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Chapter 11: Problem 139

A monorail train starts from rest on a curve of radius \(400 \mathrm{m}\) and accelerates at the constant rate \(a_{t}\). If the maximum total acceleration of the train must not exceed \(1.5 \mathrm{m} / \mathrm{s}^{2}\), determine \((a)\) the shortest distance in which the train can reach a speed of \(72 \mathrm{km} / \mathrm{h},(b)\) the corresponding constant rate of acceleration \(a_{t} .\)

Short Answer

Expert verified
(a) The shortest distance is approximately 179 m. (b) The tangential acceleration is 1.118 m/s虏.

Step by step solution

01

Convert Speed to SI Units

First, convert the speed from km/h to m/s. The conversion factor is \(1 \ \text{km/h} = \frac{1}{3.6} \ \text{m/s} \). So, \(72 \ \text{km/h} = \frac{72}{3.6} \ \text{m/s} = 20 \ \text{m/s} \).
02

Relate Total Acceleration to Tangential and Radial Components

The total acceleration \(a\) is composed of both tangential acceleration \(a_t\) and radial (or centripetal) acceleration \(a_r\). The relationship is given by \(a = \sqrt{a_t^2 + a_r^2}\). Given \(a = 1.5\ \text{m/s}^2\).
03

Calculate Radial Acceleration

Radial acceleration is given by \(a_r = \frac{v^2}{r}\), where \(v = 20\ \text{m/s}\) and \(r = 400\ \text{m}\). Substitute the values: \(a_r = \frac{20^2}{400} = 1.0 \ \text{m/s}^2\).
04

Solve for Tangential Acceleration

Using the equation from Step 2, solve for \(a_t\): \(1.5 = \sqrt{a_t^2 + 1.0^2}\). Rearrange and solve for \(a_t\):\[a_t = \sqrt{1.5^2 - 1.0^2} = \sqrt{2.25 - 1.0} = \sqrt{1.25} = 1.118 \ \text{m/s}^2\].
05

Apply the Kinematic Equation for Distance

Use the kinematic equation \(v^2 = u^2 + 2a_t s\), where \(u = 0\ \text{m/s}\) (train starts from rest), \(v = 20\ \text{m/s}\), and solve for \(s\) (distance):\[20^2 = 0 + 2(1.118)s\]\[400 = 2.236s\]\[s = \frac{400}{2.236} \approx 179 \ \text{m}\].

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Key Concepts

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

Tangential Acceleration
Tangential acceleration, usually denoted as \( a_t \), is the rate of change of velocity along the path of motion of an object. In simple terms, it is responsible for speeding up or slowing down the object in the direction of its path. For the monorail train in our exercise, tangential acceleration helps the train increase its speed along the track.

To calculate the tangential acceleration when maximum total acceleration is given, we use the formula:
  • Total acceleration \( a = \sqrt{a_t^2 + a_r^2} \)
  • Radial acceleration \( a_r = 1.0 \ \text{m/s}^2 \) (from our calculations)
  • Maximum acceleration \( a = 1.5 \ \text{m/s}^2 \)
We rearrange the formula to solve for \( a_t \) and obtain \( a_t = \sqrt{1.5^2 - 1.0^2} = 1.118 \ \text{m/s}^2 \). This value tells us how quickly the monorail can increase its speed without exceeding the total allowable acceleration.
Radial Acceleration
Radial acceleration, also known as centripetal acceleration, is directed towards the center of the circular path that an object follows. It keeps the object moving in a curve rather than off in a straight line. For circular motion, the radial acceleration can be calculated using the formula:
  • \( a_r = \frac{v^2}{r} \)
  • \( v \) is the velocity, \( r \) is the radius of the curvature
In our monorail example, we found the radial acceleration by substituting \( v = 20 \ \text{m/s} \) and \( r = 400 \ \text{m} \) into the equation, resulting in \( a_r = 1.0 \ \text{m/s}^2 \). This value represents how strongly the monorail must overcome inertia to stay on the curved track.
Kinematic Equations
Kinematic equations are essential tools in understanding the motion of objects. They relate velocity, acceleration, and displacement over time, usually without considering forces. In this exercise, we used one of the kinematic equations to find the shortest distance the monorail travels while reaching a particular speed:
  • \( v^2 = u^2 + 2a_t s \)
  • Where \( u \) is the initial velocity, \( v \) is the final velocity, \( a_t \) is the tangential acceleration, and \( s \) is the distance
Given \( u = 0 \ \text{m/s} \), \( v = 20 \ \text{m/s} \), and \( a_t = 1.118 \ \text{m/s}^2 \), we solved for \( s \) to obtain approximately \( 179 \ \text{m} \). This calculation shows how these relationships help determine distances and elapsed motion given acceleration and initial states.
SI Unit Conversion
SI Unit conversion is a crucial skill in physics and engineering. To ensure consistency and accuracy, we convert various measurements into the International System of Units (SI) standard.

