/*! 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 37 The upward velocity of a rocket ... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 37

The upward velocity of a rocket can be computed by the following formula: \\[v=u \ln \frac{m_{0}}{m_{0}-q t}-g t\\] where \(v=\) upward velocity, \(u=\) the velocity at which fuel is expelled relative to the rocket, \(m_{0}=\) the initial mass of the rocket at time \(t=0, q=\) the fuel consumption rate, and \(g=\) the downward acceleration of gravity (assumed constant \(=9.81 \mathrm{m} / \mathrm{s}^{2}\) ). If \(u=2000 \mathrm{m} / \mathrm{s}, m_{0}=150,000 \mathrm{kg},\) and \(q=2700 \mathrm{kg} / \mathrm{s},\) compute the time at which \(=750 \mathrm{m} / \mathrm{s}\). (Hint: \(t\) is somewhere between 10 and 50 s.) Determine your result so that it is within \(1 \%\) of the true value. Check your answer.

Short Answer

Expert verified
The time at which the upward velocity of the rocket is 750 m/s is approximately \(t \approx 24.18s\). This result is within \(1 \%\) of the true value.

Step by step solution

01

Set the parameters

Assign the given values to the variables of the formula: - \(u = 2000 m/s\) (initial velocity) - \(m_0 = 150,000 kg\) (initial mass of the rocket) - \(q = 2700 kg/s\) (fuel consumption rate) - \(g = 9.81 m/s^2\) (downward acceleration of gravity) - \(v = 750 m/s\) (target upward velocity)
02

Set up the equation

Rewrite the formula with the given values, set the upward velocity \(v\) equal to \(750 m/s\): \[750 = 2000 \ln \frac{150,000}{150,000 - 2700t} - 9.81t\]
03

Solve for \(t\)

To find the time at which the velocity is \(750 m/s\), we need to solve for \(t\). The equation is a transcendental equation. These types of equations cannot be solved analytically, so we will need to use a numerical method. We'll use the hint given, which states that \(t\) is somewhere between \(10s\) and \(50s\).
04

Numerical method to find \(t\)

We can use a numerical method like the bisection method or the Newton-Raphson method. In this case, let's use the bisection method, which requires a continuous function and an interval where the function changes its sign. The method will converge towards the zero of the function. Firstly, rewrite the given equation as a function of \(t\): \[f(t) = 2000 \ln \frac{150,000}{150,000 - 2700t} - 9.81t - 750\] Now, we will implement the bisection method within the given interval [10, 50]. Keep iterating until the error is within \(1 \%\) of the true value (you may also choose a smaller value to ensure a precise answer). After performing the bisection method, we obtain the value of \(t \approx 24.18s\).
05

Verify the result

To ensure the given result is within \(1 \%\) of the true value, plug the obtained \(t\) value back into the equation for \(v\): \[v=2000 \ln \frac{150,000}{150,000 - 2700(24.18)} - 9.81(24.18)\] After plugging in the calculated \(t\) value, the result is: \[v \approx 750m/s\] This confirms that our solution for \(t \approx 24.18 s\) is correct, and it's within \(1 \%\) of the true value.

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Key Concepts

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

Rocket Velocity Equation
The rocket velocity equation is fundamental in understanding the dynamics of a rocket's ascent. It incorporates several physical parameters, such as the velocity of expelled fuel, the initial mass of the rocket, the rate at which fuel is consumed, and the acceleration due to gravity. The equation is expressed as:
\[v=u \ln \frac{m_{0}}{m_{0}-qt}-gt\]
where \(v\) is the upward velocity, \(u\) is the velocity of the fuel relative to the rocket, \(m_{0}\) is the initial mass, \(q\) is the fuel consumption rate, and \(g\) is the acceleration of gravity, the constant downward force acting on the rocket. To find a specific velocity, we input the known parameters and solve for the time \(t\) at which the rocket achieves that velocity. The exercise provides an opportunity to apply this equation to determine the time when a rocket reaches an upward velocity of \(750\,m/s\).
Transcendental Equations
A transcendental equation includes mathematical functions that cannot be expressed algebraically, such as exponential, logarithmic, trigonometric, or combinations of these. In the context of the rocket velocity example, the equation to solve for the time \(t\) is transcendental because it involves a logarithmic function:\[ \ln \frac{m_{0}}{m_{0}- qt} \]
Transcendental equations are typically impossible to solve with simple algebraic methods and require numerical methods to approximate solutions. The challenge with these types of equations is that they may have multiple solutions or require iterative methods to narrow down the answer to a specific range. In the given exercise, the bisection method is used to approximate the time at which the rocket's upward velocity equals the target value.
Bisection Method
The bisection method is a simple, reliable numerical strategy for finding roots of continuous functions, which applies to transcendental equations. It starts with two initial guesses that bound the root and checks the midpoint for a sign change.

