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 3.36 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?

Answer 1

• 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.

Step-by-step 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| =
`(v)/(|r|) ^(2)`

|r| =
`(v)/(|a|) ^(2)` = (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.