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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 3.66 m/sand a centripetal acceleration of magnitude1.83m/s2. Position vectorlocates him relative to the rotation axis.(a) What is the magnitude ofr?What is the direction ofrwhen is directed (b) due east and (c) due south?

Short Answer

Expert verified
  1. The Magnitude of position vector 7.32 m.
  2. The direction of position vector whena directed toward east is at west
  3. The direction of position vector whena directed toward south is at north

Step by step solution

01

Given

The centripetal accelerationa with magnitude role="math" localid="1660900234051" 1.83m/s2, with linear velocity 3.66 m/s.

02

Understanding the concept

Find the magnitude of position vector of object by using its centripetal acceleration and linear constant speed, and it is important that the direction ofais directed towards the center of rotation.

Formulae:

Accelerationa=v2r

03

(a) Calculate the magnitude of radius of rotation

Let’s take the magnitude of acceleration to find the position of object from axis.

The equation of acceleration is

a=v2rr=v2a=3.66m/s21.83m/s2=7.32m

Therefore, the magnitude of the radius of the rotation is 7.32 m.

04

(b) Calculatethe direction of position vector when is due east.

As we know, the acceleration is directed towards the center of the circle. If the acceleration vector is directed towards east, then the center of the circle is at west from object.

05

(c) Calculatethe direction of position vector when is due south.

As discussed above, the acceleration vector is directed towards the center of the circle. If the object is at south, then the object is at north from the center of circle.

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