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For a system of \(N\) particles subject to a uniform gravitational field g acting vertically down, prove that the total gravitational potential energy is the same as if all the mass were concentrated at the center of mass of the system; that is, \(U=\sum_{\alpha} U_{\alpha}=M g Y\) where \(M=\sum m_{\alpha}\) is the total mass and \(\mathbf{R}=(X, Y, Z)\) is the position of the \(\mathrm{CM},\) with the \(y\) coordinate measured vertically up. [Hint: We know from Problem 4.5 that \(\left.U_{\alpha}=m_{\alpha} g y_{\alpha} .\right]\)

Short Answer

Expert verified
The total gravitational potential energy is \( U = M g Y \), as if all mass were at the center of mass.

Step by step solution

01

Define Potential Energy

The potential energy of a single particle in a gravitational field is given by \( U_{\alpha} = m_{\alpha} g y_{\alpha} \), where \( m_{\alpha} \) is the mass of the particle and \( y_{\alpha} \) is its vertical position.
02

Express Total Potential Energy

The total gravitational potential energy for the system is the sum of the potential energies of all particles: \( U = \sum_{\alpha} U_{\alpha} = \sum_{\alpha} m_{\alpha} g y_{\alpha} \).
03

Define Center of Mass

The vertical coordinate of the center of mass is defined as \( Y = \frac{\sum m_{\alpha} y_{\alpha}}{M} \), where \( M = \sum m_{\alpha} \) is the total mass of the system.
04

Substitute Center of Mass

Multiply both sides of the expression for \( Y \) by \( M \) to get: \( M Y = \sum m_{\alpha} y_{\alpha} \).
05

Relate Total Potential Energy to Center of Mass

Substitute \( \sum m_{\alpha} y_{\alpha} = M Y \) into the expression for total potential energy: \( U = g \sum m_{\alpha} y_{\alpha} = g M Y \).
06

Conclude

This expression \( U = M g Y \) shows that the total potential energy is the same as if all the mass were concentrated at the center of mass at height \( Y \).

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Key Concepts

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

Center of Mass
The center of mass is a crucial concept in physics and plays a pivotal role in understanding gravitational systems. Essentially, the center of mass is a point where the total mass of a system appears to be concentrated. For any system of particles, we use the center of mass to simplify complex calculations and make predictions about motion.
The vertical coordinate of the center of mass, denoted as \( Y \), is calculated using the formula: - \( Y = \frac{\sum m_{\alpha} y_{\alpha}}{M} \).
This means that the center of mass accounts for all of the individual masses \( m_\alpha \) and their respective positions \( y_\alpha \) in the system. The sum of these multiplied terms is then divided by the total mass \( M \).
In practical terms, when observing an object like a seesaw, the point at which it balances perfectly on its pivot is where its center of mass is located. This principle is applied to any system of particles to simplify the calculation of its dynamics under gravitational forces.
Uniform Gravitational Field
In physics, a uniform gravitational field is a theoretical region of space where a gravitational field has the same force acting everywhere. This means that any object within this field experiences the same gravitational pull regardless of its position.
The gravitational field strength is denoted by \( g \), commonly recognized as the acceleration due to gravity on Earth's surface, approximately \( 9.8 \, \text{m/s}^2 \). This constant force is crucial for calculating gravitational potential energy.
When dealing with systems of particles, assuming a uniform gravitational field simplifies the problem significantly. It allows us to calculate potential energy using the formula: - \( U_{\alpha} = m_\alpha g y_\alpha \). This means each particle's potential energy depends solely on its mass \( m_\alpha \), the gravitational field strength \( g \), and its vertical position \( y_\alpha \).
Understanding uniform gravitational fields helps us model real-world phenomena and develop insights into how objects interact with gravitational forces in a straightforward and effective manner.
Total Mass
Total mass is a fundamental concept that refers to the sum of all individual masses in a given system. It serves as a single value that represents the entire mass content of the system. In the context of gravitational potential energy, the total mass plays a significant role.
The total mass \( M \) is calculated using:- \( M = \sum m_{\alpha} \), where each \( m_\alpha \) represents the mass of individual particles within the system.
This collective mass is used to simplify calculations by allowing us to treat the entire system as if all mass were concentrated at a single point, specifically the center of mass. This simplifies the expression for gravitational potential energy to \( U = MgY \), where \( Y \) is the vertical position of the center of mass.
Recognizing the importance of total mass helps us understand complex interactions in multi-particle systems. It lets us analyze system behavior as a whole, rather than focusing on individual particles separately, making it easier to predict and explain physical phenomena efficiently.

