91Ó°ÊÓ

The magnitude of the angular momentum is

$$\begin{gathered} i)4.0{\text{kgm}}^{\text{2}}\text{/s} \\ ii)4.0t^2{\text{kgm}}^{\text{2}}\text{/s} \\ iii)4.0\sqrt{t}{\text{kgm}}^{\text{2}}\text{/s} \\ iv)4.0/t^2{\text{kgm}}^{\text{2}}\text{/s} \end{gathered}$$

The moment of inertia multiplied by the angular acceleration is the result of many torques acting on a rigid body about a fixed axis.

Find the torque acting on the particle by using concept of Newton’s second law in the angular form.

Formulae:

$\frac{d\stackrel{→}{l}}{dt}={\stackrel{→}{\mathrm{τ}}}_{net}$

According to Newton’s second law in angular form, the sum of all torques acting on a particle is equal to the time rate of the change of the angular momentum of that particle.

$\frac{d\stackrel{→}{l}}{dt}={\stackrel{→}{\mathrm{τ}}}_{net}$

The particle is moving in xy plane in clockwise direction; hence the angular momentum is along negative z axis.

Here the angular momentum is $\stackrel{→}{l}=\left(-4.0\text{kg.}\frac{{\text{m}}^{\text{2}}}{\text{s}}\right)\hat{\text{k}}$

The angular momentum is constant so its derivative is zero.

Hence,

$\stackrel{→}{\mathrm{τ}}=0N.m$

The particle is moving in xy plane in clockwise direction; hence the angular momentum moves along negative z axis. Here the angular momentum is $\stackrel{→}{l}=\left(-4.0t^2\text{kg.}\frac{{\text{m}}^{\text{2}}}{\text{s}}\right)\hat{\text{k}}$

$$\begin{gathered} \stackrel{→}{\mathrm{τ}}=\left(-4.0\hat{\text{k}}\right)\frac{d{t}^2}{dt} \\ =\left[2\left(-4.0\hat{\text{k}}\right)t\right]\text{N.m} \\ =\left(-8.0t\text{N.m}\right)\hat{\text{k}} \end{gathered}$$

The particle is moving in xy plane in clockwise direction; hence the angular momentum moves along negative z axis. Here the angular momentum is $\stackrel{→}{l}=\left(-4.0\sqrt{t}\text{kg.}\frac{{\text{m}}^{\text{2}}}{\text{s}}\right)\hat{\text{k}}$

$$\begin{gathered} \stackrel{→}{\mathrm{τ}}=\left(-4.0\hat{\text{k}}\right)\frac{d\sqrt{t}}{dt} \\ =\frac{\left(-4.0\hat{\text{k}}\right)}{2\sqrt{t}}\text{N.m} \\ =\left(\frac{-2.0}{\sqrt{t}}\text{N.m}\right)\hat{\text{k}} \end{gathered}$$

The particle is moving in xy plane in clockwise direction; hence the angular momentum moves along negative z axis. Here the angular momentum is $\stackrel{→}{l}=\left(-4.0/t^2\text{kg.}\frac{{\text{m}}^{\text{2}}}{\text{s}}\right)\hat{k}$

$$\begin{gathered} \stackrel{→}{\mathrm{τ}}=\left(-4.0\hat{k}\right)\frac{d{t}^{-2}}{dt} \\ =\left(\frac{8.0}{t^3}\text{N.m}\right)\hat{k} \end{gathered}$$

It indicates that torque is positive, and it is going in counterclockwise direction. Hence, the particle slows down.