Answer:
a.[tex]460.8\hat{k}[/tex]
b.Increases
Explanation:
We are given that a vector
[tex]\vec{r}=4.0t^2\hat{i}-(2.0t+6.0t^2)\hat{j}[/tex]
Mass of particle =3.0 kg
Velocity=[tex]\frac{d\vec{r}}{dt}=\frac{d(4t^2\hat{i}-(2t+6t^2)\hat{j}}{dt}=8t\hat{i}-(2+12t)\hat{j}[/tex]
[tex]\vec{r}\times \vec{v}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\4t^2&-(2t+6t^2)&0\\8t&-(2+12t)\end{vmatrix}[/tex]
[tex]\vec{r}\times\vec{r}=8.0t^2\hat{k}[/tex]
Angular momentum=l=[tex]m\times(\vec{r}\times\vec{v})[/tex]
[tex]l=3(8.0)t^2\hat{k}=24.0t^2\hat{k}[/tex]
Torque=[tex]\frac{dl}{dt}=\frac{d(24.0t^2\hat{k})}{dt}=48.0t\hat{k}[/tex]
Torque=[tex]\tau=48.0t\hat{k}[/tex]
When t=9.6 sec
a.Then, Torque=[tex]\tau=48(9.6)\hat{k}[/tex]
=[tex]460.8\hat{k}[/tex]
b. Magnitude of angular momentum=[tex]\mid\vec{l}\mid=\sqrt{(24t^2)^2}=24t^2[/tex]
Hence, Angular momentum increases.
