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Chapter 9: Problem 39

Suppose the Earth is a perfect sphere with \(R=6370 \mathrm{~km}\). If a person weighs exactly \(600.0 \mathrm{~N}\) at the North Pole, how much will the person weigh at the equator? [Hint: The upward push of the scale on the person is what the scale will read and is what we are calling the weight in this case.]

Short Answer

Expert verified
The weight is slightly less at the equator due to centrifugal force.

Step by step solution

01

Understand the Problem

The weight of a person is the gravitational force acting on them. At the equator, the rotation of the Earth causes a centrifugal force that reduces the normal force (what the scale reads) compared to at the poles. Our task is to compute this difference and find the weight at the equator.
02

Determine Centrifugal Force

The centrifugal force can be calculated with the formula: \[ F_c = m \cdot \omega^2 \cdot R \]where \( \omega \) is the angular speed of the Earth and \( R = 6370 \text{ km} \). The angular speed \( \omega = \frac{2\pi}{T} \), where \( T = 24 \text{ hours} = 86400 \text{ seconds} \).First, convert \( R \) to meters: \( R = 6370 \times 1000 \text{ m} \). Calculate \( \omega = \frac{2\pi}{86400} \text{ rad/s} \).
03

Calculate Mass of the Person

Using the given weight at the North Pole:\[ W = m \cdot g = 600\, \mathrm{N} \]Solve for mass \( m \):\[ m = \frac{W}{g} = \frac{600\, \mathrm{N}}{9.8\, \mathrm{m/s^2}} \approx 61.22\, \mathrm{kg} \]
04

Compute Centrifugal Force

Using the calculated values:\[ F_c = m \cdot \omega^2 \cdot R = 61.22 \cdot \left(\frac{2\pi}{86400}\right)^2 \cdot 6370000 \]Compute \( F_c \) to adjust the normal force at the equator.
05

Adjust Weight by Centrifugal Force

The weight at the equator is the gravitational force minus the centrifugal force:\[ W_{equator} = m \cdot g - F_c \]Substitute \( m = 61.22 \), \( g = 9.8 \), and previously calculated \( F_c \) to find \( W_{equator} \).
06

Final Calculation

Compute the final subtraction for the weight:\[ W_{equator} = 600 - F_c \].Complete the calculations to determine the weight at the equator.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravity
Gravity is a fundamental force that pulls objects toward each other. On Earth, gravity gives us weight by pulling us towards the planet's center. Our weight can be thought of as the force of gravity acting on our mass. The strength of this gravitational pull is determined by the mass of the Earth and the mass of the object. The formula to calculate this weight is given by:
  • \( W = m \cdot g \)
where:
  • \( W \) is the weight in newtons (N),
  • \( m \) is the mass in kilograms (kg),
  • \( g \) is the acceleration due to gravity, approximately \( 9.8 \mathrm{m/s^2} \) on Earth's surface.
This gravitational force keeps all objects, including humans, attached to the Earth's surface and gives us a sense of weight.
Angular Speed
Angular speed is how fast an object rotates or spins. For the Earth, it describes how quickly the planet spins around its axis. Understanding this rotation is crucial when discussing how weight can vary depending on location. The angular speed \( \omega \) of the Earth is calculated using the formula:
  • \( \omega = \frac{2\pi}{T} \)
where:
  • \( 2\pi \) represents a full circle in radians,
  • \( T \) is the period of rotation. For Earth, this is the length of a day or 24 hours (converted to seconds as needed).
This angular speed affects centrifugal force, influencing how weight is perceived at different points on the planet's surface.
Earth's Rotation
Earth's rotation impacts many physical phenomena, including how we experience weight due to centrifugal forces. As Earth rotates, it creates an outward force at the equator that slightly changes how much we weigh compared to the poles.The centrifugal force due to Earth's rotation is calculated with:
  • \( F_c = m \cdot \omega^2 \cdot R \)
where:
  • \( F_c \) is the centrifugal force,
  • \( m \) is the mass,
  • \( \omega \) is the angular speed,
  • \( R \) is the radius of the Earth at the equator.
This force is not present at the poles because there is no outward push from rotation there, only at the equator.
Weight Difference
The weight difference at various points on Earth arises primarily from the combination of gravity and centrifugal force. At the North Pole, the lack of centrifugal force means weight is purely due to gravity. On the other hand, at the equator, the added effect of the centrifugal force reduces the overall 'felt' weight.To find the weight at the equator:
  • Calculate the gravitational force: \( W = m \cdot g \)
  • Determine the centrifugal force: \( F_c = m \cdot \omega^2 \cdot R \)
  • Subtract the centrifugal force from the gravitational force to find the weight at the equator: \( W_{equator} = W - F_c \)
This accounts for how the outward force slightly lessens the weight as recorded on a scale, making a person weigh less at the equator compared to the North Pole.

