Incorrect derivation of tangential acceleration in polar coordinates

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The discussion centers on the derivation of tangential acceleration in polar coordinates, highlighting a discrepancy between derived equations. The initial derivation uses the relationship between tangential velocity, radius, and angular velocity, leading to the conclusion that tangential acceleration does not equal the time derivative of tangential velocity. It is noted that this inconsistency arises because the unit vectors in polar coordinates change with position and time. Participants question whether it is possible to reconcile the equations when unit vectors vary. The conversation emphasizes the importance of understanding the underlying mathematics in polar coordinate systems.
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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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.
 
PeroK said:
You've proved it yourself. The mathematics in your original post is the reason.
Oops... ahahaha
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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