Chapter 10, Problem 10.26P

Review. A small object with mass 4.00 kg moves counterclockwise with constant angular speed 1.50 rad/s in a circle of radius 3.00 m centered at the origin. It starts at the point with position vector $3.00\hat{i}\text{m}$. It then undergoes an angular displacement of 9.00 rad. (a) What is its new position vector? Use unit-vector notation for all vector answers. (b) In what quadrant is the particle located, and what angle does its position vector make with the positive x axis? (c) What is its velocity? (d) In what direction is it moving? (e) What is its acceleration? (f) Make a sketch of its position, velocity, and acceleration vectors. (g) What total force is exerted on the object?

To determine

The new position vector of the object.

Answer to Problem 10.26P

The new position vector of the object is $\left(-\text{2}\text{.7}\widehat{\text{i}}\text{+1}\text{.23}\widehat{\text{j}}\right)\text{m}$.

Explanation of Solution

The mass of the object is $4.00\text{kg}$ and the radius of the circle is $3.00\text{m}$ and the angular displacement of the object is $9.00\text{rad}$.

Formula to calculate the angle make by the small object is,

    $\begin{array}{c}\alpha =9\text{rad}\times \frac{180°}{\pi \text{rad}}\\ =515.66°-360°\\ =155.66°\end{array}$

Formula to calculate the position vector of the small object is,

    $\overrightarrow{\text{r}}=R\cos \alpha \widehat{\text{i}}\text{+}R\sin \alpha \widehat{\text{j}}$

Here, $R$ is the radius of the circle and $\alpha$ is the angle turn by the object.

Substitute $3.00\text{m}$ for $R$ and $155.66°$ for $\alpha$ in above vector notation.

    $\begin{array}{c}\overrightarrow{\text{r}}=\left(3.00\text{m}\right)\cos \left(155.66°\right)\widehat{\text{i}}\text{+}\left(3.00\text{m}\right)\sin \left(155.66°\right)\widehat{\text{j}}\\ \text{=}\left(-\text{2}\text{.7}\widehat{\text{i}}\text{+1}\text{.23}\widehat{\text{j}}\right)\text{m}\end{array}$

Conclusion:

Therefore, new position vector of the object is $\left(-\text{2}\text{.7}\widehat{\text{i}}\text{+1}\text{.23}\widehat{\text{j}}\right)\text{m}$.

To determine

The quadrant in which the particle is located and the angle made by its position with the positive $x$-axis.

Answer to Problem 10.26P

The object now located in the second quadrant and makes $155.66°$ angle from the positive $x$-axis.

Explanation of Solution

Form part (a), Section (1), the angle made by the position vector from the positive axis is

$155.66°$.

Since the value of angle lies between $90°$ and $180°$, the object now lies in the second quadrant.

Conclusion:

Therefore, the object now located in the second quadrant and makes $155.66°$ angle from the positive x-axis.

To determine

The velocity of the object.

Answer to Problem 10.26P

The velocity of the object is $\left(-1.8\widehat{\text{i}}-4.1\widehat{\text{j}}\right)\text{m}/\text{s}$.

Explanation of Solution

The mass of the object is $4.00\text{kg}$ and the radius of the circle is $3.00\text{m}$ and the angular displacement of the object is $9.00\text{rad}$.

Since the velocity vector always be the perpendicular to the position vector. So, the angle that the velocity vector made by the positive axis is,

    ${\theta }_{\text{v}}=\theta +90°$

Substitute $155.66°$ for ${\theta }_{\text{v}}$ in above equation.

    $\begin{array}{c}{\theta }_{\text{v}}=155.66°+90°\\ =245.66°\end{array}$

Formula to calculate the velocity of the object is,

    $v=\omega r$

Here, $\omega$ is the angular speed of the object.

Substitute $1.50\text{rad}/\text{s}$ for $\omega$ and $3.00\text{m}$ for $r$ in above equation.

    $\begin{array}{c}v=1.50\text{rad}/\text{s}\times 3.00\text{m}\\ =\text{4}\text{.5}\text{m}/\text{s}\end{array}$

Formula to calculate the velocity vector of the object is,

    $\overrightarrow{\text{v}}=v\cos {\theta }_{\text{v}}\widehat{\text{i}}\text{+}v\sin {\theta }_{\text{v}}\widehat{\text{j}}$

Substitute $\text{4}\text{.5}\text{m}/\text{s}$ for $v$ and $245.66°$ for ${\theta }_{\text{v}}$ to find velocity vector.

