Question

In Fig. 4-29, particle P is in uniform circular motion, centered on the origin of an xy coordinate system. (a) At what values of $\mathit{\theta }$is the vertical component ${\mathit{r}}_{\mathbf{y}}$of the position vector greatest in magnitude? (b) At what values of $\mathit{\theta }$is the vertical component ${\mathit{v}}_{\mathbf{y}}$of the particle’s velocity greatest in magnitude? (c) At what values of $\mathit{\theta }$is the vertical component ${\mathit{a}}_{\mathbf{y}}$
of the particle’s acceleration greatest in magnitude?

Short answer

1. $\theta =90°\mathrm{or}270°$
   

1. $\theta =0°\mathrm{or}180°$
   
2. $\theta =90°\mathrm{or}270°$
   

Step-by-step solution

Given information

Position vector $\overrightarrow{r}$is directed at an angle $\theta$from center.

To understand the concept

This problem is based on concept on uniform circular motion. It is the motion of a body moving at a constant speed on a circular path.

Formulae:

Position vector can be written as:

$\overrightarrow{\mathrm{r}}={\mathrm{r}}_{\mathrm{x}}\hat{\mathrm{i}}+{\mathrm{r}}_{\mathrm{y}}\hat{\mathrm{j}}$

Where,

$r_x=r\cos \left(\theta \right)$

$r_y=r\sin \left(\theta \right)$

(a) To find the values of at which the vertical component of the position vector greatest in magnitude?

Vertical component of position vector as:

$r_y=r\sin \left(\theta \right)$

And it would be greatest when $\theta =90°$or $270°$, because sine function has max value at$90°$and$70°$

(b) To find the values of at which the vertical component of the particle’s velocity greatest in magnitude?

Vertical component of velocity:

$$\begin{gathered} v_y=\frac{d\left(r_y\right)}{dt} \\ =r\cos \left(\theta \right) \end{gathered}$$

And it would be greatest when $\theta =0°\mathrm{or}180°$because cos function has max value at$\theta =0°\mathrm{and}180°$ .

(c) To find the values of at which the vertical component of the particle’s acceleration greatest in magnitude?

Vertical component of particle’s acceleration:

$$\begin{gathered} a_y=\frac{d\left(v_y\right)}{dt} \\ =-r\sin \left(\theta \right) \end{gathered}$$

Sine function have max value at $90°\mathrm{and}270°$. Hence, acceleration would be max at$\theta =90°\mathrm{or}270°$ .