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Power from a displacement vector-correct method?

A particle of mass m moves along the curve r⃗=acos⁡ωt i⃗+bsin⁡ωt j⃗ \vec{r}=a\cos{\omega{t}}\ \vec{i}+b\sin{\omega{t}}\ \vec{j}. What power is required to produce this motion?

I said that P=dWdt=ddt(F⃗⋅r⃗)P=\frac{dW}{dt}=\frac{d}{dt} \left(\vec{F}\cdot\vec{r}\right)

Working through all of this, I end up with P=mω3cos⁡(2ωt)(a2−b2)P=m\omega^3\cos({2\omega{t}}) \left(a^2-b^2\right)................................... is this the correct thing to do? Is my answer correct?



EDIT: This can't be correct as the force is not constant, any ideas would be very much appreciated
(edited 11 years ago)
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Avatar for paddyman4
paddyman4
14
It's a constant mass so F=ma.

Could you:

Differentiate displacement twice to give acceleration and multiply by m to give the vector force at time t. F = Fx i + Fy j, where you will know Fx and Fy (Fx is supposed to be the x-component of F, but I don't know how to do subscript).

Then dW = F.dl where dl = dx i + dy j

So dW = Fx dx + Fy dy

So P = dW/dt = Fx dx/dt + Fy dy/dt

You can get dx/dt and dy/dt because x = acoswt and y = bsinwt.

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