Question

A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the purse is $\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{00}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}\mathbf{)}\hat{\mathbf{j}}$.At that instant and in unit-vector notation, what is the acceleration of the wallet?

Short answer

The acceleration of the wallet is ${\overrightarrow{a}}_w=(3.00m/s^2)\hat{i}+(6.00m/s^2)\hat{j}$

Step-by-step solution

Given

1. The radius of the purse is$r_p=2.00m$
2. The radius of the wallet is$r_w=3.00m$

1. The acceleration of the purse is, $a=(2.00m/s^2)\hat{i}+(4.00m/s^2)\hat{j}$

Understanding the concept of circular motion

A purse and wallet travel in a uniform circular motion. They are on the same radial line; hence the angular frequency of the purse and wallet is the same. Both have centripetal acceleration only.

Formula:

$a={\omega }^{2}r$

Calculate the acceleration of the wallet

The purse and the wallet are performing uniform circular motion. The radius of purse -${\mathrm{r}}_{\mathrm{P}}$and radius of the wallet -$r_w$are acting along the same line; hence their angular velocity isthesame.

The equation of acceleration of purse is

$a_P={\omega }^2{r}_P$ …(i)

The equation of acceleration of wallet is

$a_w={\omega }^2{r}_w$ …(ii)

Dividing equation (ii) by (i),

$$\begin{gathered} \frac{a_w}{a_p}=\frac{r_w}{r_p} \\ {a}_w=\frac{r_w}{r_p}{a}_w \\ \overrightarrow{a_w}=\frac{3}{2}\left(\right(2.00m/s^2)\hat{i}+(4.00m/s^2\left)\hat{j}\right) \\ =(3.00m/s^2)\hat{i}+(6.00m/s^2)\hat{j} \end{gathered}$$

Therefore, the acceleration of the wallet is $(3.00m/s^2)\hat{i}+(6.00m/s^2)\hat{j}$.