In this problem, we converted speed from kilometers per hour (km/h) to meters per second (m/s), an essential step in working with kinematic equations. The conversion factor is:
  • \( 1 \ \text{km/h} = \frac{1}{3.6} \ \text{m/s} \)
Thus, \( 72 \ \text{km/h} = \frac{72}{3.6} = 20 \ \text{m/s} \). Understanding and applying these conversions are vital steps before using any equations or performing calculations, ensuring that our units align correctly throughout our calculations.

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Most popular questions from this chapter

Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit \(100 \mathrm{mi}\) above the surface of the planet. (See information given in Probs. \(11.153-11.154 .)\) $$ \begin{array}{l}{11.155 \text { Venus: } g=29.20 \mathrm{ft} / \mathrm{s}^{2}, R=3761 \mathrm{mi}} \\ {11.156 \mathrm{Mars}: g=12.17 \mathrm{ft} / \mathrm{s}^{2}, R=2102 \mathrm{mi}} \\ {11.157 \text { Jupiter: } g=75.35 \mathrm{ft} / \mathrm{s}^{2}, R=44,432 \mathrm{mi}}\end{array} $$

The motion of a particle is defined by the relation \(x=2 t^{3}-9 t^{2}+\) \(12 t+10,\) where \(x\) and \(t\) are expressed in feet and seconds, respectively. Determine the time, the position, and the acceleration of the particle when \(v=0\)

A race car enters the circular portion of a track that has a radius of \(70 \mathrm{m}\). When the car enters the curve at point \(P,\) it is travelling with a speed of \(120 \mathrm{km} / \mathrm{h}\) that is increasing at \(5 \mathrm{m} / \mathrm{s}^{2} .\) Three seconds later, determine the \(x\) and \(y\) components of velocity and acceleration of the car.

Instruments in airplane \(A\) indicate that, with respect to the air, the plane is headed \(30^{\circ}\) north of east with an air speed of \(300 \mathrm{mi} / \mathrm{h}\). At the same time, radar on ship \(B\) indicates that the relative velocity of the plane with respect to the ship is \(280 \mathrm{mi} / \mathrm{h}\) in the direction \(33^{\circ}\) north of east. Knowing that the ship is steaming due south at \(12 \mathrm{mi} / \mathrm{h}\), determine ( \(a\) ) the velocity of the airplane, \((b)\) the wind speed and direction.

A brass (nonmagnetic) block \(A\) and a steel magnet \(B\) are in equilibrium in a brass tube under the magnetic repelling force of another steel magnet \(C\) located at a distance \(x=0.004 \mathrm{m}\) from \(B\). The force is inversely proportional to the square of the distance between \(B\) and \(C .\) If block \(A\) is suddenly removed, the acceleration of block \(B\) is \(a=-9.81+k / x^{2},\) where \(a\) and \(x\) are expressed in \(\mathrm{m} / \mathrm{s}^{2}\) and meters, respectively, and \(k=4 \times 10^{-4} \mathrm{m}^{3} / \mathrm{s}^{2} .\) Determine the maximum velocity and acceleration of \(B .\)
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