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

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 : .......... .

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 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 on a rotating wheel is given by the equation theta = 2t^3 - 6t^2 . After what time the torque on the wheel be zeron?

The angular displacement at any time t is given by theta(t) = 2t^(3)-6^(2) . The torque on the wheel will be zero at

The angular position of a point over a rotating flywheel is changing according to the relation, theta = (2t^3 - 3t^2 - 4t - 5) radian. The angular acceleration of the flywheel at time, t = 1 s is

The angular position of a point over a rotating flywheel is changing according to the relation, theta = (2t^3 - 3t^2 - 4t - 5) radian. The angular acceleration of the flywheel at time, t = 1 s is

The angular position of a point on a rotating wheel is given by theta=2.0+4.0t^(2)+2.0t^(3) , where theta is in radians and t is in seconds. At t = 0, what are (a) the point's angular position and (b) its angular velocity? ( c) What is its angular velocity at t = 3.0 s? (d) Calculate its angular acceleration at t = 4.0 s. (e) Is its angular acceleration constant?