A particle Ptravels with constant speed on a circle of radius$\mathbf{r}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{m}$(Fig. 4-56) and completes one revolution in $\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{s}$. The particle passes through Oat time t= 0. State the following vectors in magnitude angle notation (angle relative to the positive direction of x).With respect to $\mathbf{0}$, find the particle’s position vectorat the times tof (a) $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{s}$, (b)$\mathbf{7}\mathbf{.}\mathbf{50}\mathbf{s}$, and (c)$\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{s}$. (d) For the $\mathbf{5}\mathbf{.}\mathbf{00}\mathit{s}$interval from the end of the fifth second to the end of the tenth second, find the particle’s displacement. For that interval, find (e) its average velocity and its velocity at the (f) beginning and (g) end. Next, find the acceleration at the (h) beginning and (i) end of that interval.

1) The radius of the circle is $r=3.00m$.

2) The time period of revolution is $T=20s$.

The magnitude of radius vector will be same but the direction will keep changing as it rotates. As the direction keeps changing, we will have different values of velocity and acceleration at different time. We can use trigonometric relation and find magnitude angle notation. By using expression of average velocity, we can find coordinates of the radius vector, velocity, and acceleration at different time.

At the time$t=5.0s$, the particle has traveled a fraction of the position vector.

$\begin{array}{l}\frac{t}{T}=\frac{5}{20}\\ =\frac{1}{4}\end{array}$

The angle will be for that time is

$\begin{array}{l}\Phi =\frac{1}{4}x360\\ =90°\end{array}$

This angle is measured from the vertical. Then the position coordinate from the figure $X=3.0m$and$y=3.0m$relative to the coordinate origin.

The magnitude of R is,

$$\begin{gathered} R=\sqrt{x^2+y^2} \\ =\sqrt{{\left(3.0m\right)}^2+{\left(3.0m\right)}^2} \\ =4.2m \end{gathered}$$

The angle is,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{y}{x}\right) \\ ={\tan }^{-1}\left(\frac{3.00}{3.00}\right) \\ =45° \end{gathered}$$

In magnitude the displacement vector and angle notation is$\left(4.2m,45°\right)$

The particle has traveled a fraction of$\frac{7.5}{20}=\frac{3}{8}$revolution around the circle. The angle of the particle at that instant,

$\begin{array}{l}\Phi =\frac{3}{6}x360\\ ={135}^o\end{array}$

This is the angle measured from the vertical.

From the trigonometric relation,

$$\begin{gathered} x=r\sin \Phi \\ =\left(3.0m\right)\sin \left(135°\right) \\ =2.1m \end{gathered}$$

$$\begin{gathered} y=r-r\cos \Phi \\ =\left(3.0m\right)-\left(3.0m\right)\cos \left(135\right) \\ =5.1m \end{gathered}$$

The magnitude of R is,

$$\begin{gathered} \mathrm{R}=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2} \\ =\sqrt{{\left(2.1\mathrm{m}\right)}^2+{\left(5.1\mathrm{m}\right)}^2} \\ =5.5\mathrm{m} \end{gathered}$$

The angle is,

$\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{5.1}{21}\right)\\ =68°\end{array}$

In magnitude angle notation, the displacement vector is$\left(5.5m,68°\right)$

The particle has traveled a fraction of$\frac{10}{20}=\frac{1}{2}$revolution around the circle. The angle of the particle at that instant,

$\begin{array}{l}\Phi =\frac{1}{2}x360\\ =180°\end{array}$

This is the angle measured from the vertical.

From the figure,$x=0m$and$y=6.0m$are position coordinates relative to the coordinate origin.

In magnitude angle notation, the displacement vector is$\left(6.0m,90°\right)$

We can subtract the position vector in part (a) from the position vector in (c)

$\left(6.0,90°\right)-\left(4.2,45°\right)=\left(4.2,135°\right)$

In unit vector notation,

$$\begin{gathered} ∆\overrightarrow{R}=\left(0-3.0m\right)\hat{i}+\left(6.0m-3.0m\right)\hat{j} \\ =\left(3.0m\right)\hat{i}+\left(3.0m\right)\hat{j} \end{gathered}$$

The magnitude of the displacement vector is,

$$\begin{gathered} ∆R=\sqrt{{\left(3m\right)}^2+{\left(3m\right)}^2} \\ =4.2m \end{gathered}$$

Therefore, in magnitude angle notation, the displacement vector is$\left(4.2m,135°\right)$

The time interval is $∆t=5.0s$. By using the expression of average velocity expression,

$$\begin{gathered} {\overrightarrow{V}}_{avg}=\frac{∆\overrightarrow{R}}{∆t} \\ =\frac{\left(3.0m\right)\hat{i}+\left(3.0m\right)\hat{j}}{0.5} \\ =\left(-0.60m/s\right)\hat{i}+\left(0.60m/s\right)\hat{j} \end{gathered}$$

The magnitude of the velocity vector,

$$\begin{gathered} V_{avg}=\sqrt{{\left(-0.60m/s\right)}^2+{\left(0.60m/s\right)}^2} \\ =0.85m/s \end{gathered}$$

Therefore, in magnitude angle notation, the velocity vector is$\left(0.85m/s,135°\right)$

The speed of the particle is calculated by dividing the circumference of the circle with the time of revolution.

$V=\frac{2\pi r}{T}$

Substitute the $3m$forr, and $20s$for T in the above equation.

$$\begin{gathered} =\frac{2\pi \times 3.00m}{20.00s} \\ =0.942m/s \end{gathered}$$

The velocity vector is tangent to the circle at its 3 O’ clock position at which $\overrightarrow{v}$is vertical. Therefore, the velocity at the beginning is$\left(0.94m/s,90°\right)$.

The velocity vector is tangent to the circle at its 12 O’ clock position at which$\overrightarrow{v}$is vertical. The speed will be same as calculated in part (f).

Therefore, the velocity at the end is$\left(0.94m/s,180°\right)$

At the beginning, the magnitude of the acceleration is calculated as,

$a=\frac{V^2}{r}$

Substitute the $3m$forr, and $0.94m/s^2$for Vin the above equation.

$$\begin{gathered} =\frac{{\left(0.94m/s\right)}^2}{3.00m} \\ =0.30m/s^2 \end{gathered}$$

The acceleration is horizontal (from part a) and directed towards the center of the circle. Therefore, in magnitude angle notation, the acceleration at the beginning is$\left(0.30m/s^2,180°\right)$

Similarly, at the end, the acceleration is,

$a=\frac{V^2}{r}$

Substitute the $3m$forr, and $0.94m/s^2$for Vin the above equation.

$$\begin{gathered} =\frac{{\left(0.94m/s\right)}^2}{3.00m} \\ =0.30m/s^2 \end{gathered}$$

At this instant, acceleration is vertical (from part (c)) and directed towards the center. Therefore, in magnitude angle notation, the acceleration at the beginning is $\left(0.30m/s^2,270°\right)$.