I'm learning about linear drag and I have a question with regards to the y component. The y component of the equation of motion (if we measure y downwards) is given by:

$$m \dot{v}_y = mg - bv_y$$

where b is just a coefficient of linear drag. From this, one can show that:

$$m \dot{v}_y = -b(v_y - v_{ter})$$

where $v_{ter}$ is the terminal velocity given by $v_{ter} = \frac{mg}{b}$ for linear drag. From this, we find $y(t)$ of the form:

$$y(t) = v_{ter}t + (v_{yo} - v_{ter})\tau(1-e^{-\frac{t}{\tau}})$$

where $v_{yo}$ is the intitial speed in the y direction of the projectile, and $\tau$ = $\frac{m}{b}$. My issue is that the book I'm using tells me that if we measure y vertically upwards, the sign of the terminal velocity in the final equation flips and becomes:

$$y(t) = -v_{ter}t + (v_{yo} + v_{ter})\tau(1-e^{-\frac{t}{\tau}})$$

I'm not sure why this is the case, because I thought that if the sign flips for the opposite direction of the y axis, then the sign for the initial speed would also change to a negative (I thought it was a case of flipping the signs of the equation of motion at the top) but it doesn't appear to do so. Could anyone point out where I'm going wrong?

  • $\begingroup$ Please provide information about the textbook which you are using. $\endgroup$ – Farcher Apr 8 '20 at 22:57
  • $\begingroup$ John Taylor, "Classical Mechanics" $\endgroup$ – Andrew Apr 8 '20 at 23:16

It is really all to do with which quantities you choose to be components (can be positive or negative) and which you choose to be magnitudes (can only be positive).

The y component of the equation of motion (if we measure y downwards - unit vector $\hat y$) is given by:

$m \dot{v}_y \hat y= m\,\vec g - bv_y \hat y$

Now what do you do with $\vec g$?

You say $\vec g = g \hat y$ and $g$ is the magnitude of $\vec g$ whereas $\dot{v}_y$ and $v_y$ are components in the $\hat y$ direction.

So the equation of motion becomes $m \dot{v}_y \hat y= m\,g\hat y - bv_y \hat y\Rightarrow m \dot{v}_y = m\,g - bv_y$ which is the equation that you quoted.

Now suppose you choose up a positive - unit vector $\hat Y$.
The equation of motion looks very similar $m \dot{v}_Y \hat Y= m\,\vec g - bv_Y \hat Y$ but now you would write that $\vec g = - g \,\hat Y$ where again $g$ is the magnitude of $\vec g$.
The equation of motion is now $m \dot{v}_Y \hat Y= -m\,g\hat Y - bv_Y \hat Y\Rightarrow m \dot{v}_Y = -m\,g - bv_Y.$

Now let's look at the terminal velocity with down as positive; $\vec v_{ter} =\dfrac{m\vec g}{b} \Rightarrow v_{ter} \hat y = \dfrac {mg}{b} \hat y\Rightarrow v_{ter} = \dfrac {mg}{b}$ where the magnitude of the terminal velocity is $\dfrac {mg}{b}$.

This leads to $y(t) \hat y = v_{ter}\hat yt + (v_{yo}\hat y - v_{ter}\hat y)\tau(1-e^{-\frac{t}{\tau}})\Rightarrow y(t) = v_{ter}t + (v_{yo} - v_{ter})\tau(1-e^{-\frac{t}{\tau}})$ where $y(t)$ and $v_{yo}$ are components and $v_{ter}$ is a magnitude.

With up as positive $\vec v_{ter} =\dfrac{m\vec g}{b} \Rightarrow \vec v_{ter} =\dfrac{mg(-\hat Y) }{b} = -\dfrac{mg}{b}(\hat Y)= -v_{ter} \hat Y$

This leads to $Y(t) \hat Y = \vec v_{ter}t + (v_{Yo}\hat Y - \vec v_{ter})\tau(1-e^{-\frac{t}{\tau}})\Rightarrow Y(t) \hat Y = - v_{ter}\hat Yt + (v_{Yo}\hat Y + v_{ter}\hat Y)\tau(1-e^{-\frac{t}{\tau}})$

$\Rightarrow Y(t) = -v_{ter}t + (v_{Yo} + v_{ter})\tau(1-e^{-\frac{t}{\tau}})$ where $Y(t)$ and $v_{Yo}$ are components and $v_{ter}$ is a magnitude.

  • $\begingroup$ This is a great explanation, thank you! $\endgroup$ – Andrew Apr 9 '20 at 11:13

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