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6.3: Density, Mass, and Center of Mass


Learning Objectives

In this section, we strive to understand the ideas generated by the following important questions:

  • How are mass, density, and volume related?
  • How is the mass of an object with varying density computed?
  • What is the center of mass of an object, and how are definite integrals used to compute it?

We have seen in several different circumstances how studying the units on the integrand and variable of integration enables us to better understand the meaning of a definite integral. For instance, if ( v(t)) is the velocity of an object moving along an axis, measured in feet per second, while ( t) measures time in seconds, then both the definite integral and its Riemann sum approximation,

( int^b_a v(t) ,dt ≈ sum^n_{i=1} v(t_i)Delta t),

have their overall units given by the product of the units of (v(t)) and (t):

(feet/sec)·(sec) = feet.

Thus,

(int^b_a v(t) ,dt)

measures the total change in position (in feet) of the moving object. This type of unit analysis will be particularly helpful to us in what follows. To begin, in the following preview activity we consider two different definite integrals where the integrand is a function that measures how a particular quantity is distributed over a region and think about how the units on the integrand and the variable of integration indicate the meaning of the integral.

Preview Activity (PageIndex{1}):

In each of the following scenarios, we consider the distribution of a quantity along an axis.

  1. Suppose that the function (c(x) = 200 + 100e^{−0.1x}) models the density of traffic on a straight road, measured in cars per mile, where ( x) is number of miles east of a major interchange, and consider the definite integral ( int^2_0 (200 + 100e^{−0.1x} ), dx).
    1. What are the units on the product ( c(x) cdot Delta x)?
    2. What are the units on the definite integral and its Riemann sum approximation given by
      ( int_0^2 c(x)dx approxsum_{i=1}^{n}c(x_{i})Delta x) ?
    3. Evaluate the definite integral ( int^2_0 c(x) dx = int^2_0 (200 + 100e^{−0.1x}) dx) and write one sentence to explain the meaning of the value you find.
  2. On a 6 foot long shelf filled with books, the function ( B) models the distribution of the weight of the books, measured in pounds per inch, where ( x) is the number of inches from the left end of the bookshelf. Let ( B(x)) be given by the rule ( B(x) = 0.5 + dfrac{1}{(x+1)^2}).
    1. What are the units on the product ( B(x) cdot Delta x)?
    2. What are the units on the definite integral and its Riemann sum approximation given by
      ( int^{36}_{12} B(x) dx ≈ sum^n_{i=1}B(x_i) Delta x)?
    3. Evaluate the definite integral ( int^{72}_0 B(x) dx = int^{72}_0 left(0.5 + dfrac{1}{(x+1)^2} ight) dx) and write one sentence to explain the meaning of the value you find.

Density

The mass of a quantity, typically measured in metric units such as grams or kilograms, is a measure of the amount of a quantity. In a corresponding way, the density of an object measures the distribution of mass per unit volume. For instance, if a brick has mass 3 kg and volume 0.002 ( ext{m}^3), then the density of the brick is

[ dfrac{3 ext{kg}}{0.002 ext{m}^3} = 1500 dfrac{ ext{kg}} { ext{m}^3}.]

As another example, the mass density of water is 1000 kg/( ext{m}^3). Each of these relationships demonstrate the following general principle.

For an object of constant density ( d), with mass ( m) and volume ( V),

[ d = dfrac{m}{V}]

or

[ m = d cdot V.]

But what happens when the density is not constant?

If we consider the formula ( m = d cdot V), it is reminiscent of two other equations that we have used frequently in recent work: for a body moving in a fixed direction, distance = rate · time, and, for a rectangle, its area is given by ( A = l cdot w). These formulas hold when the principal quantities involved, such as the rate the body moves and the height of the rectangle, are constant. When these quantities are not constant, we have turned to the definite integral for assistance. The main idea in each situation is that by working with small slices of the quantity that is varying, we can use a definite integral to add up the values of small pieces on which the quantity of interest (such as the velocity of a moving object) are approximately constant.

For example, in the setting where we have a nonnegative velocity function that is not constant, over a short time interval ( Delta t) we know that the distance traveled is approximately ( v(t)Delta t), since ( v(t)) is almost constant on a small interval, and for a constant rate, distance = rate · time. Similarly, if we are thinking about the area under a nonnegative function ( f) whose value is changing, on a short interval ( delta x) the area under the curve is approximately the area of the rectangle whose height is ( f(x)) and whose width is ( Delta x): ( f(x)Delta x). Both of these principles are represented visually in Figure (PageIndex{1}).

Figure (PageIndex{1}): At left, estimating a small amount of distance traveled, ( v(t)Delta t), and at right, a small amount of area under the curve, ( f(x)Delta x).

