   Chapter 7, Problem 57AP

Chapter
Section
Textbook Problem

Because of Earth’s rotation about its axis, a point on the equator has a centripetal acceleration of 0.034 0 m/s2, whereas a point at the poles has no centripetal acceleration, (a) Show that, at the equator, the gravitational force on an object (the object’s true weight) must exceed the object’s apparent weight, (b) What are the apparent weights of a 75.0-kg person at the equator and at the poles? (Assume Earth is a uniform sphere and take g = 9.800 m/s2.)

(a)

To determine

To show: The gravitational force on an object which is the true weight of the object exceeds the objects apparent weight at the equator.

Explanation

Given info: The centripetal acceleration at the equator due to the rotation of earth is 0.0340ms-2 . The poles don’t have any centripetal acceleration due to the rotation of earth.

Explanation:

The true weight of the object or the gravitational force on the object is directed downward and is given by,

Fg,true=mg

• m is the mass of the object
• g is the acceleration due to gravity

Apparent weight of a person equals the magnitude of the upward normal force exerted on the object by a scale.

The linear speed of the person due to the rotation of object is,

v=rω

• v is the linear velocity
• r is the radius of earth
• ω is the angular velocity

The centripetal acceleration will be provided by the net downward force

(b)

To determine

To show: The apparent weight of a person at equator and at poles.

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