The angularmomenta$\mathbf{l}\left(\mathbf{t}\right)$. of a particle in four situations are (1)$\text{l=3t+4}$(2)$\mathbf{}\mathbf{l}\mathbf{=}\mathbf{-}6{\mathbf{t}}^2$(3$\mathbf{l}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}$(4)$\mathbf{l}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{/}\mathbf{t}\mathbf{}$In which situation is the net torque on the particle (a) zero, (b) positive and constant, (c) negative and increasing in magnitude (t>0), and (d) negative and decreasing in magnitude (t>0)?

Angular momenta are,

1. $l=3t+4,$
   
2. $l=-6t^2,$
   
3. $l=2,$
   
4. .$l=4/t$

Torque is the rate of change of angular momentum (l). By taking the derivative of l, we find the net torque for a given situation.

Formulae are as follows:

$\tau =\frac{dl}{dt}$

Now, calculate for the first situation $l=3t+4$:

$\begin{array}{c}\tau =\frac{dl}{dt}\\ =\frac{d(3t+4)}{dt}\\ =3\end{array}$

So, the net torque is positive and constant.

Now, calculatefor the second situation$l=-6t^2$:

$\frac{d(-6t^2)}{dt}=-12t$

So, the net torque is negative and increasing.

Now, calculatethe third situation$l=2$ :

$\tau =\frac{d\left(2\right)}{dt}=0$

So, the net torque is zero.

Now, calculate for the fourthsituation$l=4/t$:

.$\tau =\frac{d\left(\frac{4}{t}\right)}{dt}=-4t^{-2}$

So, the net torque is negative and decreasing magnitude.

Therefore, find the net torque by taking the time derivative of the given angular momenta.

Then determine whether the torque is positive, negative, constant increasing or decreasing, or zero.