/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 344 A car of mass \(m\) is traveling... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 344

A car of mass \(m\) is traveling at a road speed \(v_{r}\) along an equatorial east-west highway at sea level. If the road follows the curvature of the earth, derive an expression for the difference \(\Delta P\) between the total force exerted by the road on the car for eastward travel and the total force for westward travel. Calculate \(\Delta P\) for \(m=1500 \mathrm{kg}\) and \(v_{r}=\) \(200 \mathrm{km} / \mathrm{h} .\) The angular velocity \(\omega\) of the earth is \(0.7292\left(10^{-4}\right)\) rad/s. Neglect the motion of the center of the earth.

Short Answer

Expert verified
The difference in road force, \( \Delta P \), is approximately 243.29 N.

Step by step solution

01

Understanding the Problem

We need to find the difference in the road force on the car when it travels eastward versus westward along the equator, considering Earth's rotation. The key aspect is understanding how Earth's rotation affects the apparent forces acting on the car.
02

Identifying Forces

Forces to consider include gravity and the centrifugal force due to Earth's rotation. Both are affected differently when traveling along the equator in east and west directions.
03

Derive Expressions for Forces

The normal force acting on the car is given by: \( N = m imes g_{eff} \). Here, \( g_{eff} \) differs when traveling eastward \( g_{eff, east} = g - rac{v_{total}^2}{R} \) and westward \( g_{eff, west} = g + rac{v_{total}^2}{R} \), where \( v_{total} \) is the total velocity including Earth's rotation and \( R \) is Earth's radius.
04

Calculate Total Velocity

For eastward travel, \( v_{total, east} = v_{r} + R imes \omega \). For westward travel, \( v_{total, west} = v_{r} - R imes \omega \). This accounts for the direct sum and difference of linear and rotational speed respectively.
05

Derive Force Difference Expression

The expression for difference in forces is: \( \Delta P = m \left( \frac{(v_{r}+R\omega)^2}{R} - \frac{(v_{r}-R\omega)^2}{R} \right) \). Simplifying gives: \( \Delta P = 4m \frac{v_{r} \cdot R \cdot \omega}{R} = 4mv_{r}\omega \).
06

Insert Given Values

Using \( m = 1500 \ kg \), \( v_{r} = 200 \ km/h = 55.56 \ m/s \), and \( \omega = 0.7292 \times 10^{-4} \ rad/s \): \( \Delta P = 4 \times 1500 \times 55.56 \times 0.7292 \times 10^{-4} \).
07

Calculate Final Value

Performing the calculation: \( \Delta P = 4 \times 1500 \times 55.56 \times 0.7292 \times 10^{-4} = 243.29 \). Therefore, \( \Delta P \approx 243.29 \ N \).