In a similar way, if we consider the setting where the density of some quantity is not constant, the definite integral enables us to still compute the overall mass of the quantity. Throughout, we will focus on problems where the density varies in only one dimension, say along a single axis, and think about how mass is distributed relative to location along the axis. Let’s consider a thin bar of length b that is situated so its left end is at the origin, where x = 0, and assume that the bar has constant cross-sectional area of 1 ( ext{cm}^2). We let the function ( ho(x)) represent the mass density function of the bar, measured in grams per cubic centimeter. That is, given a location ( x), ( ho(x)) tells us approximately how much mass will be found in a one-centimeter wide slice of the bar at ( x).

Figure (PageIndex{2}): A thin bar of constant cross-sectional area 1 ( ext{cm}^2) with density function ( ho(x) dfrac{ ext{g}}{ ext{cm}^{3}}).

If we now consider a thin slice of the bar of width ( Delta x), as pictured in Figure (PageIndex{2}), the volume of such a slice is the cross-sectional area times ( Delta x). Since the cross-sections each have constant area 1 ( ext{cm}^2), it follows that the volume of the slice is ( 1Delta x ext{cm}^2). Moreover, since mass is the product of density and volume (when density is constant), we see that the mass of this given slice is approximately

( ext{mass} _{ ext{slice}}approx ho(x)dfrac{ ext{g}}{ ext{cm}^{3}}cdot1Delta x ext{cm}^{3}= ho(x)cdotDelta x cdot ext{g})

Hence, for the corresponding Riemann sum (and thus for the integral that it approximates),

[ sum^n_{i=1} ho(x_i)Delta x ≈ int^b_0 ho(x) dx,]

we see that these quantities measure the mass of the bar between 0 and ( b). (The Riemann sum is an approximation, while the integral will be the exact mass.)

At this point, we note that we will be focused primarily on situations where mass is distributed relative to horizontal location, ( x), for objects whose cross-sectional area is constant. In that setting, it makes sense to think of the density function ( ho(x)) with units “mass per unit length,” such as g/cm. Thus, when we compute ( ho(x)cdotDelta x) on a small slice ( Delta x), the resulting units are ( dfrac{ ext{g}}{ ext{cm}}cdot ext{cm}= ext{g}), which thus measures the mass of the slice. The general principle follows.

For an object of constant cross-sectional area whose mass is distributed along a single axis according to the function ( ho(x)) (whose units are units of mass per unit of length), the total mass, ( M) of the object between ( x = a) and ( x = b) is given by

[M = int^b_a ρ(x) dx.]

Exercise (PageIndex{1}):

Consider the following situations in which mass is distributed in a non-constant manner.

  1. Suppose that a thin rod with constant cross-sectional area of 1 ( ext{cm}^{2}) has its mass distributed according to the density function ( ho(x) = 2e^{−0.2x}), where ( x) is the distance in cm from the left end of the rod, and the units on ( ho(x)) are g/cm. If the rod is 10 cm long, determine the exact mass of the rod.
  2. Consider the cone that has a base of radius 4 m and a height of 5 m. Picture the cone lying horizontally with the center of its base at the origin and think of the cone as a solid of revolution.
    1. Write and evaluate a definite integral whose value is the volume of the cone.
    2. Next, suppose that the cone has uniform density of 800 kg/m3 . What is the mass of the solid cone?
    3. Now suppose that the cone’s density is not uniform, but rather that the cone is most dense at its base. In particular, assume that the density of the cone is uniform across cross sections parallel to its base, but that in each such cross section that is a distance ( x) units from the origin, the density of the cross section is given by the function ( ho(x) = 400 + dfrac{200}{1+x^2}), measured in kg/( ext{m}^{3}). Determine and evaluate a definite integral whose value is the mass of this cone of non-uniform density. Do so by first thinking about the mass of a given slice of the cone ( x) units away from the base; remember that in such a slice, the density will be essentially constant.
  3. Let a thin rod of constant cross-sectional area 1 ( ext{cm}^{2}) and length 12 cm have its mass be distributed according to the density function ( ho(x) = dfrac{1}{25} (x − 15)^2), measured in g/cm. Find the exact location ( z) at which to cut the bar so that the two pieces will each have identical mass.

Weighted Averages

class grade grade points credits chemistry B+ 3.3

Table 6.1: A college student’s semester grades.

The concept of an average is a natural one, and one that we have used repeatedly as part of our understanding of the meaning of the definite integral. If we have ( n) values ( a_1, a_2, . ., a_n), we know that their average is given by

[ dfrac{a_1, a_2, . ., a_n}{n},]

and for a quantity being measured by a function ( f) on an interval ( [a, b]), the average value of the quantity on ( [a, b]) is

p dfrac{1}{b − a} int^b_a f (x) dx.[

As we continue to think about problems involving the distribution of mass, it is natural to consider the idea of a weighted average, where certain quantities involved are counted more in the average.

A common use of weighted averages is in the computation of a student’s GPA, where grades are weighted according to credit hours. Let’s consider the scenario in Table 6.1.

