeroK said:
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?

yucheng said:
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.

• yucheng
PeroK said:
You've proved it yourself. The mathematics in your original post is the reason.
Oops... ahahaha