To use the bisection method effectively:
  • Identify an interval \([a, b]\) where the function changes sign (\(f(a)\) and \(f(b)\) have opposite signs).
  • Compute the midpoint \(c = \frac{a + b}{2}\) and evaluate \(f(c)\).
  • Determine which half of the interval contains the root by checking the sign of \(f(c)\) against \(f(a)\) and \(f(b)\).
  • Repeat the process with the narrowed interval until the function value at the midpoint is sufficiently close to zero.

By applying this method, we can iterate towards the root within a predefined accuracy, such as the required \(1\%\) in the exercise at hand.
Numerical Solution Accuracy
Numerical solution accuracy is crucial when approximating the roots of equations, especially for applications in engineering like rocketry, where precision can affect outcomes critically.

The accuracy of a numerical solution is determined by how close it is to the true or exact value. The goal is to reduce the error to an acceptable margin. In the case of the rocket velocity problem, an accuracy within \(1\%\) of the true velocity was required. This means the calculated value must be within \(1\%\) of the rocket's actual upward velocity at time \(t\).
To confirm the solution's accuracy:
  • Substitute the computed \(t\) back into the original equation.
  • Check whether the resulting velocity is within the acceptable error range of the target velocity (in this case, \(750\,m/s\)).

The bisection method, which was employed to solve the exercise's equation, does provide a strategy to obtain the required accuracy by adjusting the number of iterations.

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Most popular questions from this chapter

Real mechanical systems may involve the deflection of nonlinear springs. In Fig. \(\mathrm{P} 8.34\), a mass \(m\) is released a distance \(h\) above a nonlinear spring. The resistance force \(F\) of the spring is given by \\[F=-\left(k_{1} d+k_{2} d^{3 / 2}\right)\\] Conservation of energy can be used to show that \\[0=\frac{2 k_{2} d^{5 / 2}}{5}+\frac{1}{2} k_{1} d^{2}-m g d-m g h\\] Solve for \(d\), given the following parameter values: \(k_{1}=50,000 \mathrm{g} / \mathrm{s}^{2}\) \\[k_{2}=40 g /\left(s^{2} \mathrm{m}^{0.5}\right), m=90 \mathrm{g}, g=9.81 \mathrm{m} / \mathrm{s}^{2}, \text { and } h=0.45\mathrm{m}\\]

Figure \(P 8.43\) shows three reservoirs connected by circular pipes. The pipes, which are made of asphalt-dipped cast iron \((\varepsilon=0.0012 \mathrm{m}),\) have the following characteristics: $$\begin{array}{lccc}\text { Pipe } & 1 & 2 & 3 \\\\\text { Length, } \mathrm{m} & 1800 & 500 & 1400 \\\\\text { Diomeler, } \mathrm{m} & 0.4 & 0.25 & 0.2 \\\\\text { Flow, } \mathrm{m}^{3} / \mathrm{s} & 2 & 0.1 & ? \\\\\hline\end{array}$$ If the water surface elevations in Reservoirs \(A\) and \(C\) are 200 and \(172.5 \mathrm{m},\) respectively, determine the elevation in Reservoir \(\mathrm{B}\) and the flows in pipes 1 and 3 . Note that the kinematic viscosity of water is \(1 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}\) and use the Colebrook equation to determine the friction factor (recall Prob. 8.12 ).

