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

Options

A. `t = 0.5 s`
B. `t = 0 .25 s`
C. `t = 2s`
D. `t = 1s`

Answer

D

Explanation

According to question , torque `tau = 0 ` Its means that , `alpha = 0 `
` alpha = (d^(2)theta)/(dt^(2))`
Given `theta(t) = 2t^(3)-6t^(2)`
So `(d theta)/(dt) = 6t^(2) - 12 t `
`(d^(2)theta)/(dt^(2)) = 12t - 1 2 ( :. alpha = (d^(2)theta)/(dt^(2)) = 0 ) `
`12t - 12 = 0 , t = 1s `