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Most popular questions from this chapter

A thin string wound on the rim of a wheel \(20 \mathrm{~cm}\) in diameter is pulled out at a rate of \(75 \mathrm{~cm} / \mathrm{s}\) causing the wheel to rotate about its central axis. Through how many revolutions will the wheel have turned by the time that \(9.0 \mathrm{~m}\) of string have been unwound? How long will it take?

What is the maximum speed at which a car can round a curve of \(25-\mathrm{m}\) radius on a level road if the coefficient of static friction between the tires and road is \(0.80\) ? The radial force required to keep the car in the curved path (the centripetal force) is supplied by friction between the tires and the road. If the mass of the car is \(m\), the maximum friction force (which is the centripetal force) equals \(\mu, F_{N}\) or \(0.80 \mathrm{mg}\), this arises when the car is on the verge of skidding sideways. Therefore, the maximum speed is given by $$ \frac{m v^{2}}{r}=0.80 \mathrm{mg} \quad \text { or } \quad v=\sqrt{0.80 \mathrm{gr}}=\sqrt{(0.80)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(25 \mathrm{~m})}=14 \mathrm{~m} / \mathrm{s} $$

A box rests at a point \(2.0 \mathrm{~m}\) from the central vertical axis of a horizontal circular platform that is capable of revolving in the horizontal plane. The coefficient of static friction between box and platform is \(0.25 .\) As the rate of rotation of the platform is slowly increased from zero, at what angular speed will the box begin to slide?

Express each of the following in terms of other angular measures: (a) \(28^{\circ}\), (b) \(\frac{1}{4} \mathrm{rev} / \mathrm{s}\), (c) \(2.18 \mathrm{rad} / \mathrm{s}^{2}\). (a) \(28^{\circ}=(28\) deg \()\left(\frac{1 \mathrm{rev}}{360 \mathrm{deg}}\right)=0.078 \mathrm{rev}\) \(=(28 \mathrm{deg})\left(\frac{2 \pi \mathrm{rad}}{360 \mathrm{deg}}\right)=0.49 \mathrm{rad}\) (b) \(\begin{aligned} \frac{1}{4} \frac{\mathrm{rev}}{\mathrm{s}} &=\left(0.25 \frac{\mathrm{rev}}{\mathrm{s}}\right)\left(\frac{360 \mathrm{deg}}{1 \mathrm{rev}}\right)=90 \frac{\mathrm{deg}}{\mathrm{s}} \\ &=\left(0.25 \frac{\mathrm{rev}}{\mathrm{s}}\right)\left(\frac{2 \pi \mathrm{rad}}{1 \mathrm{rev}}\right)=\frac{\pi}{2} \frac{\mathrm{rad}}{\mathrm{s}} \end{aligned}\) (c) \(\begin{aligned} 2.18 \frac{\mathrm{rad}}{\mathrm{s}^{2}} &=\left(2.18 \frac{\mathrm{rad}}{\mathrm{s}^{2}}\right)\left(\frac{360 \mathrm{deg}}{2 \pi \mathrm{rad}}\right)=125 \frac{\mathrm{deg}}{\mathrm{s}^{2}} \\ &=\left(2.18 \frac{\mathrm{rad}}{\mathrm{s}^{2}}\right)\left(\frac{1 \mathrm{rev}}{2 \pi \mathrm{rad}}\right)=0.347 \frac{\mathrm{rev}}{\mathrm{s}^{2}} \end{aligned}\)

( \(a\) ) Compute the radial acceleration of a point at the equator of the Earth. (b) Repeat for the North Pole of the Earth. Take the radius of the Earth to be \(6.37 \times 10^{6} \mathrm{~m}\).
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