Question

A 3.0 kgtoy car moves along an xaxis with a velocity given by $\overrightarrow{\mathbf{v}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}{\mathbf{t}}^{\mathbf{3}}\hat{\mathbf{i}}\mathbf{m}\mathbf{/}\mathbf{s}$, with t in seconds. For t>0, what are (a) the angular momentum $\overrightarrow{\mathbf{L}}$of the car and (b) the torque $\overrightarrow{\mathbf{\tau }}$on the car, both calculated about the origin? What are (c) $\overrightarrow{\mathbf{L}}$and (d) $\overrightarrow{\mathbf{\tau }}$about the point $\left(2.0m,5.0m,0\right)$? What are (e)$\overrightarrow{\mathbf{L}}$and (f)$\overrightarrow{\mathbf{\tau }}$about the point$\left(2.0m,-5.0m,0\right)$?

Short answer

1. Angular Momentum is zero.
2. Torque is zero.
3. Angular momentum at point (2.0,5.0) is$\left(-30{\mathrm{t}}^3\hat{\mathrm{k}}\right)\mathrm{kg}.{\mathrm{m}}^2/\mathrm{s}$.
4. Torque at point (2.0,5.0) is$\left(-90{\mathrm{t}}^2\hat{\mathrm{k}}\right)\mathrm{N}.\mathrm{m}$.
5. Angular momentum at point (2.0,-5.0) is$\left(30{\mathrm{t}}^3\hat{\mathrm{k}}\right)\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2$.
6. Torque at point (2.0,-5.0) is $\left(90{\mathrm{t}}^2\hat{\mathrm{k}}\right)\mathrm{N}.\mathrm{m}$.

Step-by-step solution

Step 1: Given Data

$$\begin{gathered} \mathrm{m}=30\mathrm{kg} \\ \overrightarrow{\mathrm{v}}=\left(-2.0{\mathrm{t}}^3\hat{\mathrm{i}}\right)\mathrm{m}/\mathrm{s} \end{gathered}$$

Determining the concept

Using the equation for position, find the velocity in terms of t by differentiating it. For the second particle, using the equation for acceleration, find the equation for the velocity for the second particle, by integrating this. Finally, equate the two equations to find the time.

Formulae are as follow:

$$\begin{gathered} \mathrm{L}=\mathrm{mr}\times \mathrm{v} \\ \mathrm{\tau }=\mathrm{r}\times \mathrm{F} \end{gathered}$$

where,$\mathrm{\tau }$is torque, F is force, r is radius, v is velocity, m is mass, L is angular momentum and ais acceleration.

(a) Determining the angular momentum

Now,

$\mathrm{L}=\mathrm{mr}\times \mathrm{v}$

As the toy is moving along x axis and the velocity vector is also along the x axis, so, the cross product is,

$\overrightarrow{\mathrm{r}}\times \overrightarrow{\mathrm{v}}=0$.

Hence, the angular momentum is zero.

(b) Determining the torque

Now,

$\overrightarrow{\mathrm{v}}=\left(-2.0{\mathrm{t}}^3\hat{\mathrm{i}}\right)\mathrm{m}/\mathrm{s}$

So, the acceleration vector can be calculated as,

$\overrightarrow{\mathrm{a}}=\frac{d\overrightarrow{\mathrm{v}}}{d\mathrm{t}}=\left(-6.0{\mathrm{t}}^2\hat{\mathrm{i}}\right)\mathrm{m}/{\mathrm{s}}^2$

From this equation, it comes to know that the acceleration vector is also along the xaxis.

So,$\overrightarrow{\mathrm{r}}\times \overrightarrow{\mathrm{a}}=0$.

Hence,$\overrightarrow{\mathrm{r}}\times \overrightarrow{\mathrm{F}}=0$.

Hence, the torque is zero.

