Question

A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3 m/s and a centripetal acceleration $\overrightarrow{a}$ of magnitude 2 m/s². Position vector $\overrightarrow{r}$ locates him relative to the rotation axis.

What is the magnitude of $\overrightarrow{r}$ ? What is the direction of $\overrightarrow{r}$ when $\overrightarrow{a}$ is directed due east?

• A. 1.5 m, east
• B. 4.5 m, east
• C. 1.5 m, west
• D. 4.5 m, west

Answer 1

The correct option is D

4.5 m, west

$v=3m/s,a=2m/s^2$
$a=\frac{v^2}{r}\Rightarrow r=\frac{v^2}{a}=\frac{3^2}{2}=\frac{9}{2}=4.5m$
$\therefore |\overrightarrow{r}|=4.5m$
When $\overrightarrow{a}$ is due east

$\overrightarrow{r}$ will be due west as r points to the particle's position W.r.t. centre
As $\overrightarrow{a}$ always points toward the centre of motion
Please watch the video if you are still doubtful about why a:$\frac{v^2}{r}$