Chapter 11, Problem 14P

A 5.00-kg particle starts from the origin at time zero. Its velocity as a function of time is given by

$\stackrel{\to }{v}=6t^2\hat{i}+2t\hat{j}$

where $\stackrel{\to }{v}$ is in meters per second and t is in seconds. (a) Find its position as a function of time. (b) Describe its motion qualitatively. Find (c) its acceleration as a function of time, (d) the net force exerted on the particle as a function of time, (e) the net torque about the origin exerted on the particle as a function of time, (f) the angular momentum of the particle as a function of time, (g) the kinetic energy of the particle as a function of time, and (h) the power injected into the system of the particle as a function of time.

To determine

The position of particle as a function of time.

Answer to Problem 14P

The position of particle as a function of time is $\left(2t^3\widehat{i}+t^2\widehat{j}\right)\text{ m}$.

Explanation of Solution

The mass of particle is $5.00\text{ kg}$ and the velocity of particle is $\overrightarrow{\text{v}}=6t^2\widehat{i}+2t\widehat{j}$.

The formula to calculate position of particle is,

    $\overrightarrow{\text{r}}={\displaystyle \int \overrightarrow{\text{v}}dt}$

Substitute $\left(6t^2\widehat{i}+2t\widehat{j}\right)$ for $\overrightarrow{\text{v}}$ in above equation to find $\overrightarrow{\text{r}}$.

    $\begin{array}{c}\overrightarrow{\text{r}}={\displaystyle \int \left(6t^2\widehat{i}+2t\widehat{j}\right)dt}\\ =\frac{6}{3}{t}^3\widehat{i}+\frac{2}{2}{t}^2\widehat{j}\\ =\left(2t^3\widehat{i}+t^2\widehat{j}\right)\text{ m}\end{array}$

Conclusion:

Therefore, the position of particle as a function of time is $\left(2t^3\widehat{i}+t^2\widehat{j}\right)\text{ m}$.

To determine

The motion of the particle qualitatively.

Answer to Problem 14P

The particle is moving in the x-y plane turning to move more parallel to the x-axis.

Explanation of Solution

The motion of a particle is described by the change in position vector (either in magnitude or orientation) with time.

The position of the particle is $\left(2t^3\widehat{i}+t^2\widehat{j}\right)$ in coordinate system. As the particle starts from origin at $t=0$ therefore, it moves in the first quadrant. The velocity of the particle gets faster and faster with the time and it is turning to the move parallel to the x-axis in the $x\text{-}y$ plane.

Conclusion:

Therefore, the particle is moving in the x-y plane turning to move more parallel to the x-axis.

To determine

The acceleration of the particle as a function of time.

Answer to Problem 14P

The acceleration of the particle as a function of time is $\left(12t\widehat{i}+2\widehat{j}\right)\text{m}/{\text{s}}^2$.

Explanation of Solution

The formula to calculate acceleration of particle is,

    $\overrightarrow{\text{a}}=\frac{d\overrightarrow{\text{v}}}{dt}$

Substitute $\left(6t^2\widehat{i}+2t\widehat{j}\right)$ for $\overrightarrow{\text{v}}$ in above equation to find $\overrightarrow{\text{a}}$.

    $\begin{array}{c}\overrightarrow{\text{a}}=\frac{d\left(6t^2\widehat{i}+2t\widehat{j}\right)}{dt}\\ =\left(12t\widehat{i}+2\widehat{j}\right)\text{m}/{\text{s}}^2\end{array}$

Conclusion:

Therefore, the acceleration of the particle as a function of time is $\left(12t\widehat{i}+2\widehat{j}\right)\text{m}/{\text{s}}^2$.

To determine

The net force exerted on the particle as a function of time.

Answer to Problem 14P

The net force exerted on the particle as a function of time is $\left(60t\widehat{i}+10\widehat{j}\right)\text{ N}$.

Explanation of Solution

The mass of particle is $5.00\text{ kg}$ and the velocity of particle is $\overrightarrow{\text{v}}=6t^2\widehat{i}+2t\widehat{j}$.

The formula to calculate force exerted on the particle is,

    $\overrightarrow{\text{F}}=m\overrightarrow{\text{a}}$

Here, $m$ is the mass of particle.

Substitute $5.00\text{ kg}$ for $m$ and $\left(12t\widehat{i}+2\widehat{j}\right)\text{m}/{\text{s}}^2$ for $\overrightarrow{\text{a}}$ in above equation to find $\overrightarrow{\text{F}}$.

    $\begin{array}{c}\overrightarrow{\text{F}}=\left(5.00\text{ kg}\right)\left(\left(12t\widehat{i}+2\widehat{j}\right)\text{m}/{\text{s}}^2\right)\\ =\left(60t\widehat{i}+10\widehat{j}\right)\text{ N}\end{array}$

Conclusion:

Therefore, the net force exerted on the particle as a function of time is $\left(60t\widehat{i}+10\widehat{j}\right)\text{ N}$.

To determine

The net torque about the origin exerted on the particle as a function of time.

Answer to Problem 14P

The net torque about the origin exerted on the particle as a function of time is $-40t^3\widehat{k}\text{N-m}$.

