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I'm not entirely sure if this is correct. I have to take the time derivative of the following:

$$\frac{d}{dt}mr^{2}\dot{\phi}$$

Now, both $r$ and $\dot{\phi}$ depends on the time $t$, so I have to use the product rule to get what I want. Now, what is bothering me is, is it only one of the $r$'s I need to do, or both of them ?

I mean, will it be

$$\frac{d}{dt}mr^{2}\dot{\phi} =mr\dot{r}\dot{\phi}+mr^{2}\ddot{\phi}$$

or do I have to take the other $r$ into account ?

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    $\begingroup$ I think this may be more appropriate on Mathematics - but don't worry, if that's the case, we'll migrate it. $\endgroup$
    – David Z
    Commented Mar 18, 2014 at 23:05
  • $\begingroup$ Well, that may be. It's from a Classical Mechanics book, so I just thought physics right away. But if so, you are welcome to put it in mathematics :) $\endgroup$ Commented Mar 18, 2014 at 23:08
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    $\begingroup$ It should be $\frac{d}{dt}mr^{2}\dot{\phi} =2mr\dot{r}\dot{\phi}+mr^{2}\ddot{\phi}$. It's just application of the chain rule. $\endgroup$
    – jinawee
    Commented Mar 18, 2014 at 23:11

2 Answers 2

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You're almost good, you just needed to use the chain rule, which you did perhaps without knowing it. It is clearer if you write it this way perhaps:

$$ \frac{d}{dt} m r(t)^2 \dot{\phi}(t) = m\frac{d}{dr}(r(t)^2)\frac{dr}{dt}\dot{\phi}(t)+mr^2\ddot{\phi} = 2mr\dot{r}\dot{\phi}+ mr^2\ddot{\phi}$$

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For the $r^2$ you will use the power rule of differentiation:

$$\frac{d}{dt}mr^{2}\dot{\phi} =2mr\dot{r}\dot{\phi}+mr^{2}\ddot{\phi}$$

with the "2" on the first term of the right hand side being the only difference between what I wrote and what you wrote.

If you are still confused, think of it this way. Instead of writing $r^2$ write is as $(r)^2$ to separate the exponent from the $r$. Then apply the chain rule:

$\frac{d}{dt}(r)^{2} = 2r \times (\frac{d}{dt}r) = 2r\dot{r}$.

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