Answer
$2.10N.m$
Work Step by Step
We can find the required torque as follows:
$\tau=I\alpha$
$\implies \tau=(\frac{1}{12}mL^2)\alpha$
We plug in the known values to obtain:
$\tau=\frac{1}{12}(8.42Kg)(3.15)^2(0.302rad/s^2)$
$\tau=2.10N.m$
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