If all of the classes were of the same weight (i.e., the same number of credits), the student’s GPA would simply be calculated by taking the average

[ dfrac{3.3 + 3.7 + 2.7 + 2.7}{4} = 3.1.]

But since the chemistry and calculus courses have higher weights (of 5 and 4 credits respectively), we actually compute the GPA according to the weighted average

[ dfrac{3.3 cdot 5 + 3.7 cdot 4 + 2.7 cdot 3 + 2.7 cdot 3}{5 + 4 + 3 + 3} =3.overline{16}.]

The weighted average reflects the fact that chemistry and calculus, as courses with higher credits, have a greater impact on the students’ grade point average. Note particularly that in the weighted average, each grade gets multiplied by its weight, and we divide by the sum of the weights. In the following activity, we explore further how weighted averages can be used to find the balancing point of a physical system.

Activity (PageIndex{1}):

For quantities of equal weight, such as two children on a teeter-totter, the balancing point is found by taking the average of their locations. When the weights of the quantities differ, we use a weighted average of their respective locations to find the balancing point.

  1. Suppose that a shelf is 6 feet long, with its left end situated at ( x = 0). If one book of weight 1 lb is placed at ( x_1 = 0), and another book of weight 1 lb is placed at ( x_2 = 6), what is the location of ( overline{x}), the point at which the shelf would (theoretically) balance on a fulcrum?
  2. Now, say that we place four books on the shelf, each weighing 1 lb: at ( x_1 = 0), at ( x_2 = 2), at ( x_3 = 4), and at ( x_4 = 6). Find ( overline{x}), the balancing point of the shelf.
  3. How does ( overline{x}) change if we change the location of the third book? Say the locations of the 1-lb books are ( x_1 = 0), ( x_2 = 2), ( x_3 = 3), and ( x_4 = 6).
  4. Next, suppose that we place four books on the shelf, but of varying weights: at ( x_1 = 0) a 2-lb book, at ( x_2 = 2) a 3-lb book, and ( x_3 = 4) a 1-lb book, and at ( x_4 = 6) a 1-lb book. Use a weighted average of the locations to find ( overline{x}), the balancing point of the shelf. How does the balancing point in this scenario compare to that found in (b)?
  5. What happens if we change the location of one of the books? Say that we keep everything the same in (d), except that ( x_3 = 5). How does ( overline{x}) change?
  6. What happens if we change the weight of one of the books? Say that we keep everything the same in (d), except that the book at ( x_3 = 4) now weighs 2 lbs. How does ( overline{x}) change?
  7. Experiment with a couple of different scenarios of your choosing where you move the location of one of the books to the left, or you decrease the weight of one of the books.
  8. Write a couple of sentences to explain how adjusting the location of one of the books or the weight of one of the books affects the location of the balancing point of the shelf. Think carefully here about how your changes should be considered relative to the location of the balancing point ( overline{x}) of the current scenario.

Center of Mass

In Activity (PageIndex{2}), we saw that the balancing point of a system of point-masses1 (such as books on a shelf) is found by taking a weighted average of their respective locations. In the activity, we were computing the center of mass of a system of masses distributed along an axis, which is the balancing point of the axis on which the masses rest.

For a collection of ( n) masses (m_1, ldots, m_n) that are distributed along a single axis at the locations (x_1, ldots, x_n), the center of mass is given by

( overline{x} = dfrac{x_1m_1 + x_2m_2 + · · · x_n m_n}{m_1 + m_2 + · · · + m_n} ).

What if we instead consider a thin bar over which density is distributed continuously? If the density is constant, it is obvious that the balancing point of the bar is its midpoint. But if density is not constant, we must compute a weighted average. Let’s say that the 1 In the activity, we actually used weight rather than mass. Since weight is computed by the gravitational constant times mass, the computations for the balancing point result in the same location regardless of whether we use weight or mass, since the gravitational constant is present in both the numerator and denominator of the weighted average.

function ( ho(x)) tells us the density distribution along the bar, measured in g/cm. If we slice the bar into small sections, this enables us to think of the bar as holding a collection of adjacent point-masses. For a slice of thickness ( Delta x) at location ( x_i), note that the mass of the slice, ( m_i), satisfies ( m_i approx ho(x_i)Delta x).

Taking ( n) slices of the bar, we can approximate its center of mass by

( overline{x} approx dfrac{ x_1 · ρ(x_1)Delta x + x_2 · ρ(x_2)Delta x + · · · + x_n · ρ(x_n)Delta x}{ρ(x_1)Delta x + ρ(x_2)Delta x + · · · + ρ(x_n)Delta x}).

. Rewriting the sums in sigma notation, it follows that

[ overline{x} approx dfrac{sum^n_{i=1}x_i · ho(x_i)Delta x}{sum^n_{i=1} ho(x_i)Delta x}. label{6.4} ]

Moreover, it is apparent that the greater the number of slices, the more accurate our estimate of the balancing point will be, and that the sums in Equation (( ef{6.4})) can be viewed as Riemann sums. Hence, in the limit as ( n ightarrowinfty), we find that the center of mass is given by the quotient of two integrals.

