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 3.66 m/s and a centripetal acceleration :a of magnitude 1.83 m/s2. Position vector :r locates him relative to the rotation axis. (a) What is the magnitude of :r? What is the direction of :r when :a is di- rected (b) due east and (c) due south?

Answer 1

Answer:

a) 7.32 m

b) Due west

c) Due north

Explanation:

a) The centripetal acceleration on a circular movement is:

$a_{rad}=\frac{v^{2}}{r}$ (1)

with v the tangential velocity and r the magnitude of the position vector from the center of the movement, so because we have the v and a values we can solve (1) for r:

$r=\frac{v^{2}}{a_{rad}}=r=\frac{(3.66)^{2}}{1.83}$

$r=7.32 m$

B) Centripetal acceleration of an object always points towards the center of the movement (the merry-go-round center in our case) and the position vector r point from the center of the movement (the merry-go-round center in our case) towards the outside of the merry-go-ground, that is $\overrightarrow{a}$ and $\overrightarrow{r}$ are always opposite, so if $\overrightarrow{a}$ is directed due east $\overrightarrow{r}$ is directed due west, and if $\overrightarrow{a}$ is directed due south $\overrightarrow{r}$ is directed due north.