Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 15

A test car is traveling along a level road at \(88 \mathrm{km} / \mathrm{hr}\). In order to study the acceleration characteristics of a newly installed engine, the car accelerates at its maximum possible rate. The test crew records the following velocities at various locations along the level road: $$\begin{array}{lll} x=0 & V=88.5 \mathrm{km} / \mathrm{hr} \\ x=0.1 \mathrm{km} & V=93.1 \mathrm{km} / \mathrm{hr} \\ x=0.2 \mathrm{km} & V=98.3 \mathrm{km} / \mathrm{hr} \\ x=0.3 \mathrm{km} & V=104.0 \mathrm{km} / \mathrm{hr} \\ x=0.4 \mathrm{km} & V=110.3 \mathrm{km} / \mathrm{hr} \\ x=0.5 \mathrm{km} & V=117.2 \mathrm{km} / \mathrm{hr} \\ x=1.0 \mathrm{km} & V=164.5 \mathrm{km} / \mathrm{hr} \end{array}$$ A preliminary study shows that these data follow an equation of the form \(V=A\left(1+e^{B x}\right),\) where \(A\) and \(B\) are positive constants. Find \(A\) and \(B\) and a Lagrangian expression \(V=V\left(V_{0}, t\right),\) where \(V_{0}\) is the car velocity at time \(t=0\) when \(x=0\)

Short Answer

Expert verified
The constants A and B are calculated by substituting two values of x and V into the given equation and solving it simultaneously. The Lagrangian expression is : \(\frac{1}{2} m v^2 - A(1+e^{Bx})\).

Step by step solution

01

Calculating Constants A and B

Using the given equation \(V=A\left(1+e^{Bx}\right)\), we can substitute the data for the known values of V and x to obtain an equation for A and B. For instance, using x=0.1, V=93.1 and x=0.5, V=117.2, we will end up with 2 equations with 2 variables A and B. Solving these equations simultaneously, we get A and B.
02

Using the Lagrangian

Lagrangian mechanics uses its principle of first varying the action. Here we can treat \(V=A\left(1+e^{Bx}\right)\) as a system with potential energy V(x) and use the Lagrange's equation or Hamilton's equations.
03

Writing the Lagrangian formula

The Lagrangian expression \(L = T - V\) where T is the kinetic energy and V is the potential energy. For the given system, we have:\n\(T = \frac{1}{2} m v^2\) and \(V = A(1+e^{Bx})\)\nSo, the Lagrangian is given by:\n\(L = \frac{1}{2} m v^2 - A(1+e^{Bx})\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For any steady flow the streamlines and streaklines are the same. For most unsteady flows this is not true. However, there are unsteady flows for which the streamlines and streaklines are the same. Describe a flow field for which this is true.

As is indicated in Fig. \(P 4.42,\) the speed of exhaust in a car's cxhaust pipe varies in time and distance because of the periodic nature of the engine's operation and the damping effect with distance from the engine. Assume that the speed is given \\[ \text { by } \quad V=V_{0}\left[1+a e^{-b x} \sin (\omega t)\right], \quad \text { where } \quad V_{0}=8 \text { fps, } a=0.05 \\] \(b=0.2 \mathrm{ft}^{-1},\) and \(\omega=50 \mathrm{rad} / \mathrm{s} .\) Calculate and plot the fluid acceleration at \(x=0,1,2,3,4,\) and 5 ft for \(0 \leq t \leq \pi / 25\) s.

As a valve is opened, water flows through the diffuser shown in Fig. \(\mathrm{P} 4.33\) at an increasing flowrate so that the velocity along the centerline is given by \(\mathbf{V}=u \hat{\mathbf{i}}=V_{0}\left(1-e^{-\alpha}\right)(1-x / \ell) \hat{\mathbf{i}},\) where \(u_{0}, c,\) and \(\ell\) are constants. Determine the acceleration as a function of \(x\) and \(t .\) If \(V_{0}=10 \mathrm{ft} / \mathrm{s}\) and \(\ell=5 \mathrm{ft},\) what value of \(c\) (other than \(c=0\) ) is needed to make the acceleration zero for any \(x\) at \(t=1 \mathrm{s} ?\) Explain how the acceleration can be zero if the flowrate is increasing with time.

Assume the temperature of the exhaust in an exhaust pipe can be approximated by \(T=T_{0}\left(1+a e^{-b x}\right)[1+c \cos (\omega t)],\) where \\[ T_{0}=100^{\circ} \mathrm{C}, a=3, b=0.03 \mathrm{m}^{-1}, c=0.05, \text { and } \omega=100 \mathrm{rad} / \mathrm{s} \\] If the exhaust speed is a constant \(3 \mathrm{m} / \mathrm{s}\), determine the time rate of change of temperature of the fluid particles at \(x=0\) and \(x=4 \mathrm{m}\) when \(t=0\)

From calculus, one obtains the following formula (Leibnitz rule) for the time derivative of an integral that contains time in both the integrand and the limits of the integration: \\[ \frac{d}{d t} \int_{x_{1}(t)}^{x_{2}(t)} f(x, t) d x=\int_{x_{1}}^{x_{2}} \frac{\partial f}{\partial t} d x+f\left(x_{2}, t\right) \frac{d x_{2}}{d t}-f\left(x_{1}, t\right) \frac{d x_{1}}{d t} \\] Discuss how this formula is related to the time derivative of the total amount of a property in a system and to the Reynolds transport theorem.
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Wave Optics

Read Explanation

Kinematics Physics

Read Explanation

Space Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Geometrical and Physical Optics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free