Answer
The angular momentum at $t = 7.0~s$ is $~~24~kg~m^2/s$
Work Step by Step
The area under a torque versus time graph is the change in angular momentum.
From $t = 0$ to $t = 7.0~s$, we can divide the area under the graph into three sections, including a rectangle (t = 0 to t = 2.0), a triangle (t = 2.0 to t = 4.0), and a rectangle (t = 2.0 to t = 7.0).
We can find the area of each section separately:
$A_1 = (4.0~N\cdot m)(2.0~s) = 8.0~kg~m^2/s$
$A_1 = \frac{1}{2}(1.0~N\cdot m)(2.0~s) = 1.0~kg~m^2/s$
$A_3 = (3.0~N\cdot m)(5.0~s) = 15~kg~m^2/s$
We can find the total area:
$A = A_1+A_2+A_3$
$A = (8.0~kg~m^2/s)+(1.0~kg~m^2/s)+(15~kg~m^2/s)$
$A = 24~kg~m^2/s$
Since the disk is stationary initially, the angular momentum at $t = 7.0~s$ is $~~24~kg~m^2/s$
