MCQ
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 $t=$ ...... $\sec$
- ✓$1$
- B$0.5$
- C$0.25$
- D$2$
✓
Answer
Correct option: A.
$1$
a
Given : $\theta \left( t \right) = 2{t^3} - 6{t^2}$
Given : $\theta \left( t \right) = 2{t^3} - 6{t^2}$
$\therefore \,\frac{{d\theta }}{{dt}} = 6{t^2} - 12t \Rightarrow \frac{{{d^2}\theta }}{{d{t^2}}} = 12t - 12$
Angular acceleration, $\alpha = \frac{{{d^2}\theta }}{{d{t^2}}} = 12t - 12$
When angular acceleration $(\alpha )$ is zero, then the torque on the wheel becomes zero $\left( {\,\tau = I\alpha } \right)$
$ \Rightarrow \,12t - 12 = 0\,\,or\,t = 1\,s$
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