Question

Figure P4.40 represents the W total acceleration of a particle moving clockwise in a circle of radius 2.50 m at a certain instant of time. For that instant, find

(a) the radial acceleration of the particle,

(b) the speed of the particle, and

(c) its tangential acceleration.

Short answer

(a) The radial acceleration of the particle is,$13.0 {\text{m/s}}^{\text{2}}$ .

(b) The speed of the particle is,$5.70 \text{m/s}$ .
(c) The tangential acceleration of the particle is $7.54 {\text{m/s}}^{\text{2}}$.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The radius of the circular path is, $r=2.50 \text{m}$.
• The acceleration of the particle is, $a=15.0 {\text{m/s}}^{\text{2}}$.

The relation between the total acceleration, tangential acceleration and the radial acceleration

The relation between the total acceleration, tangential acceleration and the radial acceleration is,

${\left(a_{\text{r}}\right)}^{\mathbf{2}}\mathbf{=}{\mathit{a}}^{\mathbf{2}}\mathbf{-}{\left(a_{\text{t}}\right)}^{\mathbf{2}}$ (1)

Here,

${\mathit{a}}_{\mathbf{\text{t}}}$is the tangential acceleration.

${\mathit{a}}_{\mathbf{r}}$is the radial acceleration.

The radial acceleration of the particle

From the Figure the relation between the tangential acceleration and the radial acceleration is,

$\begin{array}{rcl}\tan \left(30°\right)& =& \frac{a_t}{a_r}\\ 0.58& =& \frac{a_t}{a_r}\\ a_t& =& 0.58a_r\\ & & \end{array}$

Substitute the values in equation 1 and we get,

$\begin{array}{rcl}{\left(a_{\text{r}}\right)}^2& =& {\left(15.0 {\text{m/s}}^{\text{2}}\right)}^2-{\left(0.58a_{\text{r}}\right)}^2\\ 1.336{\left(a_{\text{r}}\right)}^2& =& {\left(15.0 {\text{m/s}}^{\text{2}}\right)}^2\\ a_{\text{r}}& =& 12.97 {\text{m/s}}^{\text{2}}\\ & \approx & 13.0 {\text{m/s}}^{\text{2}}\\ & & \end{array}$

Therefore, the radial acceleration of the particle is 13 m/s².

The speed of the particle

(b)

The formula to calculate the speed off the particle is,

$v=\sqrt{a_{\text{r}}r}$

Substitute the values in the above equation and we get,

$\begin{array}{rcl}v& =& \sqrt{\left(13.0 {\text{m/s}}^{\text{2}}\right)(2.50 \text{m})}\\ & =& \sqrt{32.5} \text{m/s}\\ & =& 5.70 \text{m/s}\\ & & \end{array}$

Therefore, the speed of the particle is .5.70 m/s.

The tangential acceleration of the particle

(c)

From part (a) the relation between the radial acceleration and the tangential acceleration is,

$a_{\text{t}}=0.58a_{\text{r}}$

And also, from part (a) the radial acceleration if the article is 13 m/s².

Substitute the value in the above equation.

role="math" localid="1663794434331" $\begin{array}{rcl}{a}_{\text{t}}& =& 0.58\left(13.0 {\text{m/s}}^{\text{2}}\right)\\ & =& 7.54 {\text{m/s}}^{\text{2}}\end{array}$

Therefore, the tangential acceleration of the particle is $\begin{array}{rcl}& & 7.54 {\text{m/s}}^{\text{2}}\end{array}$.