/*! 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 35 Gravity in Two Dimensions. Two p... [FREE SOLUTION] | 91Ó°ÊÓ

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Gravity in Two Dimensions. Two point masses, \(m_{1}\) and \(m_{2},\) lie in the \(x y\) -plane, with \(m_{1}\) held in place at the origin and \(m_{2}\) free to move a distance \(r\) away at a point \(P\) having coordinates \(x\) and \(y\) (Fig. 7.27\()\) . The gravitational potential energy of these masses is found to be \(U(r)=-G m_{1} m_{2} / r,\) where \(G\) is the gravitational constant. (a) Show that the components of the force on \(m_{2}\) due to \(m_{1}\) are $$ F_{x}=-\frac{G m_{1} m_{2} x}{\left(x^{2}+y^{2}\right)^{3 / 2}} \quad \text { and } \quad F_{y}=-\frac{G m_{1} m_{2} y}{\left(x^{2}+y^{2}\right)^{3 / 2}} $$ (Hint: First write \(r\) in terms of \(x\) and \(y . )(\text { b) Show that the magnitude}\) of the force on \(m_{2}\) is \(F=G m_{1} m_{2} / r^{2} .\) (c) Does \(m_{1}\) attract or repel \(m_{2} ?\) How do you know?

Short Answer

Expert verified
(a) Use negative gradient to find components. (b) Resultant forms expected inverse-square law. (c) m1 attracts m2; force is attractive.

Step by step solution

01

Express r in Terms of x and y

Given that the point mass \(m_2\) is located at \((x, y)\), we can use the Pythagorean theorem to express the distance \(r\) between \(m_1\) and \(m_2\). Thus, \(r = \sqrt{x^2 + y^2}\).
02

Write Gravitational Potential Energy in Terms of r

The given gravitational potential energy is \(U(r) = -\frac{G m_1 m_2}{r}\). Substitute \(r = \sqrt{x^2 + y^2}\) into this equation to relate \(U\) with \(x\) and \(y\). The expression does not change form because \(r\) is in the denominator.
03

Derive Force Components from Potential Energy

The force can be found as the negative gradient of potential energy. Calculate the partial derivatives: \(F_x = -\frac{\partial U}{\partial x}\) and \(F_y = -\frac{\partial U}{\partial y}\).
04

Calculate Fx

To find \(F_x\):\(\frac{\partial U}{\partial x} = \frac{\partial}{\partial x}\left(-\frac{G m_1 m_2}{\sqrt{x^2 + y^2}}\right) = \frac{G m_1 m_2 x}{(x^2 + y^2)^{3/2}}\).Thus, \(F_x = -\frac{G m_1 m_2 x}{(x^2 + y^2)^{3/2}}\).
05

Calculate Fy

Similarly, to find \(F_y\):\(\frac{\partial U}{\partial y} = \frac{\partial}{\partial y}\left(-\frac{G m_1 m_2}{\sqrt{x^2 + y^2}}\right) = \frac{G m_1 m_2 y}{(x^2 + y^2)^{3/2}}\).Thus, \(F_y = -\frac{G m_1 m_2 y}{(x^2 + y^2)^{3/2}}\).
06

Calculate the Magnitude of the Force

The magnitude of the force on \(m_2\) is the resultant force derived from the components \(F_x\) and \(F_y\):\[ F = \sqrt{F_x^2 + F_y^2} = \sqrt{\left(-\frac{G m_1 m_2 x}{(x^2 + y^2)^{3/2}}\right)^2 + \left(-\frac{G m_1 m_2 y}{(x^2 + y^2)^{3/2}}\right)^2} \].Simplifying, this results in \(F = \frac{G m_1 m_2}{r^2}\), as expected.
07

Determine Whether m1 Attracts or Repels m2

The force \(F\) is directed towards \(m_1\) as the force expression has a negative sign, indicating attraction. Thus, \(m_1\) attracts \(m_2\).

