Question

(a) Can a particle moving with instantaneous speed $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\text{m/s}}$on a path with radius of curvature 2.00 mhave an acceleration of magnitude$\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{ }{\mathbf{\text{m/s}}}^{\mathbf{\text{2}}}$? (b) Can it have an acceleration of magnitude$\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{ }{\mathbf{\text{m/s}}}^{\mathbf{\text{2}}}$? In each case, if the answer is yes, explain how it can happen; if the answer is no, explain why not.

Short answer

(a) The particle can have an acceleration of magnitude of$6 {\text{m/s}}^{\text{2}}$since it is greater than centripetal acceleration role="math" $4.5 {\text{m/s}}^{\text{2}}$ .

(b) The particle cannot have an acceleration of magnitude of$4 {\text{m/s}}^{\text{2}}$ .

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The instantaneous speed of moving particle is v = 3.00 m/s.
• The radius of curvature of the path is r = 2.00 m .

The centripetal acceleration

The centripetal acceleration is given by

${\mathit{a}}_{\mathbf{\text{c}}}\mathbf{=}\frac{{\mathbf{v}}^{\mathbf{2}}}{\mathbf{r}}$

Here, vis the speed of particle and ris the radius of curvature of the path.

Determination of the particle can have acceleration of magnitude 6.00 m/s2

(a)

The total acceleration of a particle moving in a circular path is always greater than the centripetal acceleration.

Substitute the value in the above expression, and we get,

$\begin{array}{rcl}{a}_{\text{c}}& =& \frac{{(3 \text{m/s})}^2}{2 \text{m}}\\ & =& 4.5 {\text{m/s}}^{\text{2}}\end{array}$

Since the particle stay in the circular motion, the tangential acceleration $a_t$is also there for the particle. So the magnitude of total acceleration $\sqrt{a_{\text{c}}^2+a_{\text{t}}^2}$is greater than the centripetal acceleration.

$$\begin{gathered} \left|a\right|⩾a_{\text{c}} \\ \left|a\right|⩾\left(4.5 {\text{m/s}}^{\text{2}}\right) \end{gathered}$$

Here, a is the total acceleration.

Therefore, the particle can have an acceleration of magnitude of $6 {\text{m/s}}^{\text{2}}$ since it is greater than centripetal acceleration $4.5 {\text{m/s}}^{\text{2}}$.

The acceleration magnitude of particle

(b)

The total acceleration of a particle moving in a circular path is always greater than the centripetal acceleration.

From the part (a), the particle cannot have an acceleration of magnitude of $4 {\text{m/s}}^{\text{2}}$since it is less than the centripetal acceleration.

Therefore, the particle cannot have an acceleration of magnitude of $4 {\text{m/s}}^{\text{2}}$.