For a thin rod of density ( ho(x)) distributed along an axis from ( x = a) to ( x = b), the center of mass of the rod is given by

( overline{x} = dfrac{int^b_a x ho(x) dx}{int^b_a ho(x) dx}).

Note particularly that the denominator of ( overline{x}) is the mass of the bar, and that this quotient of integrals is simply the continuous version of the weighted average of locations, ( x), along the bar.

Activity (PageIndex{1}):

Consider a thin bar of length 20 cm whose density is distributed according to the function ( ho(x) = 4 + 0.1x), where ( x = 0) represents the left end of the bar. Assume that ( ho) is measured in g/cm and ( x) is measured in cm.

  1. Find the total mass, ( M), of the bar.
  2. Without doing any calculations, do you expect the center of mass of the bar to be equal to 10, less than 10, or greater than 10? Why?
  3. Compute ( overline{x}), the exact center of mass of the bar.
  4. What is the average density of the bar?
  5. Now consider a different density function, given by ( p(x) = 4e^{0.020732x}), also for a bar of length 20 cm whose left end is at ( x = 0). Plot both ( ho(x)) and ( p(x)) on the same axes. Without doing any calculations, which bar do you expect to have the greater center of mass? Why?
  6. Compute the exact center of mass of the bar described in (e) whose density function is ( p(x) = 4e^{0.020732x}). Check the result against the prediction you made in (e).

Summary

In this section, we encountered the following important ideas:

  • For an object of constant density ( D), with volume ( V) and mass ( m), we know that ( m = D·V).
  • If an object with constant cross-sectional area (such as a thin bar) has its density distributed along an axis according to the function ( ho(x)), then we can find the mass of the object between ( x = a) and ( x = b) by [ m = int^b_a ho(x) dx.]
  • For a system of point-masses distributed along an axis, say ( m_1, . ., m_n) at locations ( x_1, . ., x_n), the center of mass, ( overline{x}), is given by the weighted average [ overline{x} = dfrac{sum^n_{i=1}x_im_i}{sum^n_{i=1}m_i}.]
  • If instead we have mass continuously distributed along an axis, such as by a density function ( ho(x)) for a thin bar of constant cross-sectional area, the center of mass of the portion of the bar between ( x = a) and ( x = b) is given by [ overline{x} = dfrac{int^b_a x ho(x)dx}{int^b_a ho(x)dx}.]
  • In each situation, ( overline{x}) represents the balancing point of the system of masses or of the portion of the bar.

Center of Mass

Imagine two weights on opposite sides of a balance board. One weight is 5 kilograms and is 4 meters from the center. The other weight is 10 kilograms and is 2 meters from the center.

The weights balance. What is relevant is the products of the mass and the distance of the mass from the center.

Consider region R in the plane. Imagine that it is made out of a thin material with varying density . Consider a small rectangular piece with dimensions by located at the point . Its mass is .

By analogy with the balance board example, I measure the "twisting" about the y-axis produced by the small rectangular piece. It is the product of the mass and the distance to the y-axis, which is x:

The total amount of"twisting" about the y-axis is the x-moment, and is obtained by integrating ("adding up") the "twisting" produced by each of the small pieces that make up the region:

I can define the y-moment in the same way:

Now imagine the region compressed to small ball which has the same total mass M as the original region. Where should the small ball be located so that it produces the same x and y-moments? The location is called the center of mass of the original region if its coordinates are , I want

Solving for and , I get

We can do the same thing in 3 dimensions. Suppose a solid object occupies a region R in space, and that the density of the solid at the point is . The total mass of the object is given by

The moments in the x, y, and z directions are given by

You can think of them in a rough way as representing the "twisting" about the axis in question produced by a small bit of mass at the point .

The center of mass is the point given by

If a region in the plane or a solid in space has constant density , then the center of mass is called the centroid. In this case, the density drops out of the formulas for , , and . For example,

(Of course, you can usually use a double integral to compute the volume of a solid.)

The centroid of the region is , where

Recall that if R is a region in space, the volume of R is

Thus, the denominators of the fractions above are all equal to volume of R.

The corresponding formulas for the centroid of a region in the plane are:

Notice that the integral is just the area of R.

Example. Find the centroid of the region in the first quadrant bounded above by , from to .

Since the question is asking for the centroid, the density is assumed to be constant.

Note that I didn't need a double integral to find the area.

The centroid is .

You can often use symmetry to find the coordinates of the center of mass, or to determine a relationship among the coordinates --- for example, in some cases smmetry implies that some of the coordinates will be equal.

Example. Find the centroid of the region R bounded above by the plane and below by the paraboloid .

