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Chapter 10: Problem 3

Five equal point masses are placed at the five corners of a square pyramid whose square base is centered on the origin in the \(x y\) plane, with side \(L,\) and whose apex is on the \(z\) axis at a height \(H\) above the origin. Find the CM of the five-mass system.

Short Answer

Expert verified
The center of mass is at \((0, 0, \frac{H}{5})\).

Step by step solution

01

Identify Positions of Masses

The pyramid's base is a square located in the \(xy\)-plane and centered at the origin. The corners of the square are \((\frac{L}{2}, \frac{L}{2}, 0)\), \((-\frac{L}{2}, \frac{L}{2}, 0)\), \((-\frac{L}{2}, -\frac{L}{2}, 0)\), and \((\frac{L}{2}, -\frac{L}{2}, 0)\). The apex is above the origin at \((0, 0, H)\).
02

Calculate Center of Mass for X-Axis

The \(x\)-coordinates for the masses are \(\frac{L}{2}, -\frac{L}{2}, -\frac{L}{2}, \frac{L}{2},\) and \(0\). The total mass is \(5m\). The center of mass in the x-direction is calculated as:\[\bar{x} = \frac{m(\frac{L}{2}) + m(-\frac{L}{2}) + m(-\frac{L}{2}) + m(\frac{L}{2}) + m(0)}{5m} = 0\]
03

Calculate Center of Mass for Y-Axis

The \(y\)-coordinates for the masses are \(\frac{L}{2}, \frac{L}{2}, -\frac{L}{2}, -\frac{L}{2}, 0\). The center of mass in the y-direction is:\[\bar{y} = \frac{m(\frac{L}{2}) + m(\frac{L}{2}) + m(-\frac{L}{2}) + m(-\frac{L}{2}) + m(0)}{5m} = 0\]
04

Calculate Center of Mass for Z-Axis

The \(z\)-coordinates for the masses are \(0, 0, 0, 0, H\). The center of mass in the z-direction is:\[\bar{z} = \frac{m(0) + m(0) + m(0) + m(0) + m(H)}{5m} = \frac{H}{5}\]
05

Compile Results for Center of Mass

The overall center of mass for the system is:\[(\bar{x}, \bar{y}, \bar{z}) = (0, 0, \frac{H}{5})\].

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Key Concepts

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

Coordinate Geometry
Coordinate geometry helps us map different geometrical figures and understand their positions in space using a coordinate system. In this exercise, we deal with a square pyramid based on the Cartesian coordinate system.
The coordinates are important for pinpointing where each mass lies. The square base is on the xy-plane, represented by the coordinates
  • \((\frac{L}{2}, \frac{L}{2}, 0)\)
  • \((-\frac{L}{2}, \frac{L}{2}, 0)\)
  • \((-\frac{L}{2}, -\frac{L}{2}, 0)\)
  • \((\frac{L}{2}, -\frac{L}{2}, 0)\)
All four points are equidistant from the center, which is the origin (0, 0, 0).
The apex, which is above the base, lies on the z-axis at \(0, 0, H\). Placing these points in the coordinate system allows us to understand and visualize how the mass is distributed across the structure.
Mass Distribution
Understanding how mass is distributed is crucial in analyzing the center of mass. In our pyramid, all five point masses are equal, which simplifies calculations.
The masses at the four corners of the base contribute equally in the xy-plane, which is why the center of mass in both the x-direction and y-direction is zero. The apex mass lies above the origin, affecting the center of mass in the z-direction.
In essence,
  • The equal mass at each vertex simplifies the symmetry.
  • All points on the base contribute to a balanced mass distribution around the origin.
  • The apex鈥檚 mass distinguishes the system鈥檚 center of mass more in the vertical (z) direction.
This understanding underscores how different points in a structure affect the overall position of the center of mass, especially when symmetry and equal mass distribution are involved.
Pyramid Structure
A pyramid is a fascinating geometrical shape characterized by its polygonal base and triangular faces that converge at a single point, the apex. In this particular problem, the base is a square where equal point masses are placed.
The apex is vertically above the center of the base, which introduces a unique point compared to the planar distribution of the base. The structure:
  • Has a central apex above the origin making it three-dimensional.
  • Forms a balance between the base and apex, contributing to how the center of mass is calculated.
  • Helps visualize the balance of mass in relation to height, showing how it shifts based on the structure not being flat but rather elevated.
Understanding the structure is essential in predicting how mass and height affect the center of mass in three-dimensional figures.
Symmetry in Physics
Symmetry plays a pivotal role in physics, simplifying complex problems using uniformity and predictability principles. In our pyramid problem, symmetry is quite evident, especially given the equal distribution of masses at symmetric points.
This symmetry:
  • Means that masses on the base counterbalance each other's x and y coordinates, resulting in a center of mass at the origin for these dimensions.
  • The symmetry helps simplify calculations because the predictable distribution negates any influence in lateral directions, thanks to canceling effects.
  • The symmetry in mass distribution aids in focusing on the z-axis for the center of mass shift.
By understanding symmetry, not only does the calculation become straightforward, but it also allows us to predict the system behavior, even before doing the math.

