Answer
The magnitude of the total angular momentum is $~~6.5~kg~m^2/s$
Work Step by Step
We can find the angular momentum of the arm that sweeps through half a revolution:
$L_1 = I~\omega$
$L_1 = \frac{1}{3}ML^2~\omega$
$L_1 = \frac{1}{3}(4.0~kg)(0.60~m)^2~(\frac{\pi~rad}{0.700~s})$
$L_1 = 2.154~kg~m^2/s$
We can find the angular momentum of the arm that sweeps through a full revolution:
$L_2 = I~\omega$
$L_2 = \frac{1}{3}ML^2~\omega$
$L_2 = \frac{1}{3}(4.0~kg)(0.60~m)^2~(\frac{2\pi~rad}{0.700~s})$
$L_2 = 4.308~kg~m^2/s$
We can find the total angular momentum:
$L = L_1+L_2$
$L = (2.154~kg~m^2/s)+(4.308~kg~m^2/s)$
$L = 6.5~kg~m^2/s$
The magnitude of the total angular momentum is $~~6.5~kg~m^2/s$
