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 3.36 m/s and a centripetal acceleration "a" of magnitude 1.86 m/s^2. Position vector "r" locates him relative to the rotation axis.
(a) What is the magnitude of r?
(b) What is the direction of r when a :
i) is directed (ii) due east and (iii) due south?

The magnitude of r is 7.32 m.If the centripetal acceleration is directed due East, the direction of r is West.If the centripetal acceleration is directed due South, the direction of r is North.

Explanation:

Given,    constant speed v = 3.36 m / s, centripetal acceleration a = 1.86 m/s^2.

a) The magnitude of position vector r is relative to the rotational axis is the  radius of circular path.

                          |a| =  

                           |r| =   = (3.66)^2 / 1.86

                           |r| = 7.32 m.

The centripetal acceleration and radius vector are in opposite directions and the direction of centripetal acceleration is always inwards along radius vector.

b) If the centripetal acceleration is directed due East, the direction of r is West.If the centripetal acceleration is directed due South, the direction of r is North.

                 

Ididn’t pass this test sorry