Question

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

• A. t = 1 s
• B. t = 0.5 s
• C. t = 0.25 s
• D. t = 2 s

Answer 1

Answer: (A)

t = 1 s

Answer 2

$\theta=2 t^{3}-6 t^{2}$
$\omega=\frac{ d \theta}{d t}=6 t^{2}-12 t$
$\alpha=\frac{ d \omega}{d t}=12 t-12$
$\tau=I \alpha$
Torque will be zero when $\alpha$ is zero
so $\alpha=12 t -12=0$
$t =1 \,sec$