Answer
$\int\Sigma Mdt=0.833Kg\cdot m^2/s$
Work Step by Step
We can determine the required angular impulse as follows:
$\Sigma \int Mdt=I\omega~~~$[eq(1)]
We know that
$I=\frac{1}{12}mr_r L_r^2+2(\frac{1}{2}mr^2+mr^2_{A/G})$
We plug in the known values to obtain:
$I=\frac{1}{12}(1)(0.58)^2+2[\frac{1}{2}(1)(0.01)^2+1(0.3)^2]$
This simplifies to:
$I=0.2081kg\cdot m^2$
We plug in the known values in eq(1) to obtain:
$\int\Sigma Mdt=I_{axle}\omega=(0.2081)(4)=0.833Kg\cdot m^2/s$