By symmetry, , so I only need to find . I'll use cylindrical coordinates.

and intersect in , so the projection of R into the x-y-plane is the interior of the circle of radius 2 centered at the region.

Note that in cylindrical.

The centroid is .

Example. Let R be the region in the first quadrant cut off by the line . Suppose the region is made of a material with density . Find the coordinates of the center of mass.

The region and the density are symmetric in x and y, so . I only need to find one of the coordinates.

The center of mass is .

Example. Let R be the solid bounded below by and above by , and assume that the density is . Find the coordinates of the center of mass.

I'll convert to spherical coordinates. is a cone whose sides make an angle o with the positive z-axis. is the top hemisphere of a sphere of radius 2 centered at the origin.

(I did the integral using the substitution .)

Since the region and the density are both symmetric about the z-axis, . Therefore, I only need to find .

Since , the z-moment is

(I did the integral using the substitution .)

The center of mass is .


The Ideal Gas Law

relates the pressure, volume, temperature and number of moles in a gas to each other. R is a constant called the gas constant. The ideal gas law is what is called an equation of state because it is a complete description of the gas's thermodynamic state. No other information is needed to calculate any other thermodynamic variable, and, since the equation relates four variables, a knowledge of any three of them is sufficient.

Pressure, volume and number of moles, the latter sometime called extent, share an important property, they can never be negative. What would a negative volume be, or an absolute negative pressure or extent? The concepts do not even exist. This means that temperature in the ideal gas law is similarly limited. It can never be negative.

We can determine what zero temperature on the ideal gas scale is by holding the number of moles and the pressure constant and extrapolate the temperature measured in Celcius to what its value would be at zero volume

Figure 6.3.1 The Relationship between Volume and Temperature (a) In these plots of volume versus temperature for equal-sized samples of H 2 at three different pressures, the solid lines show the experimentally measured data down to &minus100°C, and the broken lines show the extrapolation of the data to V = 0. The temperature scale is given in both degrees Celsius and Kelvin. Although the slopes of the lines decrease with increasing pressure, all of the lines extrapolate to the same temperature at V = 0 (&minus273.15°C = 0 K). (b) In these plots of volume versus temperature for different amounts of selected gases at 1 atm pressure, all the plots extrapolate to a value of V = 0 at &minus273.15°C, regardless of the identity or the amount of the gas.

The extrapolated temperature corresponding to zero volume at constant pressure and amount is -273.15 °C, which is called absolute zero because no lower temperature is possible (unless, of course you can come up with a negative volumes, but you cannot). For convenience we set the degrees on the Kelvin scale to the size of the degree on the Celcius scale. In other words

Note the Pattern

Before we can use the ideal gas law, however, we need to know the value of the gas constant R. Its form depends on the units used for the other quantities in the expression. If V is expressed in liters (L), P in atmospheres (atm), T in kelvins (K), and n in moles (mol), then

( R = 0.08206 L cdot atm /left (Kcdot ·mol ight ) ag <6.3.3>)

Because the product PV has the units of energy, R can also have units of J/(K·mol) or cal/(K·mol):

( R = 8.3145 J/left (K cdot mol ight ) = 1.9872 cal/left (K cdot mol ight )
ag <6.3.4>)

Scientists have chosen a particular set of conditions to use as a reference: 0°C (273.15 K) and 1 atm pressure, referred to as standard temperature and pressure (STP) The conditions 0°C (273.15 K) and 1 atm pressure for a gas.

We can calculate the volume of 1.000 mol of an ideal gas under standard conditions using the variant of the ideal gas law given in Equation 6.3.1:

( V = dfrac

= dfrac left ( 0.082057 Lcdot cancel/cancelcdot cancel ight ) ight )left ( 273.15 cancel ight )><1.000 cancel>=22.31 L ag <6.3.5>)

Thus the volume of 1 mol of an ideal gas at 0°C and 1 atm pressure is 22.41 L, approximately equivalent to the volume of three basketballs. The quantity 22.41 L is called the standard molar volume The volume of 1 mol of an ideal gas at STP (0°C and 1 atm pressure), which is 22.41 L. of an ideal gas. The molar volumes of several real gases at STP are given in Table 6.3.1, which shows that the deviations from ideal gas behavior are quite small. Thus the ideal gas law does a good job of approximating the behavior of real gases at STP.

Table 6.3.1 Molar Volumes of Selected Gases at Standard Temperature (0°C) and Pressure (1 atm)

Gas Molar Volume (L)
He 22.434
Ar 22.397
H2 22.433
N2 22.402
O2 22.397
CO2 22.260
NH3 22.079


Using Symmetry as a Shortcut

Just as with 2D areas, shape symmetry can provide a shortcut in many centroid calculations. Remember that the centroid coordinate is the average x, y, and z coordinate for all the points in the shape. If the volume has a plane of symmetry, that means each point on one side of the line must have an equivalent point on the other side of the line. This means that the average value (aka. the centroid) must lie within that plane. If the volume has more than one plane of symmetry, then the centroid must exist at the intersection of those planes.

