A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre . It is subjected to a torque which produces a constant angular acceleration of 2.0 rad s^(-2) . Its net acceleration in m s^(-2) at the end of 2.0 s is approximately

A uniform circular disc of radius 50 cm at rest is free to turn about

1.	A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre . It is subjected to a torque which produces a constant angular acceleration of 2.0 rad s^(-2) . Its net acceleration in m s^(-2) at the end of 2.0 s is approximately
Answer» `6.0` `3.0` `8.0` `7.0` Solution :Given , r = 50 CM = 0.5 m `alpha = 2.0` rad `s^(-2) , omega_(0) = 0` At the end of 2 s , Tangential acceleration , `a_(t) = r alpha = 0.5 xx 2 = 1 MS^(-2)` RADIAL acceleration , `a_(r) = omega_(2) r = (omega_(0) + alpha t)^(2) r` `= (0 + 2 xx 2)^(2) xx 0.5 = 8 m s^(-2)` `THEREFORE` Net acceleration , `a = sqrt(a_(t)^(2) + a_(r)^(2)) = sqrt(1^(2) + 8^(2)) = sqrt(65) = 8 m s^(-2)`