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My lecturer gave us some exercises on Dynamics with explained solutions in case we get stuck. One of the exercises is this:

A particle of mass $\space m \space$ is attached to the end of a light inelastic string of length $\space a \space$, the other end being fixed.

Initially, the particle hangs freely vertically below the fixed end, and is given a horizontal velocity $\space u \space$. After the string has turned through an angle $\space \theta \space$, show that the tension in the string is:

$$ m \left[g \left (3 \cos{\theta} - 2 \right) + {{u^2}\over a} \right]$$

My lecturer handwrites the solutions. Here is a snippet of what he wrote and what confuses me:

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Notice how he subbed the $\space \ddot{r} \space$ with velocity $\space u \space$, but also subbed $\space {d \dot{\theta} \over dt} \space$ with velocity $\space u \space$.

Also, I can't tell whether he wrote $\space u \space$ or $\space \dot{u} \space$. If it's the latter, isn't $\space u \space$ just a constant and not a velocity function of time? Hence making the act of dotting it pointless?

Sorry for the overload of questions, I'm really stuck. I much rather prefer typed notes to avoid issues like this.

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    $\begingroup$ $u$ is not a constant, it changes in time though acceleration due to gravity $\endgroup$
    – John Doe
    Oct 5, 2017 at 16:50
  • $\begingroup$ @JohnDoe - Ahh thank you, I feel so stupid haha. So did my lecturer write $\space \dot{u} \space$? $\endgroup$ Oct 5, 2017 at 16:52

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As I pointed out in another answer, your lecturer seems to be writing "i.e."

You claimed that in the second of your confusions that he replaced $\frac{d\dot\theta}{dt}$ with $u$. This is not what happens in that step. Here is what happens: $$\require{cancel}\begin{align}\frac {\cancel{m}}{r}\frac d{dt} (r^2\dot\theta)\hat\theta&=\cancel mg\sin(\theta)\hat\theta\\\frac 1r(2r\dot r\dot\theta+r^2\ddot\theta)&=g\sin\theta\\2\dot r\dot \theta+r\ddot\theta&=g\sin\theta\\\end{align}$$ Then if you use the fact that $r=a$ does not chance then $\dot r=0$, and this gives $$a\ddot\theta=g\sin\theta$$ which is what your lecturer looks to have written.

Also as I wrote in a comment of that answer, later in the text, the lecturer uses 9.15 in 9.13, and there is no trace of $u$ or $\dot u$. Finally, looking at the way your lecturer wrote $u$ in the last line suggests that he does not write it with a curl at the end, as all the circled ones have. So the circled ones appear to say i.e.

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  • $\begingroup$ Thank you. Really helped clear things up. $\endgroup$ Oct 5, 2017 at 17:22
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Unfortunately I cannot comment but I think that what you circled in red is simply the Latin abbreviation ie which corresponds to id est. It means something like that is to say and is used to explain or clarify something. See the Wikitionary for more informations.

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  • $\begingroup$ It is clearly not that @M.P. In eq. (9.13) he clearly uses $u : = \dot{r}$ $\endgroup$
    – R.W
    Oct 5, 2017 at 17:03
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    $\begingroup$ Actually @RafaelWagner I changed my mind. Later in the text, the lecturer uses 9.15 in 9.13, and there is no trace of $u$ or $\dot u$ $\endgroup$
    – John Doe
    Oct 5, 2017 at 17:12
  • $\begingroup$ I never said that all the time you have the letter u it corresponds to ie. I only said that it corresponds to ie when it is circled in red. $\endgroup$
    – M. P.
    Oct 5, 2017 at 17:14
  • $\begingroup$ Yes, sorry, there is no sense using $u := \dot{r}$, if it is velocity should be $u := a\dot{\theta}$ and then the lecturer would be wrong. $\endgroup$
    – R.W
    Oct 5, 2017 at 18:32
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    $\begingroup$ This is the worst i.e. ever $\endgroup$
    – R.W
    Oct 5, 2017 at 18:32

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