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

Options

A. 1
B. `0.5`
C. `0.25`
D. 2

Answer

A

Explanation

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`