Acarnival merry-go-round rotates about a vertical axis at a constant rate. a man standing on the edge has a constant speed of 3.40 m/s and a centripetal acceleration of magnitude 2.37 m/s2. position vector locates him relative to the rotation axis. (a) what is the magnitude of ? what is the direction of when is directed (b) due east and (c) due south?

a.) Magnitude of position vector = 4.877 m

b.) Direction of position vector = West

c.) Direction of position vector = North

Explanation:

a.)  This problem is related to centripetal acceleration. The magnitude of the position vector relative to rotation axis is the radius of the circle.

Given data :

Velocity = v = 3.40 m/s

Centripetal acceleration = 2.37 m/s²

magnitude of position vector = r = ?

Using centripetal acceleration (a) formula

a = v²/r

r = v²/a

r = (3.4)²/2.37

r = 4.877 m

Hence, the magnitude of position vector relative to rotational axis is  4.877 m.

b.) & c.)

When an object is rotating about a rotational axis, the direction of the centripetal acceleration is always directed towards the rotational axis and the direction of position vector is always in opposite to the centripetal acceleration i.e outward from rotational axis.b.)When the direction of the centripetal acceleration is due east, its mean that direction of position vector is due west. Because direction of position vector is always in opposite to the centripetal acceleration. c.)When the direction of the centripetal acceleration is due south, its mean that direction of position vector is due north. Because direction of position vector is always in opposite to the centripetal acceleration.

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