Answer:
a) 7.32 m
b) Due west
c) Due north
Explanation:
a) The centripetal acceleration on a circular movement is:
[tex]a_{rad}=\frac{v^{2}}{r}[/tex] (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:
[tex]r=\frac{v^{2}}{a_{rad}}=r=\frac{(3.66)^{2}}{1.83} [/tex]
[tex]r=7.32 m [/tex]
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 [tex] \overrightarrow{a}[/tex] and [tex] \overrightarrow{r}[/tex] are always opposite, so if [tex]\overrightarrow{a} [/tex] is directed due east [tex]\overrightarrow{r} [/tex] is directed due west, and if [tex]\overrightarrow{a} [/tex] is directed due south [tex]\overrightarrow{r} [/tex] is directed due north.
