Question

Question: At time t=0 , a 3.00 kg particle with velocity$\mathit{v}\mathbf{=}\left(5.0\text{m/s}\right)\widehat{\mathbf{\text{i}}}\mathbf{-}\left(6.0\text{m/s}\right)\widehat{\mathbf{\text{j}}}$ is at x =3.0 m and y= 8.0m. It pulled by a 7.0 N force in the negative xdirection. About the origin, what are (a) the particle’s angular momentum, (b) the torque acting on the particle, and (c) the rate at which the angular momentum is changing?

Short answer

Answer

1. The particle’s angular momentum is$\overrightarrow{l}=\left(-174\text{kg.}\frac{{\text{m}}^{\text{2}}}{\text{s}}\right)\hat{\text{k}}$
2. The torque acting on the particle is$\overrightarrow{\tau }=\left(56\text{N.m}\right)\hat{\text{k}}$
3. The rate of change of angular momentum is$\frac{d\overrightarrow{l}}{dt}=\left(56\text{N.m}\right)\hat{\text{k}}$

Step-by-step solution

Identification of given data

The mass of the particle is$m=3.0\text{kg}$ .

ii) The velocity of the particle is$\overrightarrow{v}=\left(5.0\text{m/s}\right)\hat{\text{i}}-\left(6.0\text{m/s}\right)\hat{\text{j}}$

iii) The position of the particle is$\left(x,y\right)=\left(3.0,8.0\right)\text{m}$

iv) The force acting on the particle is$F=\left(-7.0\text{N}\right)\text{i}$

To understand the concept

Use the expression of angular momentum and torque acting on the particle.Find the rate of angular momentum by using the concept of Newton’s second law in angular form.

(a) Determining the particle’s angular momentum

et position vector be$\overrightarrow{r}=x\hat{\text{i}}+y\hat{\text{j}}+z\hat{\text{k}}$ and velocity vector be $\overrightarrow{v}=v_x\hat{\text{i}}+v_y\hat{j}+v_z\hat{\text{k}}$. The cross product of the position vector and velocity vector is,

.$\overrightarrow{r}\times \overrightarrow{v}=\left(y{v}_z-z{v}_y\right)\hat{\text{i}}+\left(z{v}_x-x{v}_z\right)\hat{\text{j}}+\left(x{v}_y-y{v}_x\right)\hat{\text{k}}$

In the given position and velocity vector, a . Then,

.

$\overrightarrow{r}\times \overrightarrow{v}=\left(x{v}_y-y{v}_x\right)\hat{\text{k}}$

The angular momentum of the object with position vector and velocity vector is

.$$\begin{gathered} \overrightarrow{l}=m\left(x{v}_y-y{v}_x\right)\hat{\text{k}} \\ =3.0\text{kg}\left[\left(3.0\text{m}\right)\left(-6.0\text{m/s}\right)-\left(8.0\text{m}\right)\left(5.0\text{m/}s\right)\right] \\ =\left(-174\text{kg.}\frac{{\text{m}}^{\text{2}}}{\text{s}}\right)\hat{\text{k}} \end{gathered}$$

(b) Determining the torque acting on the particle

To find torque acting on the object, force acting on it will be $\overrightarrow{F}=\overrightarrow{F_x}\hat{\text{i}}+\overrightarrow{F_y}\hat{\text{j}}+\overrightarrow{F_z}\hat{\text{k}}$. The expression of torque is,

$$\begin{gathered} \overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F} \\ =\left(y{F}_z-z{F}_y\right)\hat{i}+\left(z{F}_x-x{F}_z\right)\hat{j}+\left(x{F}_y-y{F}_x\right)\hat{k} \end{gathered}$$

With the and then the torque acting on the object is,
$$\begin{gathered} \overrightarrow{\tau }=\left(-y{F}_x\right)\hat{\text{k}} \\ =\left(-8.0\text{m}\times -7.0\text{N}\right)\hat{\text{k}} \\ =\left(56\text{N.m}\right)\hat{\text{k}} \end{gathered}$$

(c) Determining the rate of change of angular momentum

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.

$$\begin{gathered} \frac{d\overrightarrow{l}}{dt}={\overrightarrow{\tau }}_{net} \\ \Rightarrow \frac{d\overrightarrow{l}}{dt}=\left(56\text{N.m}\right)\hat{\text{k}} \end{gathered}$$

The rate of change of angular momentum is acting along positive z axis.