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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?

Respuesta :

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  • 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| =  [tex]\frac{v}{|r|} ^{2}[/tex]

                           |r| =  [tex]\frac{v}{|a|} ^{2}[/tex] = (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.

                 

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