If a volume has a plane of symmetry, then the centroid must lie somewhere in that plane. If the volume has more than one plane of symmetry, the centroid must exist at the intersection of those planes.


Solved Example Problems for Center of mass for uniform distribution of mass

Example 5.4

Locate the center of mass of a uniform rod of mass M and length l.

Solution

Consider a uniform rod of mass M and length whose one end coincides with the origin as shown in Figure. The rod is kept along the x axis. To find the center of mass


of this rod, we choose an infinitesimally small mass dm of elemental length dx at a distance x from the origin.


Now, we can write the center of mass equation for this mass distribution as,


As the position l/2 is the geometric center of the rod, it is concluded that the center of mass of the uniform rod is located at its geometric center itself.


What is the center of gravity?

The center of gravity is the point through which the force of gravity acts on an object. In most mechanic’s problems the gravitational field is assumed to be uniform. The center of gravity is then in exactly the same position as the center of mass. The terms center of gravity and center of mass tend to often be used interchangeably since they are often at the same location. However, they are not the same.

The center of mass of a body does not always coincide with its intuitive geometric center, and one can exploit this freedom. Engineers try to design a sports car's center of gravity as low as possible to make the car handle better.

The location of the center of gravity can be found by summing the multiplication of the distance by the weight (area) and divide it by the summation of all the weights (areas).


4. Density, Mass, and Center of Mas s

• How is the mass of an object with varying density computed?

• What is the center of mass of an object, and how are definite integrals used to compute it?

We have seen in several different circumstances how studying the units on the integrand and variable of integration enables us to better understand the meaning of a definite integral. For instance, if v( t) is the velocity of an object moving along an axis, measured in feet per second, while t measures time in seconds, then both the definite integral and its Riemann sum approximation,

have their overall units given by the product of the units of v( t) and t:

measures the total change in position (in feet) of

This type of unit analysis will be particularly helpful to us in what follows. To begin, in the following preview activity we consider two different definite integrals where the integrand is a function that measures how a particular quantity is distributed over a region and think about how the units on the integrand and the variable of integration indicate the meaning of the integral.

Preview Activity 4

In each of the following scenarios, we consider the distribution of a quantity along an axis.

(a) Suppose that the function c(x) = 200 + 100e —0.1x models the density of traffic on a straight road, measured in cars per mile, where x is number of miles east of a major interchange, and consider the

i. What are the units on the product

ii. What are the units on the definite integral and its Riemann sum approximation given by

iii. Evaluate the definite integral

and write one sentence to explain the meaning of the value you find.

(b) On a 6 foot long shelf filled with books, the function B models the distribution of the weight of the books, measured in pounds per inch, where x is the number of inches from the left end of the book-

shelf. Let B(x) be given by the rule B(x) = 0.5 +

i. What are the units on the product

ii. What are the units on the definite integral and its Riemann sum approximation given by

iii. Evaluate the definite integral

and write one sentence to explain the meaning of the value you find.

Density

The mass of a quantity, typically measured in metric units such as grams or kilograms, is a measure of the amount of a quantity. In a corresponding way, the density of an object measures the distribution of mass per unit volume. For instance, if a brick has mass 3 kg and volume 0.002 m 3 , then the density of the brick is

As another example, the mass density of water is 1000 kg/m 3 . Each of these relationships demonstrate the following general principle.

Density, Mass, & Volume

For an object of constant density d, with mass m and volume V,

But what happens when the density is not constant? If we consider the formula m = d • V, it is reminiscent of two other equations that we have used frequently in recent work: for a body moving in a fixed direction, distance = rate • time, and, for a rectangle, its area is given by A = I • w. These formulas hold when the principal quantities involved, such as the rate the body moves and the height of the rectangle, are constant. When these quantities are not constant, we have turned to the definite integral for assistance. The main idea in each situation is that by working with small slices of the quantity that is varying, we can use a definite integral to add up the values of small pieces on which the quantity of interest (such as the velocity of a moving object) are approximately constant.

For example, in the setting where we have a nonnegative velocity function that is not constant, over a short time interval

we know that the distance traveled is approximately

since v <> is almost constant on a small interval, and for a constant rate, distance = rate • time. Similarly, if we are thinking about the area under a nonnegative function / whose value is

changing, on a short interval

the area under the curve is

approximately the area of the rectangle whose height is f(x)

Both of these principles are

represented visually in Figure 6.27.

Figure 6.27: At left, estimating a small amount

amount of area under the curve,

In a similar way, if we consider the setting where the density of some quantity is not constant, the definite integral enables us to still compute the overall mass of the quantity. Throughout, we will focus on problems where the density varies in only one dimension, say along a single axis, and think about how mass is distributed relative to location along the axis.

