Question

Question: A particle is to move in an xyplane, clockwise around the origin as seen from the positive side of the zaxis. In unit-vector notation, what torque acts on the particle (aIf the magnitude of its angular momentum about the origin is$\mathbf{4}\mathbf{.}\mathbf{0}{\mathbf{\text{kgm}}}^{\mathbf{\text{2}}}\mathbf{\text{/s?}}$? (b) If the magnitude of its angular momentum about the origin is$\mathbf{4}\mathbf{.}\mathbf{0}{\mathit{t}}^{\mathbf{2}}{\mathbf{\text{kgm}}}^{\mathbf{\text{2}}}\mathbf{\text{/s?}}$(b) If the magnitude of its angular momentum about the origin is $\mathbf{4}\mathbf{.}\mathbf{0}\sqrt{\mathbf{t}}{\mathbf{\text{kgm}}}^{\mathbf{\text{2}}}\mathbf{\text{/s?}}$(d)If the magnitude of its angular momentum about the origin is $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{/}{\mathit{t}}^{\mathbf{2}}{\mathbf{\text{kgm}}}^{\mathbf{\text{2}}}\mathbf{\text{/s?}}$

Short answer

Answer

The torque acts on the particle bout the origin is

$$\begin{gathered} \left(a\right)\overrightarrow{\tau }=0\text{N.m} \\ \left(b\right)\overrightarrow{\tau }=\left(-8.0t\hat{\text{k}}\right)\text{N.m} \\ \left(c\right)\overrightarrow{\tau }=\left(\frac{-2.0}{\sqrt{t}}\hat{k}\right)\text{N.m} \\ \left(d\right)\overrightarrow{\tau }=\left(\frac{8.0}{t^3}\hat{k}\right)\text{N.m} \end{gathered}$$

Step-by-step solution

Identification of given data

The magnitude of the angular momentum is

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

Significance of Newton’s second law

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:

$\frac{d\overrightarrow{l}}{dt}={\overrightarrow{\tau }}_{net}$

(a) Determining the torque acting on particle if angular momentum is  4.0 kg.m2s

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.

$\frac{d\overrightarrow{l}}{dt}={\overrightarrow{\tau }}_{net}$

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

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

The angular momentum is constant so its derivative is zero.

Hence,

$\overrightarrow{\tau }=0N.m$

(b) Determining the torque acting on particle if angular momentum is  4.0t2kg.m2s

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

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

(c) Determining the torque acting on particle if angular momentum is  4.0t kg.m2s

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

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

(d) Determining the torque acting on particle if angular momentum is  4.0/t2kg.m2s

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

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

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