A particle moves along an arc of a circle of radius R according to the law `l=a sin omegat`, where l is the displacement from the initial position measured along the arc, and a and `omega` are constants. Assuming `R=1.00m`, `a=0.80m`, and `omega=2.00rad//s`, find:
(a) the magnitude of the total acceleration of the particle at the points `l=0` and `l=+-a`,
(b) the minimum value of the total acceleration `w_(min)` and the corresponding displacement `l_m`.

Answer

(a) `w_0=a^2omega^2//R=2.6m//s^2`, `w_a=aomega^2=3.2m//s^2`; (b) `w_(min)=aomega^2sqrt(1-(R//2a)^2)=2.5m//s^2`, `l_m=+-asqrt(1-R^2//2a^2)=+-0.37m`.

Explanation

From the equation `l=a sin omega t`
`(dl)/(dt)=v=a omega cos omega t`
So, `w_t=(dv)/(dt)=-a omega^2sin omega t`, and
`w_n=(V^2)/(R)=(a^2omega^2cos^2omegat)/(R)`
(a) At the point `l=0`, `sin omega t=0` and `cos omega t=+-1` so, `omega t=0`, `pi` etc.
Hence `w=w_n=(a^2omega^2)/(R)`
Similarly at `l=+-a`, `sin omega t=+-1` and `cos omega t=0`, so, `w_n=0`
Hence `w=|w_t|=a omega^2`