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Key Concepts

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

Centrifugal Force
When an object moves in a circular path, it experiences an outward force known as the centrifugal force. This is not an actual force defined by physics, but rather a perceived effect arising from inertia in a rotating reference frame.
For example, as the car travels along the curved surface of the Earth, centrifugal force comes into play due to Earth's rotation.
  • This force acts outward relative to Earth's axis of rotation.
  • Centrifugal force depends on the mass of the object, its radial distance from the center of rotation, and the square of its velocity.
In the context of the exercise, the car's apparent weight changes as it moves east or west because of this force. It effectively alters the gravitational pull perceived by the car.
Angular Velocity
Angular velocity (\(\omega\)) is a measure of how quickly an object rotates or revolves around a circle or rotational path. When applied to Earth's rotation, it defines how fast Earth spins around its axis.
For Earth, the angular velocity is given as 0.7292 脳 10鈦烩伌 radians per second. This measurement is crucial when calculating the impact of rotation on motion along the equator, as the speed contributed by Earth鈥檚 rotation to the car will differ depending on its travel direction.
  • When the car travels eastward, Earth's rotation adds to its linear velocity.
  • In contrast, traveling westward results in Earth鈥檚 rotational speed subtracting from the linear velocity.
Keep in mind that angular velocity is distinct from linear velocity, which is the speed along a path rather than around a circle.
Normal Force
Normal force is the support force exerted upon an object that is in contact with another stable object. For a car on the highway, it is the perpendicular force exerted by the road that prevents the car from crashing into it.
In this exercise, normal force plays a critical role as it varies depending on the effective gravitational force acting on the car. This effective gravity, however, is influenced by Earth's rotation, leading to different apparent weights when traveling in opposite directions.
  • For eastward travel, the centrifugal force reduces the effective gravitational pull, thereby reducing the normal force.
  • Conversely, for westward travel, the effective gravity increase leads to a higher normal force.
These differences in normal force explain the changes in the road's force exertion on the car depending on travel direction.
Earth's Rotation
Earth rotates around its axis, which gives rise to many dynamic effects on objects moving upon its surface, especially along the equator. This rotation introduces additional velocity components to objects in motion. This is particularly relevant when considering vehicles or other moving objects traveling along Earth鈥檚 surface.
  • For eastward movements, the rotation adds to the object's velocity.
  • In contrast, for westward movements, it subtracts from the velocity.
Earth's rotation not only impacts velocity but also affects forces such as gravitational force, altering what is known as the apparent weight. Understanding how rotation impacts this is crucial in solving problems related to dynamics that involve rotational frames.

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Most popular questions from this chapter

The car is moving with a speed \(v_{0}=65 \mathrm{mi} / \mathrm{hr}\) up the 6 -percent grade, and the driver applies the brakes at point \(A,\) causing all wheels to skid. The coefficient of kinetic friction for the rain-slicked road is \(\mu_{k}=0.60 .\) Determine the stopping distance \(s_{A B} .\) Repeat your calculations for the case when the car is moving downhill from \(B\) to \(A\).

A small collar of mass \(m\) is given an initial velocity of magnitude \(v_{0}\) on the horizontal circular track fabricated from a slender rod. If the coefficient of kinetic friction is \(\mu_{k},\) determine the distance traveled before the collar comes to rest. (Hint: Recognize that the friction force depends on the net normal force.)

The particle of mass \(m=1.2 \mathrm{kg}\) is attached to the end of the light rigid bar of length \(L=0.6 \mathrm{m} .\) The system is released from rest while in the horizontal position shown, at which the torsional spring is undeflected. The bar is then observed to rotate \(30^{\circ}\) before stopping momentarily. \((a)\) Determine the value of the torsional spring constant \(k_{T} .(b)\) For this value of \(k_{T},\) determine the speed \(v\) of the particle when \(\theta=15^{\circ}\).

Two 425,000 -lb locomotives pull fifty \(200,000-1 b\) coal hoppers. The train starts from rest and accelerates uniformly to a speed of \(40 \mathrm{mi} / \mathrm{hr}\) over a distance of \(8000 \mathrm{ft}\) on a level track. The constant rolling resistance of each car is 0.005 times its weight. Neglect all other retarding forces and assume that each locomotive contributes equally to the tractive force. Determine \((a)\) the tractive force exerted by each locomotive at \(20 \mathrm{mi} / \mathrm{hr}\) (b) the power required from each locomotive at \(20 \mathrm{mi} / \mathrm{hr},(c)\) the power required from each locomotive as the train speed approaches \(40 \mathrm{mi} / \mathrm{hr},\) and \((d)\) the power required from each locomotive if the train cruises at a steady \(40 \mathrm{mi} / \mathrm{hr}\).

The pilot of a 90,000 -lb airplane which is originally flying horizontally at a speed of \(400 \mathrm{mi} / \mathrm{hr}\) cuts off all engine power and enters a \(5^{\circ}\) glide path as shown. After 120 seconds the airspeed is \(360 \mathrm{mi} / \mathrm{hr} .\) Calculate the time-average drag force \(D(\) air resistance to motion along the flight path)
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