Question

Please solve this question ?

A particle's position vector as function of time ,in polar coordinates, is given by `vecr = 2k cos( (ωt)/2)hat r` & where `dotθ = ω` is constant. Find the ratio `a_r/a_θ` for this particle ,where `a_theta` and `a_r` are components of acceleration in polar co-ordinates ?

Answer 1

See below.

Explanation:

Making `hat r = (costheta,sintheta)` and `f(t)=2kcos((omega t)/2)`

we have

`(d hat r)/(dt) = dot theta(-sintheta,costheta) = dot theta hat tau`

with `hat tau = (-sintheta,costheta)`

Now

`(d^2 hat r)/(d^2t) = d/(dt)(dot theta hat tau) = ddot theta hat tau + dot theta (d hat tau)/(dt)`

but `(d hat tau)/(dt) = -dot theta (costheta,sintheta) = -dot theta hat r` so

`(d^2 hat r)/(d^2t) = ddot theta hat tau - (dot theta)^2 hat r`

and then

`(d vec r)/(dt) = dot( f)(t) hat r + dot theta f(t)hat tau` and

`(d^2 vec r)/(d^2t) =ddot(f)(t) hat r + dot theta dot(f)(t) hat tau +ddot(theta)f(t) hat tau+dot theta dot(f)(t) hat tau -(dot theta)^2f(t) hat r = (ddot(f)(t)-(dot theta)^2f(t) )hat r+(2dot theta dot(f)(t)+ddot(theta)f(t))hat tau`

but

`dot(f)(t) = -komegasin((omega t)/2)` and
`ddot(f)(t) = -1/2k omega^2 cos((omega t)/2)`

and

`dot theta = omega` and also `ddot theta = 0`

Here `hat tau` can also be called `hat theta` so

`a_r = ddot(f)(t)-(dot theta)^2f(t)`
`a_theta = 2dot theta dot(f)(t)+ddot(theta)f(t)`

The final substitutions are left to the reader.