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» <html><body><p>1<br/>`0.5`<br/>`0.25`<br/><a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a></p>Solution :We know that `tau=1.alpha` when the angular acceleration of a <a href="https://interviewquestions.tuteehub.com/tag/wheel-736315" style="font-weight:bold;" target="_blank" title="Click to know more about WHEEL">WHEEL</a> `(alpha)` <a href="https://interviewquestions.tuteehub.com/tag/becomes-1994370" style="font-weight:bold;" target="_blank" title="Click to know more about BECOMES">BECOMES</a> <a href="https://interviewquestions.tuteehub.com/tag/zero-751093" style="font-weight:bold;" target="_blank" title="Click to know more about ZERO">ZERO</a>, <a href="https://interviewquestions.tuteehub.com/tag/torque-1422441" style="font-weight:bold;" target="_blank" title="Click to know more about TORQUE">TORQUE</a> is also zero <br/> Now `theta_((t))=2t^(3)-6t^(2)` <br/> `therefore` Angular velocity `omega=(d theta_((t)))/(dt^(2))=6t^(2)-12t` and angular accelertion `alpha=(domega)/(dt)=(d)/(dt)(6t^(2)-12t)` <br/> `alpha=12t-12` <br/> putting `t=1 s, alpha=12-2=0` <br/> `therefore t=1 sec`</body></html>