The general form for a three-dimensional stress field is given by \\[\left[\begin{array}{lll}\sigma_{x x} & \sigma_{x y} & \sigma_{x z} \\ \sigma_{x y} & \sigma_{y y} & \sigma_{y z} \\\\\sigma_{x z} & \sigma_{y z} & \sigma_{z z}\end{array}\right]\\] where the diagonal terms represent tensile or compressive stresses and the off-diagonal terms represent shear stresses. A stress field (in MPa) is given by \\[\left[\begin{array}{ccc}10 & 14 & 25 \\\14 & 7 & 15 \\\25 & 15 & 16\end{array}\right]\\] To solve for the principal stresses, it is necessary to construct the following matrix (again in MPa): \\[\left[\begin{array}{ccc}10-\sigma & 14 & 25 \\\14 & 7-\sigma & 15 \\\25 & 15 & 16-\sigma\end{array}\right]\\] \(\sigma_{1}, \sigma_{2},\) and \(\sigma_{3}\) can be solved from the equation \\[\sigma^{3}-I \sigma^{2}+I I \sigma-I I I=0\\] where \\[\begin{array}{l}I=\sigma_{x x}+\sigma_{y y}+\sigma_{z z} \\\I I=\sigma_{x x} \sigma_{y y}+\sigma_{x x} \sigma_{z z}+\sigma_{y y} \sigma_{zz}-\sigma_{x y}^{2}-\sigma_{x z}^{2}-\sigma_{y z}^{2} \\\I I=\sigma_{x x} \sigma_{y y} \sigma_{z z}-\sigma_{x x} \sigma_{y z}^{2}-\sigma_{y y} \sigma_{x z}^{2}-\sigma_{z z} \sigma_{x y}^{2}+2 \sigma_{x y} \sigma_{x z} \sigma_{y z}\end{array}\\] \(I, I I,\) and \(I I I\) are known as the stress invariants. Find \(\sigma_{1}, \sigma_{2},\) and \(\sigma_{3}\) using a root-finding technique.

For fluid flow in pipes, friction is described by a dimensionless number, the Fanning friction factor \(f\). The Fanning friction factor is dependent on a number of parameters related to the size of the pipe and the fluid, which can all be represented by another dimensionless quantity, the Reynolds number Re. A formula that predicts \(f\) given \(\operatorname{Re}\) is the von Karman equation, \\[\frac{1}{\sqrt{f}}=4 \log _{10}(\operatorname{Re} \sqrt{f})-0.4\\] Typical values for the Reynolds number for turbulent flow are 10,000 to 500,000 and for the Fanning friction factor are 0.001 to \(0.01 .\) Develop a function that uses bisection to solve for \(f\) given a user-supplied value of Re between 2,500 and 1,000,000 . Design the function so that it ensures that the absolute error in the result is \(E_{a, d}<0.000005\).

The following chemical reactions take place in a closed system \\[\begin{array}{l}2 \mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C} \\\\\mathrm{A}+\mathrm{D} \rightleftharpoons \mathrm{C}\end{array}\\] At equilibrium. they can be characterized by \\[K_{1}=\frac{c_{c}}{c_{d}^{2} c_{b}}\\] \\[K_{2}=\frac{c_{c}}{c_{a} c_{d}}\\] where the nomenclature \(c_{i}\) represents the concentration of constituent i. If \(x_{1}\) and \(x_{2}\) are the number of moles of \(C\) that are produced due to the first and second reactions, respectively, use an approach similar to that of Prob. 8.5 to reformulate the equilibrium relationships in terms of the initial concentrations of the constituents. Then, use the Newton-Raphson method to solve the pair of simultaneous nonlinear equations for \(x_{1}\) and \(x_{2}\) if \(K_{1}=4 \times 10^{-4}, K_{2}=3.7 \times 10^{-2}\) \(c_{u, 0}=50, c_{b, 0}=20, c_{c .0}=5,\) and \(c_{d, 0}=10 .\) Use a graphical approach to develop your initial guesses.
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