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

The instantaneous angular position of a point on a rotating wheel is g

1.	The instantaneous angular position of a point on a rotating wheel is given by the equation theta_((t))=2t^(3)-6t^(2). The torque on the wheel becomes zero at ………. s
Answer» 1 `0.5` `0.25` 2 Solution :We know that `tau=1.alpha` when the angular acceleration of a WHEEL `(alpha)` BECOMES ZERO, TORQUE is also zero Now `theta_((t))=2t^(3)-6t^(2)` `therefore` Angular velocity `omega=(d theta_((t)))/(dt^(2))=6t^(2)-12t` and angular accelertion `alpha=(domega)/(dt)=(d)/(dt)(6t^(2)-12t)` `alpha=12t-12` putting `t=1 s, alpha=12-2=0` `therefore t=1 sec`