- #1
marksyncm
- 100
- 5
Homework Statement
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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
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None
The Attempt at a Solution
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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.