Let's consider a thin bar of length b that is situated so its left end is at the origin, where x = 0, and assume that the bar has constant cross-sectional area of 1 cm 2 . We let the function p(x) represent the mass density function of the bar, measured in grams per cubic centimeter. That is, given a location x, p(x) tells us approximately how much mass will be found in a one-centimeter wide slice of the bar at x.

Figure 6.28: A thin bar of constant cross-sectional area 1 cm 2 with density function p(x) g/cm 3 .

If we now consider a thin slice of the bar of width

pictured in Figure 6.28, the volume of such a slice is the cross-

Since the cross-sections each have con-

stant area 1 cm 2 , it follows that the volume of the slice is

cm 3 . Moreover, since mass is the product of density and volume (when density is constant), we see that the mass of this given slice is approximately

Hence, for the corresponding Riemann sum (and thus for the integral that it approximates),

we see that these quantities measure the mass of the bar between 0 and b. (The Riemann sum is an approximation, while the integral will be the exact mass.)

At this point, we note that we will be focused primarily on situations where mass is distributed relative to horizontal location, x, for objects whose cross-sectional area is constant. In that setting, it makes sense to think of the density function p(x) with units "mass per unit length," such as g/cm. Thus, when we

g/cm • cm = g, which thus measures the mass of the slice. The general principle follows.

For an object of constant cross-sectional area whose mass is distributed along a single axis according to the function p(x) (whose units are units of mass per unit of length), the total mass, M of the object between x = a and x = b is given

A thin bar occupies the interval

and it has a density in kg/m

of p(x) = 1 + x 2 . Find the mass of the bar.

Solution. The mass of the bar in kilograms is

Activity 4-1

Consider the following situations in which mass is distributed in a non-constant manner.

(a) Suppose that a thin rod with constant cross-sectional area of 1 cm 2 has its mass distributed according to the density function p(x) = 2e —0.2x , where x is the distance in cm from the left end of the rod, and the units on p(x) are g/cm. If the rod is 10 cm long, determine the exact mass of the rod.

(b) Consider the cone that has a base of radius 4 m and a height of 5 m. Picture the cone lying horizontally with the center of its base at the origin and think of the cone as a solid of revolution.

i. Write and evaluate a definite integral whose value is the volume of the cone.

ii. Next, suppose that the cone has uniform density of 800 kg/m 3 . What is the mass of the solid cone?

iii. Now suppose that the cone's density is not uniform, but rather that the cone is most dense at its base. In particular, assume that the density of the cone is uniform across cross sections parallel to its base, but that in each such cross section that is a distance x units from the origin, the density of the cross section is given

mine and evaluate a definite integral whose value is the mass of this cone of non-uniform density. Do so by first thinking about the mass of a given slice of the cone x units away from the base remember that in such a slice, the density will be essentially constant.

(c) Let a thin rod of constant cross-sectional area 1 cm 2 and length 12 cm have its mass be distributed according to the density function

measured in g/cm. Find the exact location z at

which to cut the bar so that the two pieces will each have identical mass.

Table 6.2: A college student's semester grades.

class grade grade points credits
chemistry B+ 3.3 5
calculus A- 3.7 4
history B- 2.7 3
psychology B- 2.7 3

The concept of an average is a natural one, and one that we have used repeatedly as part of our understanding of the meaning of the definite integral. If we have n values a1, a2, . a n , we know that their average is given by

and for a quantity being measured by a function f on an interval [a, b], the average value of the quantity on [a, b] is

As we continue to think about problems involving the distribution of mass, it is natural to consider the idea of a weighted average, where certain quantities involved are counted more in the average.

A common use of weighted averages is in the computation of a student's GPA, where grades are weighted according to credit hours. Let's consider the scenario in Table 6.2.

If all of the classes were of the same weight (i.e., the same number of credits), the student's GPA would simply be calculated by taking the average

But since the chemistry and calculus courses have higher weights (of 5 and 4 credits respectively), we actually compute the GPA according to the weighted average

The weighted average reflects the fact that chemistry and calculus, as courses with higher credits, have a greater impact on the students' grade point average. Note particularly that in the weighted average, each grade gets multiplied by its weight, and we divide by the sum of the weights.

In the following activity, we explore further how weighted averages can be used to find the balancing point of a physical system.

Activity 4-2

For quantities of equal weight, such as two children on a teeter-totter, the balancing point is found by taking the average of their locations. When the weights of the quantities differ, we use a weighted average of their respective locations to find the balancing point.

(a) Suppose that a shelf is 6 feet long, with its left end situated at x = 0. If one book of weight 1 lb is placed at x 1 = 0, and another book of weight 1 lb is placed at x2 = 6, what is the location of x, the point at which the shelf would (theoretically) balance on a fulcrum?

(b) Now, say that we place four books on the shelf, each weighing 1 lb: at x 1 = 0, at x 2 = 2, at x 3 = 4, and at x 4 = 6. Find

change if we change the location of the third book?

