At${\mathit{t}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathit{s}$, the acceleration of a particle in counterclockwise circular motion is$\left(6.00m/s^2\right)\hat{\mathbf{i}}\mathbf{+}\left(4.00m/s^2\right)\hat{\mathbf{j}}$. It moves at constant speed. At time${\mathit{t}}_{\mathbf{2}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathit{s}$, the particle’s acceleration is$\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\mathbf{)}\hat{\mathbf{j}}$. What is the radius of the path taken by the particle if${\mathit{t}}_{\mathbf{2}}\mathbf{-}{\mathit{t}}_{\mathbf{1}}$is less than one period?

1. Initial time for particle is$t_1=2.00s$
2. Atthe acceleration of the particle is${\overrightarrow{a}}_1=\left(6.00m/s^2\right)\hat{i}+\left(4.00m/s^2\right)\hat{j}$
3. The final time for particle is$t_2=5.00s$
4. Atthe acceleration of the particle is${\overrightarrow{a}}_2=\left(4.00m/s^2\right)\hat{i}+\left(-6.00m/s^2\right)\hat{j}$
5. $t_2-t_1<period\left(T\right)$
   

The particle is moving counterclockwise in a circular motion. It moves with constant speed; hence, the acceleration is of constant magnitude. When we draw the sketch, the particle travels three quarters of the circumference. We can find the period from this condition and find radius of the path taken by the particle if${\mathit{t}}_{\mathbf{2}}\mathbf{-}{\mathit{t}}_{\mathbf{1}}$is less than one period by using the relation between acceleration, period and radius.

Formula:

The magnitude of the acceleration is,

$a=\sqrt{a_x^2+a_y^2}$

The acceleration of the particle is,

role="math" localid="1661147688917" $a={\omega }^{2}r$

The angular frequency of the particle is,

$\omega =\frac{2\tau \tau }{T}$

The particle is moving counterclockwise in a circular motion. It moves with constant speed, so the acceleration is of constant magnitude.

The magnitude of the acceleration is,

$$\begin{gathered} a=\sqrt{a_x^2+a_y^2} \\ =\sqrt{{\left(6.00m/s^2\right)}^2+\left(4.00m/s^2\right)} \\ =7.21m/s^2 \end{gathered}$$

If the particle is moving counterclockwise, from$t_2$to$t_1$ , it travels three quarters of circumference as shown in the figure.So, time$t_2-t_1$ is three quarters of period T.

$$\begin{gathered} t_2-t_1=\frac{3T}{4} \\ 5.00s-2.00s=\frac{3T}{4} \\ 3.00=\frac{3T}{4} \\ T=4.00s \end{gathered}$$

The particle has uniform circular motion; hence it has radial acceleration as

$$\begin{gathered} a={\omega }^{2}r \\ ={\left(\frac{2\mathrm{\pi }}{T}\right)}^{2}r \\ r=\frac{a{T}^2}{4{\mathrm{\pi }}^2} \\ =\frac{7.21m/s^2\times {\left(4.00s\right)}^2}{4{\mathrm{\pi }}^2} \\ =2.92m \end{gathered}$$

The radius of the path taken by the particle is$r=2.92m$ .