A particle is moving in a circle of radius 0.3 . Calculate its linear acceleration at the instant when its angular velocity is `1rad//s` and angular acceleration is `4rad//s^(2)`

Explanation

Given, radius , r=0.3
Angular velocity, `omega=1 rad//s`
Angular acceleration, `alpha=4 rad//s^(2)`
Tangential acceleration
`a_(T)=ralpha=0.3xx4=1.2m//s^(2)`
Centripetal acceleration
`a_(c)=romega^(2)=0.3m//s^(2)`
`therefore` Linear acceleration,
`alpha=sqrt(alpha_(T^(2))+alpha_(c)^(2))=sqrt((1.2)^(2)+(0.3)^(2))=1.24m//s^(2)`
If resultant acceleration `vec a` make an angle `beta` with tangential acceleration `veca_(T)`, then
`tan beta=(a_(c))/(a_(T))=(0.3)/(1.2)=0.25`
`beta=14.04^(@)`