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Chapter 12: Problem 21

The potential energy associated with a particle at position \(x\) is given by \(U=2 x^{3}-2 x^{2}-7 x+10\), with \(x\) in meters and \(U\) in joules. Find the positions of any stable and unstable equilibria.

Short Answer

Expert verified
The actual x-values found via the quadratic formula must be inserted into the short answer, as well as whether those x-values indicate a stable or unstable equilibrium.

Step by step solution

01

Calculate First Derivative

Begin by calculating the first derivative of the potential energy. The derivative uses the power rule, where the derivative of a term \(a x^{n}\) is \(a n x^{n-1}\). So the first derivative \(U'\) of \(U=2x^{3}-2x^{2}-7x+10\) is \(U'=6x^{2}-4x-7\).
02

Solve for Equilibrium

To find the equilibrium points, set the first derivative equal to zero and solve for \(x\), i.e.,\(6x^{2}-4x-7=0\). This is a quadratic equation in the form \(ax^{2} + bx + c = 0\) and can be solved using the quadratic formula \(x = [-b 卤 sqrt(b^2 -4ac)]/(2a)\).
03

Check Stability of Equilibrium

To determine whether an equilibrium is stable or unstable, compute the second derivative \(U''\) (once again applying the power rule). In this case, \(U'' = 12x-4\). Evaluate \(U''\) at the x-values found in Step 2. If \(U''>0\), then the point is a stable equilibrium. If \(U''<0\) then the point is an unstable equilibrium.
04

Substitute Values in the Stability Check

Plug the obtained equilibrium values, \(x\), into the second derivative and check their sign. The outcome will reveal whether the equilibrium is stable or unstable.

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Key Concepts

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

Understanding Equilibrium
In physics, equilibrium refers to a state where a particle is at rest or moving with constant velocity because the net force acting on it is zero.
This means that the potential energy of the system is neither increasing nor decreasing; often, it's at a local maximum or minimum.
In mathematical terms, equilibrium points occur where the first derivative of the potential energy function equals zero.
These are the points where the slope of the tangent line to the curve of the potential energy function is zero.
The Role of the First Derivative
The first derivative of a function gives us the slope of the tangent to the curve at any given point.
For potential energy, the first derivative helps us find equilibrium by setting it to zero and solving for x.
This is because the equilibrium occurs where there is no change in the potential energy; hence, the derivative is zero.
  • Find the derivative using power rule: \(6x^{2} - 4x - 7\).
  • Set the derivative to zero: \(6x^{2} - 4x - 7 = 0\).
Exploring the Second Derivative
The second derivative provides us with information about the concavity of the graph of the function.
It is calculated by differentiating the first derivative.
  • If the second derivative is positive at an equilibrium point, the graph is concave up, indicating a stable equilibrium.
  • If negative, the graph is concave down, suggesting an unstable equilibrium.
In this exercise, the second derivative is \(12x - 4\). Evaluating this at each equilibrium point will determine their stability.
Conducting Stability Analysis
Stability analysis is crucial for understanding whether an equilibrium point will hold unless disturbed.
This involves substituting the x-values of equilibrium points into the second derivative.For our equation:
  • If \(U'' > 0\) after substitution, any small disturbance will return the system to equilibrium, making it stable.
  • If \(U'' < 0\), any slight displacement will cause the system to move away from equilibrium, making it unstable.
By evaluating the second derivative at the x-values found, you can ascertain the nature of the equilibrium.

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Most popular questions from this chapter

You're investigating ladder safety for the Consumer Product Safety Commission. Your test case is a uniform ladder of mass \(m\) leaning against a frictionless vertical wall with which it makes an angle \(\theta\). The coefficient of static friction at the floor is \(\mu\). Your job is to find an expression for the maximum mass of a person who can climb to the top of the ladder without its slipping. With that result, you're to show that anyone can climb to the top if \(\mu \geq \tan \theta\) but that no one can if \(\mu<\frac{1}{2} \tan \theta\).

A uniform, solid cube of mass \(m\) and side \(s\) is in stable equilibrium when sitting on a level tabletop. How much energy is required to bring it to an unstable equilibrium where it's resting on its corner?

Figure \(12.31\) shows a popular system for mounting bookshelves. An aluminum bracket is mounted on a vertical aluminum support by small tabs inserted into vertical slots. Contact between the bracket and support occurs only at the upper tab and at the bottom of the bracket, \(4.5 \mathrm{~cm}\) below the upper tab. If each bracket in the shelf system supports \(35 \mathrm{~kg}\) of books, with the center of gravity \(12 \mathrm{~cm}\) out from the vertical support, what is the horizontal component of the force exerted on the upper bracket tab?

The potential energy as a function of position for a particle is given by $$ U(x)=U_{0}\left(\frac{x^{3}}{x_{0}{ }^{3}}+a \frac{x^{2}}{x_{0}{ }^{2}}+4 \frac{x}{x_{0}}\right) $$ where \(x_{0}\) and \(a\) are constants. For what values of \(a\) will there be two static equilibria? Comment on the stability of these equilibria.

If you secure the boom at a fixed angle and lower the sampling apparatus at constant speed, the boom rope tension will a. increase. b. decrease. c. remain the same. d. increase only if the sampling apparatus is more massive than the boom.
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