(c) Determining the angular momentum at point (2.0,5.0)

For this case, calculate the position vector first,

$\overrightarrow{{\mathrm{r}}^{\text{'}}}=\overrightarrow{\mathrm{r}}-\overrightarrow{{\mathrm{r}}_0}$

Where, $\overrightarrow{{\mathrm{r}}_0}=2.0\mathrm{i}+5.0\mathrm{j}$

Now,

$$\begin{gathered} \mathrm{L}={\mathrm{mr}}^{\text{'}}\times \mathrm{v} \\ \mathrm{L}=\mathrm{m}\left(\mathrm{r}-{\mathrm{r}}_0\right)\times \mathrm{v} \\ \mathrm{L}=\mathrm{m}\left(-{\mathrm{r}}_0\times \mathrm{v}\right) \\ \mathrm{L}=3.0\left(-2.0\hat{\mathrm{i}}-5.0\hat{\mathrm{j}}\right)\times 2.0{\mathrm{t}}^3 \\ \mathrm{L}=\left(-30{\mathrm{t}}^3\hat{\mathrm{k}}\right)\mathrm{kg}.{\mathrm{m}}^2/\mathrm{s} \end{gathered}$$.

Hence, angular momentum at point $\left(2.0,5.0\right)$is $\left(-30{\mathrm{t}}^3\hat{\mathrm{k}}\right)\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2$.

(d) Determining the torque at point (2.0,5.0)

Now,

$\overrightarrow{\mathrm{\tau }}=\overrightarrow{\mathrm{r}}\times \overrightarrow{\mathrm{F}}$

Also,

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

So,

$$\begin{gathered} \overrightarrow{\mathrm{\tau }}=\mathrm{m}\left(\overrightarrow{{\mathrm{r}}^{\text{'}}}\times \overrightarrow{\mathrm{a}}\right)=-\mathrm{m}\left(\overrightarrow{{\mathrm{r}}_0}\times \overrightarrow{\mathrm{a}}\right)=-\left(3.0\right)\left(\left(2.0\right)\left(0\right)-\left(5.0\right)\left(-6.0{\mathrm{t}}^2\right)\right)\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{\tau }}=\left(-90{\mathrm{t}}^2\hat{\mathrm{k}}\right)\mathrm{N}.\mathrm{m} \end{gathered}$$

Hence, torque at point $\left(2.0,5.0\right)$is $\left(-90{\mathrm{t}}^2\hat{\mathrm{k}}\right)\mathrm{N}.\mathrm{m}$.

(e) Determining the angular momentum at point (2.0,-5.0)

$$\begin{gathered} \overrightarrow{{\mathrm{r}}^{\text{'}}}=\overrightarrow{\mathrm{r}}-\overrightarrow{{\mathrm{r}}_0} \\ \overrightarrow{{\mathrm{r}}_0}=2.0\hat{\mathrm{i}}-5.0\hat{\mathrm{j}} \\ \overrightarrow{\mathrm{l}}=\mathrm{m}\left({\overrightarrow{\mathrm{r}}}^{\text{'}}\times \overrightarrow{\mathrm{v}}\right)=\left(-3.0\right)\left(\left(2.0\right)\left(0\right)-\left(-5.0\right)\left(-2.0{\mathrm{t}}^2\right)\right)\hat{\mathrm{k}} \end{gathered}$$

That gives,

data-custom-editor="chemistry" $\overrightarrow{\mathrm{l}}=\left(30{\mathrm{t}}^3\hat{\mathrm{k}}\right)\mathrm{kg}.{\mathrm{m}}^2/\mathrm{s}$

Hence, angular momentum at point $\left(2.0,-5.0\right)$is data-custom-editor="chemistry" $\left(30{\mathrm{t}}^3\hat{\mathrm{k}}\right)\mathrm{kg}.{\mathrm{m}}^2/\mathrm{s}$.

(f) Determining the torque at point (2.0,-5.0)

$$\begin{gathered} \overrightarrow{\mathrm{\tau }}=\mathrm{m}\left({\overrightarrow{\mathrm{r}}}^{\text{'}}\times \overrightarrow{\mathrm{a}}\right)=-\mathrm{m}\left(\overrightarrow{{\mathrm{r}}_0}\times \overrightarrow{\mathrm{a}}\right)=-\left(3.0\right)\left(\left(2.0\right)\left(0\right)-\left(-5.0\right)\left(-6.0{\mathrm{t}}^2\right)\right)\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{\tau }}=\left(90{\mathrm{t}}^2\hat{\mathrm{k}}\right)\mathrm{N}.\mathrm{m} \end{gathered}$$

Hence, torque at point$\left(2.0,-5.0\right)$is$\left(90{\mathrm{t}}^2\hat{\mathrm{k}}\right)\mathrm{N}.\mathrm{m}$.

Therefore, using the concept of differentiation and integration, the velocity from displacement and acceleration equations can be found respectively. Using these equations of velocity, it is possible to find the required answers.