What is the magnitude and direction of total linears acceleration of a particle moving in a circular path of radius 0.7 m having an instantaneous angular velocity of 3.5 rad s^(-1) and angular acceleration of 4.2 rad s^(-2).

What is the magnitude and direction of total linears acceleration of a

1.	What is the magnitude and direction of total linears acceleration of a particle moving in a circular path of radius 0.7 m having an instantaneous angular velocity of 3.5 rad s^(-1) and angular acceleration of 4.2 rad s^(-2).
Answer» Solution :`r = 0.7m, OMEGA = 3.5 rad s^(-1), PROP = 4.2 rad s^(-2)` `a_(r) = romega^(2) = 0.7 xx 3.5^(2) = 8.57 MS^(-2)` `a_(L) = r prop = 0.4 xx 4.2 = 1.68 ms^(-2)` Linear ACCELERATION `=a = sqrt(a_(r)^(2) + a_(l)^(2)) = sqrt(8.57^(2) + 1.68^(2)) = 8.73 ms^(-2)` Let `theta` be the angle made by a with the direction of tangential acceleration. Then, `tan theta = a_(r)/a_(l) = 8.57/1.68 = 5.1 theta = tan^(-1) 5.1 = 78.9^(@)`