A particle is in uniform circular motion about the origin of an x-ycoordinate system, moving clockwise with a period of 7.00s. At one instant, its position vector (measured from the origin) is$\overrightarrow{\mathbf{r}}\mathbf{=}\left(2.00m\right)\hat{\mathbf{i}}\mathbf{.}\left(3.00m\right)\hat{\mathbf{j}}$ . At that instant, what is its velocity in unit-vector notation?

a) The time period of the particle, T=7s

b) The position vector of the particle, $\overrightarrow{\mathrm{r}}=\left(2.00\mathrm{m}\right)\hat{\mathrm{i}}-\left(3.00\mathrm{m}\right)\hat{\mathrm{j}}$

For the Uniform Circular motion, velocity is given by the circumference of a circular path divided by period. As the position of a particle is in the fourth quadrant and the motion is in a clockwise direction, both velocity components should be negative.

Formula:

The velocity of a particle in uniform circular motion, $\mathrm{V}=\frac{2\mathrm{\pi r}}{\mathrm{T}}$ ...(i)

Using the given data in equation (i), the x-component of the velocity of the particle can be given as:

$$\begin{gathered} {\mathrm{V}}_{\mathrm{x}}=\frac{2\mathrm{\pi }\times \left(2.00\mathrm{m}\right)}{7.0\mathrm{s}} \\ =1.80\mathrm{m}/\mathrm{s} \end{gathered}$$

Similarly, the y-component is given as:

$$\begin{gathered} {\mathrm{V}}_{\mathrm{y}}=\frac{-2\mathrm{\pi }\times \left(3.00\mathrm{m}\right)}{7.0\mathrm{s}} \\ =-2.69\mathrm{m}/\mathrm{s} \end{gathered}$$

Since, the position vector $\stackrel{\rightharpoonup }{r}$is in the fourth quadrant, the motion is clockwise. So$V_x$and$V_y$must be negative.

So, the velocity of a particle is given in unit-vector notion as:

$\mathrm{V}=\left(-1.80\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{i}}+\left(2.69\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{j}}$

Hence, the velocity value is$\left(-1.80\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{i}}+\left(2.69\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{j}}$