Question

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

• A. $t=0.5\text{s}$
• B. $t=0.25\text{s}$
• C. $t=2\text{s}$
• D. $t=1\text{s}$

Answer 1

The correct option is D $t=1\text{s}$

Since torque, $\tau =I\alpha$

Torque is zero only when $\alpha =0$

$\theta \to \frac{d\theta }{dt}\to \omega \to \frac{d{\omega }^2}{dt}\to \alpha$

$\theta \left(t\right)=2t^3-6t^2$

Since $\frac{d}{dt}{t}^n=n{t}^{n-1}$

Therefore, $\frac{d\theta }{dt}=2\times 3t^{3-1}-6\times 2t^{2-1}$

$\omega =6t^2-12t$

$\frac{d\omega }{dt}=12t-12=\alpha$

$12t-12=0$

$⟹t=1s$