Figure gives the torque t that acts on an initially stationary disk that can rotate about its centre like a merry-go-round. The scale on the $\mathbf{\tau }$ axis is set by${\mathbf{\tau }}_{\mathbf{s}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4.0}\mathbf{}\mathbf{Nm}\mathbf{.}$

(a) what is the angular momentum of the disk about the rotation axis at times$\mathbf{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{7}\mathbf{.0}\mathbf{}\mathbf{s}$?

(b) What is the angular momentum of the disk about the rotation axis at times$\mathbf{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{20}\mathbf{}\mathbf{s}\mathbf{?}\mathbf{}$

The graph of the torque of the disc vs time.

To find the angular momentum, use the relation between torque and angular momentum. Angular momentum is the time integral of the torque. So, this can be found from the graph of torque vs time by calculating the area under the curve up to that instant.

Formula:

$\mathbf{\tau }\mathbf{=}\frac{\mathbf{dL}}{\mathbf{dt}}$

(a)

We know that, $\text{τ}=\frac{\text{dL}}{\text{dt}}$.

So, the angular momentum is, $\text{L}\left(\text{t}\right)=\int \text{τdt}$.

Therefore, the angular momentum of the disk can be calculated from the area under the curve.

The angular momentum of the disk about the rotation axis at $\text{t}=7.00\text{sis},$

$\text{L}(\text{t}=7\text{s})=$Area of the triangle with base 2 and height 1.

Area of the rectangle with length 7 and width 3.

Area of the rectangle with length 2 and width 1.

$\text{L}(\text{t}=7\text{s})=\left(\frac{1}{2}\right)(2\times 1)+(7\times 3)+(2\times 1)$

$\Rightarrow \text{L}(\text{t}=7\text{s})=1+21+2$

$\Rightarrow \mathrm{L}(\mathrm{t}=7\mathrm{s})=24{\mathrm{kgm}}^2/\mathrm{s}$

(b)

Similarly, the angular momentum of the disk about the rotation axis at $\text{t}=20.00\text{sis}$

$\begin{array}{c}\mathrm{L}(\mathrm{t}=20\mathrm{s})=\left[24+\left(\frac{1}{2}\right)(2\times 3)\right]-\left[\left(\frac{1}{2}\right)(3\times 3)+(6\times 3)+\left(\frac{1}{2}\right)(2\times 3)\right]\\ =1.5{\mathrm{kgm}}^2/\mathrm{s}\end{array}$