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?

A solid body starts rotating about a stationary axis with an angular a

1.	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» 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)`

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