Explanation of Solution

The formula to calculate torque exerted on the particle is,

    $\overrightarrow{\text{τ}}=\overrightarrow{\text{r}}\times \overrightarrow{\text{F}}$

Substitute $\left(2t^3\widehat{i}+t^2\widehat{j}\right)\text{ m}$ for $\overrightarrow{\text{r}}$ and $\left(60t\widehat{i}+10\widehat{j}\right)\text{ N}$ for $\overrightarrow{\text{F}}$ in above equation to find $\overrightarrow{\text{τ}}$.

    $\begin{array}{c}\overrightarrow{\text{τ}}=\left(2t^3\widehat{i}+t^2\widehat{j}\right)\text{ m}\times \left(60t\widehat{i}+10\widehat{j}\right)\text{ N}\\ =\left(2t^3\widehat{i}\times 60t\widehat{i}\right)+\left(2t^3\widehat{i}\times 10\widehat{j}\right)+\left(t^2\widehat{j}\times 60t\widehat{i}\right)+\left(t^2\widehat{j}\times 10\widehat{j}\right)\text{ N-m}\\ =\left(20t^3\widehat{k}-60t^3\widehat{k}\right)\text{ N-m}\\ =-40t^3\widehat{k}\text{N-m}\end{array}$

Conclusion:

Therefore, the net torque about the origin exerted on the particle as a function of time is $-40t^3\widehat{k}\text{N-m}$.

To determine

The angular momentum of the particle as a function of time.

Answer to Problem 14P

The angular momentum of the particle as a function of time is $-10t^4\widehat{k}\text{kg}\cdot {\text{m}}^2/\text{s}$.

Explanation of Solution

The mass of particle is $5.00\text{ kg}$ and the velocity of particle is $\overrightarrow{\text{v}}=6t^2\widehat{i}+2t\widehat{j}$.

The formula to calculate angular momentum of the particle is,

    $\overrightarrow{\text{L}}=m\left(\overrightarrow{\text{r}}\times \overrightarrow{\text{v}}\right)$

Substitute $5.00\text{ kg}$ for $m$, $\left(2t^3\widehat{i}+t^2\widehat{j}\right)\text{ m}$ for $\overrightarrow{\text{r}}$ and $\left(6t^2\widehat{i}+2t\widehat{j}\right)$ for $\overrightarrow{\text{v}}$ in above equation to find $\overrightarrow{\text{L}}$.

    $\begin{array}{c}\overrightarrow{\text{L}}=\left(5.00\text{ kg}\right)\left[\left(2t^3\widehat{i}+t^2\widehat{j}\right)\text{ m}\times \left(6t^2\widehat{i}+2t\widehat{j}\right)\text{m}/\text{s}\right]\\ =\left(5.00\text{ kg}\right)\left[\left(2t^3\widehat{i}\times 6t^2\widehat{i}\right)+\left(2t^3\widehat{i}\times 2t\widehat{j}\right)+\left(t^2\widehat{j}\times 6t^2\widehat{i}\right)+\left(t^2\widehat{j}\times 2t\widehat{j}\right)\right]\text{kg}\cdot {\text{m}}^2/\text{s}\\ =\left(20t^4\widehat{k}-30t^4\widehat{k}\right)\text{kg}\cdot {\text{m}}^2/\text{s}\\ =-10t^4\widehat{k}\text{kg}\cdot {\text{m}}^2/\text{s}\end{array}$

Conclusion:

Therefore, the angular momentum of the particle as a function of time is $-10t^4\widehat{k}\text{kg}\cdot {\text{m}}^2/\text{s}$.

To determine

The kinetic energy of the particle as a function of time.

Answer to Problem 14P

The kinetic energy of the particle as a function of time is $\left(90t^4+10t^2\right)\text{J}$.

Explanation of Solution

The formula to calculate kinetic energy of the particle is,

    $KE=\frac{1}{2}m\left(\overrightarrow{\text{v}}\cdot \overrightarrow{\text{v}}\right)$

Substitute $5.00\text{ kg}$ for $m$ and $\left(6t^2\widehat{i}+2t\widehat{j}\right)$ for $\overrightarrow{\text{v}}$ in above equation to find $KE$.

    $\begin{array}{c}KE=\frac{1}{2}\left(5.00\text{ kg}\right)\left(\left(6t^2\widehat{i}+2t\widehat{j}\right)\text{m}/\text{s}\cdot \left(6t^2\widehat{i}+2t\widehat{j}\right)\text{m}/\text{s}\right)\\ =\frac{1}{2}\left(5.00\text{kg}\right)\left(36t^4+4t^2\right){{\text{m}}^2/\text{s}}^2\\ =\left(90t^4+10t^2\right)\text{J}\end{array}$

Conclusion:

Therefore, the kinetic energy of the particle as a function of time is $\left(90t^4+10t^2\right)\text{J}$.

To determine

The power injected into the system as a function of time.

Answer to Problem 14P

The power injected into the system as a function of time is $\left(360t^3+20t\right)\text{W}$.

Explanation of Solution

The mass of particle is $5.00\text{ kg}$ and the velocity of particle is $\overrightarrow{\text{v}}=6t^2\widehat{i}+2t\widehat{j}$.

The formula to calculate power of the particle is,

    $\begin{array}{c}P=\frac{d}{dt}\left(90t^4+10t^2\right)\text{J}\\ =\left(360t^3+20t\right)\text{W}\end{array}$

Conclusion:

Therefore, the power of the particle as a function of time is $\left(360t^3+20t\right)\text{W}$.