The instantaneous angular position of a point ona rotating wheel is given by the equation theta (t) = 2t^(3) - 6t^(2) . The torque on the wheel becomes zero at

The instantaneous angular position of a point ona rotating wheel is gi

1.	The instantaneous angular position of a point ona rotating wheel is given by the equation theta (t) = 2t^(3) - 6t^(2) . The torque on the wheel becomes zero at
Answer» t = 1 s t = 0.5 s t= 0.25 s t = 2s SOLUTION :GIVEN : 0 `(t) = 2 t^(3) 6t^(2) , (d theta)/(dt) = 6t^(2) - 12 t` Angular acceleration , `ALPHA = (d^(2) theta)/(dt^(2)) = 12 t -12` When angular acceleration `(alpha)` is zero , then the torque on the WHEEL becomes zero `(becausetau = I alpha)` `implies 12 t - 12 = 0` or `t = 1` s