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Key Concepts

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

Potential Energy
In the realm of gravitational forces, potential energy is a vital concept. It represents the energy stored due to the position of two masses in a gravitational field. For two point masses, such as in our exercise with masses \(m_1\) and \(m_2\), the gravitational potential energy is given by the formula:\[ U(r) = -\frac{G m_1 m_2}{r} \]Here, \(U(r)\) is the potential energy dependent on the distance \(r\) between the masses, and \(G\) is the universal gravitational constant. This equation highlights that when two masses are closer, the potential energy becomes more negative. This concept tells us that increasing the distance between the masses reduces the potential energy, thus indicating that work is needed to separate them further. It is crucial to understand how potential energy links with force. The force can be derived as the negative gradient of potential energy, leading to force components acting on the masses.
Force Components
Force components are segments of the total force acting in specific directions, split into horizontal and vertical parts, typically aligned with the axes of a coordinate system. In two-dimensional problems like ours, these are particularly important.
  • For the \(x\)-component, the force \(F_x\) can be calculated using partial derivatives of the potential energy with respect to \(x\):\[ F_x = -\frac{G m_1 m_2 x}{(x^2 + y^2)^{3/2}} \]
  • Similarly, for the \(y\)-component, the force \(F_y\) is:\[ F_y = -\frac{G m_1 m_2 y}{(x^2 + y^2)^{3/2}} \]
These components explain how \(m_2\) experiences force along both axes due to the gravitational pull from \(m_1\). Calculate these by observing changes in potential energy as they relate to either horizontal or vertical displacement. By finding the magnitude of these components, we can understand the total force acting on the mass in the plane.
Negative Gradient
In physics, the negative gradient of a potential function helps us to determine force. This principle links the change in potential energy to the force experienced by an object in a field.The gradient, in mathematics, provides the rate of change of a function with respect to its variables. For gravitational forces, by taking the negative gradient of the potential energy \(U(r)\), we can find the force components. This is because force is inherently an action of change that counters this potential.
  • The gradient is computed through partial derivatives; hence, \(F_x = -\frac{\partial U}{\partial x}\).
  • This process reveals how energy changes impact force direction and magnitude in a gravitational interaction.
Understanding negative gradients not only helps in finding the force direction (always seeking lower potential energy) but also provides insights into how energy landscapes shape motion in physics problems.
Two-Dimensional Motion
Two-dimensional motion addresses how objects move within a plane. In this exercise, the point mass \(m_2\) moves within the \(xy\)-plane, influenced by the gravitational pull from \(m_1\).
  • The point \((x, y)\) represents \(m_2\)'s position, dictating how gravitational force components come into play.
  • Understanding forces in this bi-planar motion requires breaking down the interplay of gravitational forces — achieved by calculating both \(F_x\) and \(F_y\).
This breakdown into two axes helps to analyze separately how forces affect horizontal and vertical motion. The complexity of two-dimensional motion is simplified by using vectors, which are resultant forces derived from both \(x\) and \(y\) components. Grasping this fully ensures comprehensive insight into how gravitational interactions guide object movements across planes.

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Most popular questions from this chapter

An object has several forces acting on it. One force is \(\overrightarrow{\boldsymbol{F}}=\alpha x \hat{\imath},\) a force in the \(x\) -direction whose magnitude depends on the position of the object. (See Problem \(6.96 . )\) The constant is \(\alpha=2.00 \mathrm{N} / \mathrm{m}^{2} .\) The object moves along the following path: (1) It starts at the origin and moves along the \(y\) -axis to the point \(x=0\) , \(y=1.50 \mathrm{m} ;(2)\) it moves parallel to the \(x\) -axis to the point \(x=1.50 \mathrm{m}, y=1.50 \mathrm{m} ;(3)\) it moves parallel to the \(y\) -axis to the point \(x=1.50 \mathrm{m}, y=0 ;(4)\) it moves parallel to the \(x\) -axis back to the origin. (a) Sketch this path in the \(x y\) -plane. (b) Calculate the work done on the object by \(\overrightarrow{\boldsymbol{F}}\) for each leg of the path and for the complete round trip. (c) Is \(\overrightarrow{\boldsymbol{F}}\) conservative or nonconservative? Explain.

A truck with mass \(m\) has a brake failure while going down an ?cy mountain road of constant downward slope angle \(\alpha\) (Fig. 7.40\()\) Initially the truck is moving downhill at speed \(v_{0}\) - After careening downhill a distance \(L\) with negligible friction, the truck driver steers the runaway vehicle onto a runaway truck ramp of constant upward slope angle \(\boldsymbol{\beta}\) . The truck rump has a soft sand suffice for which the coefficient of rolling friction is \(\mu_{r}\) What is the distance that the truck moves up the rump before coming to a halt? Solve using energy methods.

Pendulum. A small rock with mass 0.12 \(\mathrm{kg}\) is fastened to a massless string with length 0.80 \(\mathrm{m}\) to form a pendulum. The pendulum is swinging so as to make a maximum angle of \(45^{\circ}\) with the vertical. Air resistance is negligible. (a) What is the speed of the rock when the string passes through the vertical position? \((b)\) What is the tension in the string when it makes an angle of \(45^{\circ}\) with the vertical? (c) What is the tension in the string as it passes through the vertical?

A 0.60\(\cdot \mathrm{kg}\) book slides on a horizontal table. The kinetic friction force on the book has magnitude 1.2 \(\mathrm{N}\) . (a) How much work is done on the book by friction during a displacement of 3.0 \(\mathrm{m}\) to the left? (b) The book now slides 3.0 \(\mathrm{m}\) to the right, returning to its starting point. During this second 3.0 \(\mathrm{m}\) displacement, how much work is done on the book by friction? (c) What is the total work done on the book by friction during the complete round trip? (d) On the basis of your answer to part (c), would you say that the friction force is conservative or nonconservative? Explain.

Up and Down the Hill. A \(28-\mathrm{kg}\) rock approaches the foot of a hill with a speed of 15 \(\mathrm{m} / \mathrm{s}\) . This hill slopes upward at a constant angle of \(40.0^{\circ}\) above the horizontal. The coefficients of static and kinetic friction between the hill and the rock are 0.75 and 0.20 , respectively. (a) Use energy conservation to find the maximum height above the foot of the hill reached by the rock. (b) Will the rock remain at rest at its highest point, or will it slide back down the hill? (c) If the rock does slide back down, find its speed when it returns to the bottom of the hill.

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