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Most popular questions from this chapter

(a) The force exerted by a one-dimensional spring, fixed at one end, is \(F=-k x,\) where \(x\) is the displacement of the other end from its equilibrium position. Assuming that this force is conservative (which it is) show that the corresponding potential energy is \(U=\frac{1}{2} k x^{2},\) if we choose \(U\) to be zero at the equilibrium position. (b) Suppose that this spring is hung vertically from the ceiling with a mass \(m\) suspended from the other end and constrained to move in the vertical direction only. Find the extension \(x_{\mathrm{o}}\) of the new equilibrium position with the suspended mass. Show that the total potential energy (spring plus gravity) has the same form \(\frac{1}{2} k y^{2}\) if we use the coordinate \(y\) equal to the displacement measured from the new equilibrium position at \(x=x_{\mathrm{o}}\) (and redefine our reference point so that \(U=0\) at \(y=0\) ).

A mass \(m\) is in a uniform gravitational field, which exerts the usual force \(F=m g\) vertically down, but with \(g\) varying with time, \(g=g(t) .\) Choosing axes with \(y\) measured vertically up and defining \(U=m g y\) as usual, show that \(\mathbf{F}=-\nabla U\) as usual, but, by differentiating \(E=\frac{1}{2} m v^{2}+U\) with respect to \(t,\) show that \(E\) is not conserved.

Both the Coulomb and gravitational forces lead to potential energies of the form \(U=\gamma / | \mathbf{r}_{1}-\) \(\mathbf{r}_{2} |,\) where \(\gamma\) denotes \(k q_{1} q_{2}\) in the case of the Coulomb force and \(-G m_{1} m_{2}\) for gravity, and \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\) are the positions of the two particles. Show in detail that \(-\nabla_{1} U\) is the force on particle 1 and \(-\nabla_{2} U\) that on particle 2.

Consider the Atwood machine of Figure \(4.15,\) but suppose that the pulley has radius \(R\) and moment of inertia \(I\). (a) Write down the total energy of the two masses and the pulley in terms of the coordinate \(x\) and \(\dot{x}\). (Remember that the kinetic energy of a spinning wheel is \(\frac{1}{2} I \omega^{2}\).) (b) Show (what is true for any conservative one-dimensional system) that you can obtain the equation of motion for the coordinate \(x\) by differentiating the equation \(E=\) const. Check that the equation of motion is the same as you would obtain by applying Newton's second law separately to the two masses and the pulley, and then eliminating the two unknown tensions from the three resulting equations.

(a) Consider a mass \(m\) in a uniform gravitational field \(\mathbf{g},\) so that the force on \(m\) is \(m \mathbf{g},\) where \(\mathbf{g}\) is a constant vector pointing vertically down. If the mass moves by an arbitrary path from point 1 to point \(2,\) show that the work done by gravity is \(W_{\mathrm{grav}}(1 \rightarrow 2)=-m g h\) where \(h\) is the vertical height gained between points 1 and 2. Use this result to prove that the force of gravity is conservative (at least in a region small enough so that \(\mathrm{g}\) can be considered constant). (b) Show that, if we choose axes with \(y\) measured vertically up, the gravitational potential energy is \(U=m g y\) (if we choose \(U=0\) at the origin).

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