Transverse acceleration in polar coordinates

[marksyncm]

Homework Statement

[/B]
A particle is moving along a curve described by $p(t) = Re^{\omega t}$ and $\varphi (t) = \omega t$. What is the particles transverse acceleration?

Homework Equations

[/B]
None

The Attempt at a Solution

[/B]
The position vector is $Re^{\omega t} \vec{e_p}$. Differentiating once to get the velocity:

$$\vec{v} = \frac{d\vec{p}}{dt} = \omega Re^{\omega t}\vec{e_p} + R\omega e^{\omega t}\vec{e_\varphi}$$

And again to get the acceleration:

$$\vec{a} = \frac{d\vec{v}}{dt} = \omega^2 Re^{\omega t}\vec{e_p} + \omega^2 Re^{\omega t}\vec{e_\varphi} + \omega^2 Re^{\omega t}\vec{e_\varphi} - \omega^2 Re^{\omega t}\vec{e_p} = 2\omega^2Re^{\omega t}\vec{e_\varphi}$$

This means that the transverse acceleration is $2\omega^2Re^{\omega t}$. This answer is in line with the solution provided in my textbook, but I have a question: why did I never need to use the fact that $\varphi (t) = \omega t$? I feel like my understanding is incomplete here and I'm not sure why.

 

[marksyncm]

Nevermind; it was a silly question :) (It was necessary to differentiate the unit vector $\vec{e_p}$.)

 

[marksyncm]

Actually, I have another related problem that's similar enough I thought I'd add it here rather than start a new thread - hope that's OK:

Homework Statement

A particle's movement along a curve is described in polar coordinates by $r(t) = bt$ and $\varphi (t) = \frac{c}{t}$ ($b$ and $c$ are constants). Find the velocity and acceleration of the particle as functions of time.

Homework Equations

[/B]
None

The Attempt at a Solution

[/B]
$\varphi(t) = \frac{c}{t} \rightarrow t = \frac{c}{\varphi}$. Substituting, we get that $r(\varphi) = \frac{bc}{\varphi}$. To get the velocity as a function of time, we differentiate and get:

1) $v = \frac{-bc}{\varphi^2} \frac{d\varphi}{dt}$
2) $\frac{d\varphi}{dt} = \frac{-c}{t^2}$
3) $v = \frac{-bc}{\varphi^2} \frac{-c}{t^2} = \frac{bc^2}{\varphi^2 t^2} = \frac{bc^2}{\frac{c^2}{t^2} t^2} = b$

This doesn't sound right. The velocity of the particle is just $b$ (and, therefore, the acceleration is $0$)?

 

[marksyncm]

Just a quick thread bump, hoping someone can still help. Thank you.