Author Topic: Finding the velocity vector from the normal and tangential coordinates (Read 3946 times) Tweet Share

BigAl · « **on:** October 28, 2013, 09:36:21 pm »

Can you please help me with this question?

b^3 · « **Reply #1 on:** October 28, 2013, 10:07:26 pm »

Firstly, remember that $r$ and $\phi$ are functions of $t$ . ( $r$ won't be constant and will be a function of $t$ as you have that velocity component that is normal to the circle).

$\begin{alignedat}{1}\underset{\sim}{r}\left(t\right) & =r\underset{\sim}{e_{r}} \\ & =r\left(\cos\left(\phi\right)\underset{\sim}{i}+\sin\left(\phi\right)\underset{\sim}{j}\right) \\ \underset{\sim}{\dot{r}}\left(t\right) & =\frac{d}{dt}\left(r\left(\cos\left(\phi\right)\underset{\sim}{i}+\sin\left(\phi\right)\underset{\sim}{j}\right)\right) \\ & \text{As \ensuremath{r} and \ensuremath{\phi} are functions of \ensuremath{t} we apply the product rule} \\ \underset{\sim}{\dot{r}}\left(t\right) & =\left(r\frac{d}{dt}\left(\cos\left(\phi\right)\right)+\frac{dr}{dt}\cos\left(\phi\right)\right)\underset{\sim}{i}+\left(r\frac{d}{dt}\left(\sin\left(\phi\right)\right)+\frac{dr}{dt}\sin\left(\phi\right)\right)\underset{\sim}{j} \\ & \text{Since we have a function of \ensuremath{\phi} which is a function of \ensuremath{t} we can't differentiate directly, so we apply the chain rule} \\ \underset{\sim}{\dot{r}}\left(t\right) & =\left(r\frac{d}{d\phi}\left(\cos\left(\phi\right)\right)\frac{d\phi}{dt}+\cos\left(\phi\right)\frac{dr}{dt}\right)\underset{\sim}{i}+\left(r\frac{d}{d\phi}\left(\sin\left(\phi\right)\right)\frac{d\phi}{dt}+\sin\left(\phi\right)\frac{dr}{dt}\right)\underset{\sim}{j} \\ & =\left(-r\sin\phi\frac{d\phi}{dt}+\cos\left(\phi\right)\frac{dr}{dt}\right)\underset{\sim}{i}+\left(r\cos\phi+\sin\left(\phi\right)\frac{dr}{dt}\right)\frac{d\phi}{dt}\underset{\sim}{j} \\ & =\frac{d\phi}{dt}r\left(-\sin\phi\underset{\sim}{i}+\cos\phi\underset{\sim}{j}\right)+\frac{dr}{dt}\left(\cos\left(\phi\right)\underset{\sim}{i}+\sin\left(\phi\right)\underset{\sim}{j}\right) \\ & \text{As }\underset{\sim}{e_{r}}=\cos\left(\phi\right)\underset{\sim}{i}+\sin\left(\phi\right)\underset{\sim}{j}\text{ and }\underset{\sim}{e_{\phi}}=-\sin\left(\phi\right)\underset{\sim}{i}+\cos\left(\phi\right)\underset{\sim}{j} \\ & =\frac{d\phi}{dt}r\underset{\sim}{e_{\phi}}+\frac{dr}{dt}\underset{\sim}{e_{r}} \\ \therefore\underset{\sim}{\dot{r}}\left(t\right) & =\dot{r}\underset{\sim}{e_{r}}+r\dot{\phi}\underset{\sim}{e_{\phi}}\text{, as required} \end{alignedat}$

EDIT: Was referring to the wrong rule. Fixed.
EDIT2: Also guessing this is for MAE1042 rather than maths?

BigAl · « **Reply #2 on:** October 28, 2013, 10:23:23 pm »

Quote from: b^3 on October 28, 2013, 10:07:26 pm

Firstly, remember that $r$ and $\phi$ are functions of $t$ . ( $r$ won't be constant and will be a function of $t$ as you have that velocity component that is normal to the circle).

$\begin{alignedat}{1}\underset{\sim}{r}\left(t\right) & =r\underset{\sim}{e_{r}} \\ & =r\left(\cos\left(\phi\right)\underset{\sim}{i}+\sin\left(\phi\right)\underset{\sim}{j}\right) \\ \underset{\sim}{\dot{r}}\left(t\right) & =\frac{d}{dt}\left(r\left(\cos\left(\phi\right)\underset{\sim}{i}+\sin\left(\phi\right)\underset{\sim}{j}\right)\right) \\ & \text{As \ensuremath{r} and \ensuremath{\phi} are functions of \ensuremath{t} we apply the product rule} \\ \underset{\sim}{\dot{r}}\left(t\right) & =\left(r\frac{d}{dt}\left(\cos\left(\phi\right)\right)+\frac{dr}{dt}\cos\left(\phi\right)\right)\underset{\sim}{i}+\left(r\frac{d}{dt}\left(\sin\left(\phi\right)\right)+\frac{dr}{dt}\sin\left(\phi\right)\right)\underset{\sim}{j} \\ & \text{Since we have a function of \ensuremath{\phi} which is a function of \ensuremath{t} we can't differentiate directly, so we apply the chain rule} \\ \underset{\sim}{\dot{r}}\left(t\right) & =\left(r\frac{d}{d\phi}\left(\cos\left(\phi\right)\right)\frac{d\phi}{dr}+\cos\left(\phi\right)\frac{dr}{dt}\right)\underset{\sim}{i}+\left(r\frac{d}{d\phi}\left(\sin\left(\phi\right)\right)\frac{d\phi}{dt}+\sin\left(\phi\right)\frac{dr}{dt}\right)\underset{\sim}{j} \\ & =\left(-r\sin\phi\frac{d\phi}{dt}+\cos\left(\phi\right)\frac{dr}{dt}\right)\underset{\sim}{i}+\left(r\cos\phi+\sin\left(\phi\right)\frac{dr}{dt}\right)\frac{d\phi}{dt}\underset{\sim}{j} \\ & =\frac{d\phi}{dt}r\left(-\sin\phi\underset{\sim}{i}+\cos\phi\underset{\sim}{j}\right)+\frac{dr}{dt}\left(\cos\left(\phi\right)\underset{\sim}{i}+\sin\left(\phi\right)\underset{\sim}{j}\right) \\ & \text{As }\underset{\sim}{e_{r}}=\cos\left(\phi\right)\underset{\sim}{i}+\sin\left(\phi\right)\underset{\sim}{j}\text{ and }\underset{\sim}{e_{\phi}}=-\sin\left(\phi\right)\underset{\sim}{i}+\cos\left(\phi\right)\underset{\sim}{j} \\ & =\frac{d\phi}{dt}r\underset{\sim}{e_{\phi}}+\frac{dr}{dt}\underset{\sim}{e_{r}} \\ \therefore\underset{\sim}{\dot{r}}\left(t\right) & =\dot{r}\underset{\sim}{e_{r}}+r\dot{\phi}\underset{\sim}{e_{\phi}}\text{, as required} \end{alignedat}$

EDIT: Was referring to the wrong rule. Fixed.
EDIT2: Also guessing this is for MAE1042 rather than maths?

no this is for asp2062 introduction to astrophysics...it's really great those two units have some cross-over

thanks for the solution

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Author Topic: Finding the velocity vector from the normal and tangential coordinates (Read 3946 times) Tweet Share

BigAl

Finding the velocity vector from the normal and tangential coordinates

b^3

Re: Finding the velocity vector from the normal and tangential coordinates

BigAl

Re: Finding the velocity vector from the normal and tangential coordinates

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