Question

A solid body starts rotating about a stationary axis with an angular acceleration `alpha=(2.0xx10^(-2))trad//s^(2)` here `t` is in seconds. How soon after the beginning of rotation will the total acceleration vector of an arbitrary point of the body form an angle `theta=60^(@)` with its velocity vector?

Answer 1

Angle `alpha` is related with `|w_t|` and `w_n` by means of the formula:
`tan alpha=(w_n)/(|w_t|)`, where `w_n=omega^2R` and `|w_t|=betaR` (1)
where R is the radius of the circle which an arbitrary point of the body circumscribes. From the given equation `beta=(domega)/(dt)=at` (here `beta=(domega)/(dt)`, as `beta` is positive for all values of t)
Integrating within the limit `underset(0)overset(omega)int domega=a underset(0)overset(t)int dt`, or, `omega=1/2at^2`
So, `w_n=omega^2R=((at^2)/(2))^2 R=(a^2t^4)/(4)R`
and `|w_t|=betaR=atR`
Putting the values of `|w_t|` and `w_n` in Eq. (1), we get,
`tan alpha=(a^2t^4R//4)/(atR)=(at^3)/(4)` or, `t=[(4/a)tanalpha]^(1//3)`