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Most popular questions from this chapter

A thin rod (of width zero, but not necessarily uniform) is pivoted freely at one end about the horizontal \(z\) axis, being free to swing in the \(x y\) plane ( \(x\) horizontal, \(y\) vertically down). Its mass is \(m,\) its \(\mathrm{CM}\) is a distance \(a\) from the pivot, and its moment of inertia (about the \(z\) axis) is \(I\). (a) Write down the equation of motion \(\dot{L}_{z}=\Gamma_{z}\) and, assuming the motion is confined to small angles (measured from the downward vertical), find the period of this compound pendulum. ("Compound pendulum" is traditionally used to mean any pendulum whose mass is distributed \(-\) as contrasted with a "simple pendulum," whose mass is concentrated at a single point on a massless arm.) (b) What is the length of the "equivalent" simple pendulum, that is, the simple pendulum with the same period?

Suppose that you have found three independent principal axes (directions \(\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}\) ) and corresponding principal moments \(\lambda_{1}, \lambda_{2}, \lambda_{3}\) of a rigid body whose moment of inertia tensor \(\mathbf{I}\) (not diagonal) you had calculated. (You may assume, what is actually fairly easy to prove, that all of the quantities concerned are real.) (a) Prove that if \(\lambda_{i} \neq \lambda_{j}\) then it is automatically the case that \(\mathbf{e}_{i} \cdot \mathbf{e}_{j}=0\) (It may help to introduce a notation that distinguishes between vectors and matrices. For example, you could use an underline to indicate a matrix, so that \(\underline{\mathbf{a}}\) is the \(3 \times 1\) matrix that represents the vector a, and the vector scalar product a \(\cdot \mathbf{b}\) is the same as the matrix product \(\tilde{\mathbf{a}} \mathbf{b}\) or \(\underline{\mathbf{b}}\) a. Then consider the number \(\tilde{\mathbf{e}}_{i} \mathbf{I} \mathbf{e}_{j},\) which can be evaluated in two ways using the fact that both \(\mathbf{e}_{i}\) and \(\mathbf{e}_{j}\) are eigenvectors of I.) (b) Use the result of part (a) to show that if the three principal moments are all different, then the directions of three principal axes are uniquely determined. (c) Prove that if two of the principal moments are equal, \(\lambda_{1}=\lambda_{2}\) say, then any direction in the plane of \(\mathbf{e}_{1}\) and \(\mathbf{e}_{2}\) is also a principal axis with the same principal moment. In other words, when \(\lambda_{1}=\lambda_{2}\) the corresponding principal axes are not uniquely determined. (d) Prove that if all three principal moments are equal, then any axis is a principal axis with the same principal moment.

To illustrate the result (10.18), that the total KE of a body is just the rotational KE relative to any point that is instantaneously at rest, do the following: Write down the KE of a uniform wheel (mass M) rolling with speed \(v\) along a flat road, as the sum of the energies of the CM motion and the rotation about the CM. Now write it as the energy of the rotation about the instantaneous point of contact with the road and show that you get the same answer. (The energy of rotation is \(\frac{1}{2} I \omega^{2} .\) The moment of inertia of a uniform wheel about its center is \(I=\frac{1}{2} M R^{2} .\) That about a point on the rim is \(I^{\prime}=\frac{3}{2} M R^{2}\).)

The moment of inertia of a continuous mass distribution with density \(\varrho\) is obtained by converting the sum of (10.25) into the volume integral \(\int \rho^{2} d m=\int \rho^{2} \varrho d V\). (Note the two forms of the Greek "rho": \(\rho=\) distance from \(z\) axis, \(\varrho=\) mass density.) Find the moment of inertia of a uniform circular cylinder of radius \(R\) and mass \(M\) for rotation about its axis. Explain why the products of inertia are zero.

Consider an arbitrary rigid body with an axis of rotational symmetry, which we'll call \(\hat{\mathbf{z}}\). (a) Prove that the axis of symmetry is a principal axis. (b) Prove that any two directions \(\hat{\mathbf{x}}\) and \(\hat{\mathbf{y}}\) perpendicular to \(\hat{\mathbf{z}}\) and each other are also principal axes. (c) Prove that the principal moments corresponding to these two axes are equal: \(\lambda_{1}=\lambda_{2}\).
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