Answer
The angular momentum at $t=3.00~s$ is $~~23.0~kg~m^2/s$
Work Step by Step
The area under the torque versus time graph is equal to the change in momentum.
We can find the area under the torque versus time graph between $t = 1.00~s$ and $t=3.00~s$:
$\Delta L = \int_{1.00~s}^{3.00~s}(5.00+2.00~t)~dt$
$\Delta L = (5.00~t+t^2)\Big\vert_{1.00~s}^{3.00~s}$
$\Delta L = [(5.00)(3.00)+(3.00)^2]-[(5.00)(1.00)+(1.00)^2]$
$\Delta L = (24.0)-(6.00)$
$\Delta L = 18.0~kg~m^2/s$
We can find the angular momentum at $t=3.00~s$:
$L = L_i+\Delta L$
$L = (5.00~kg~m^2/s)+(18.0~kg~m^2/s)$
$L = 23.0~kg~m^2/s$
The angular momentum at $t=3.00~s$ is $~~23.0~kg~m^2/s$.