    $\begin{array}{c}\overrightarrow{\text{v}}=\left(\text{4}\text{.5}\text{m}/\text{s}\right)\cos \left(245.66°\right)\widehat{\text{i}}+\left(\text{4}\text{.5}\text{m}/\text{s}\right)\sin \left(245.66°\right)\widehat{\text{j}}\\ \text{=}\left(-1.8\widehat{\text{i}}-4.1\widehat{\text{j}}\right)\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the velocity vector of the object is $\left(-1.8\widehat{\text{i}}-4.1\widehat{\text{j}}\right)\text{m}/\text{s}$.

To determine

The quadrant in which the particle is moving.

Answer to Problem 10.26P

The object is moving in third quadrant in anticlockwise direction.

Explanation of Solution

The mass of the object is $4.00\text{kg}$ and the radius of the circle is $3.00\text{m}$ and the angular displacement of the object is $9.00\text{rad}$.

Form part (c), Section (1), the angle made by the velocity vector from the positive axis is

$245.66°$.

Since the value of angle lies between $180°$ and $270°$, the object lies in the third quadrant.

Conclusion:

Therefore, the object is moving in third quadrant in anticlockwise direction.

To determine

The acceleration of the object.

Answer to Problem 10.26P

The acceleration of the object is

Explanation of Solution

Since the acceleration vector always be the perpendicular to the velocity vector. So, the angle that the acceleration vector made by the positive axis is,

    ${\theta }_{\text{a}}={\theta }_{\text{v}}+90°$

Substitute $245.66°$ for ${\theta }_{\text{v}}$ in above equation.

    $\begin{array}{c}{\theta }_{\text{v}}=245.66°+90°\\ =335.66°\end{array}$

Formula to calculate the acceleration of the object is,

    $a={\omega }^{2}r$

Here, $\omega$ is the angular speed of the object.

Substitute $1.50\text{rad}/\text{s}$ for $\omega$ and $3.00\text{m}$ for $r$ in above equation.

    $\begin{array}{c}a={\left(1.50\text{rad}/\text{s}\right)}^2\times 3.00\text{m}\\ =\text{6}\text{.75}\text{m}/{\text{s}}^2\end{array}$

Formula to calculate the acceleration vector of the object is,

    $\overrightarrow{\text{a}}=a\cos {\theta }_{\text{a}}\widehat{\text{i}}\text{+}a\sin {\theta }_{\text{a}}\widehat{\text{j}}$

Substitute $\text{6}\text{.75}\text{m}/{\text{s}}^2$ for $a$ and $335.66°$ for ${\theta }_{\text{a}}$ to find acceleration vector.

    $\begin{array}{c}\overrightarrow{\text{a}}=\left(\text{6}\text{.75}\text{m}/{\text{s}}^2\right)\cos \left(335.66°\right)\widehat{\text{i}}+\left(\text{6}\text{.75}\text{m}/{\text{s}}^2\right)\sin \left(335.66°\right)\widehat{\text{j}}\\ \text{=}\left(6.15\widehat{\text{i}}-2.78\widehat{\text{j}}\right)\text{m}/{\text{s}}^2\end{array}$

Conclusion:

Therefore, the acceleration of the object is $\left(6.15\widehat{\text{i}}-2.78\widehat{\text{j}}\right)\text{m}/{\text{s}}^2$.

To determine

The position, acceleration and velocity vector.

Answer to Problem 10.26P

position, acceleration and velocity vector is shown in figure (I).

Explanation of Solution

Position, velocity and acceleration is a vector quantity hence they represented by both magnitude and direction in a vector diagram.

Draw the vector diagram of the position, velocity and the acceleration.

Figure (I)

Conclusion:

Therefore, the position, acceleration and velocity vector is shown in figure (I).

To determine

The total force vector exerted on the object.

Answer to Problem 10.26P

The total force vector exerted on the object is $\left(\text{24}\text{.6}\widehat{\text{i}}-11.12\widehat{\text{j}}\right)\text{N}$.

Explanation of Solution

Formula to calculate the force on the object is,

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

Conclusion:

Substitute $\left(6.15\widehat{\text{i}}-2.78\widehat{\text{j}}\right)\text{m}/{\text{s}}^2$ for $\overrightarrow{a}$ and $4.00\text{kg}$ for $m$ to find force vector.

    $\begin{array}{c}F=4.00\text{kg}\left(\left(6.15\widehat{\text{i}}-2.78\widehat{\text{j}}\right)\text{m}/{\text{s}}^2\right)\\ =\left(\text{24}\text{.6}\widehat{\text{i}}-11.12\widehat{\text{j}}\right)\text{N}\end{array}$

Therefore, the total force vector exerted on the object is $\left(\text{24}\text{.6}\widehat{\text{i}}-11.12\widehat{\text{j}}\right)\text{N}$.