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 ?

1 Answer
Dec 13, 2017

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.