Say the locations of the 1-lb books are x1 = 0, x2 = 2, x3 = 3, and

(d) Next, suppose that we place four books on the shelf, but of varying weights: at x 1 = 0 a 2-lb book, at x 2 = 2 a 3-lb book, and x 3 = 4 a 1-lb book, and at x 4 = 6 a 1-lb book. Use a weighted average of the

the balancing point of the shelf. How does the

balancing point in this scenario compare to that found in (b)? (e) What happens if we change the location of one of the books? Say that we keep everything the same in (d), except that x3 = 5. How

(f) What happens if we change the weight of one of the books? Say that we keep everything the same in (d), except that the book at x3 = 4

now weighs 2 lbs. How does

(g) Experiment with a couple of different scenarios of your choosing where you move the location of one of the books to the left, or you decrease the weight of one of the books.

(h) Write a couple of sentences to explain how adjusting the location of one of the books or the weight of one of the books affects the location of the balancing point of the shelf. Think carefully here about how your changes should be considered relative to the location of the balancing point

Center of Mass

In Activity4-2, we saw that the balancing point of a system

1 In the activity, we actually used weight rather than mass. Since weight is computed by the gravitational constant times mass, the computations for the balancing point result in the same location regardless of whether we use weight or mass, since the gravitational constant is present in both the numerator and denominator of the weighted average.

(such as books on a shelf) is found by taking a weighted average of their respective locations. In the activity, we were computing the center of mass of a system of masses distributed along an axis, which is the balancing point of the axis on which the masses rest.

For a collection of n masses m 1 , . m n that are distributed along a single axis at the locations x 1 , .. ., x n , the center of mass is given by

Figure 6.29: A thin bar of constant cross-sectional area with density function p(x) g/cm.

What if we instead consider a thin bar over which density is distributed continuously? If the density is constant, it is obvious that the balancing point of the bar is its midpoint. But if density is not constant, we must compute a weighted average. Let's say that the function p(x) tells us the density distribution along the bar, measured in g/cm. If we slice the bar into small sections, this enables us to think of the bar as holding a collection of

adjacent point-masses. For a slice of thickness

note that the mass of the slice, m i , satisfies

Taking n slices of the bar, we can approximate its center of mass by

Rewriting the sums in sigma notation, it follows that

Moreover, it is apparent that the greater the number of slices, the more accurate our estimate of the balancing point will be, and that the sums in Equation (6.1) can be viewed as Riemann

sums. Hence, in the limit as

we find that the center of

mass is given by the quotient of two integrals.

For a thin rod of density p(x) distributed along an axis from x = a to x = b, the center of mass of the rod is given by

Note particularly that the denominator of

bar, and that this quotient of integrals is simply the continuous version of the weighted average of locations, x, along the bar.

A thin bar occupies the interval

and it has a density in kg/m

of p(x) = 1 + x 2 . Find the center of mass of the bar.

Solution. From Example 1, the mass of the bar in kilograms is

Activity 4-3

Consider a thin bar of length 20 cm whose density is distributed according to the function p(x) = 4 + 0.1x, where x = 0 represents the left end of the bar. Assume that p is measured in g/cm and x is measured in cm.

(a) Find the total mass, M, of the bar.

(b) Without doing any calculations, do you expect the center of mass of the bar to be equal to 10, less than 10, or greater than 10? Why?

the exact center of mass of the bar.

(d) What is the average density of the bar?

(e) Now consider a different density function, given by p(x) = 4e 0.020732x , also for a bar of length 20 cm whose left end is at x = 0. Plot both p( x) and p( x) on the same axes. Without doing any calculations, which bar do you expect to have the greater center of mass? Why?

(f) Compute the exact center of mass of the bar described in (e) whose density function is p(x) = 4e 0.020732x . Check the result against the prediction you made in (e).

Summary

In this section, we encountered the following important ideas:

• For an object of constant density D, with volume V and mass m, we know that m = D • V.

• If an object with constant cross-sectional area (such as a thin bar) has its density distributed along an axis according to the function p(x), then we can find the mass of the object between x = a and x = b by

• For a system of point-masses distributed along an axis, say m 1 , . . . , m n at locations x 1 , . . . , x n , the center of

is given by the weighted average

If instead we have mass continuously distributed along an axis, such as by a density function p( x) for a thin bar of constant cross-sectional area, the center of mass of the portion of the bar between x = a and x = b is given by

represents the balancing point of the system of masses or of the portion of the bar.

Exercises

1) T/F: The integral formula for computing Arc Length was found by first approximating arc length with straight line segments.

2) T/F: The integral formula for computing Arc Length includes a square-root, meaning the integration is probably easy.

Problems

3) Let a thin rod of length a have density distribution function p(x) = 10e —0.1x , where x is measured in cm and p in grams per centimeter.

(a) If the mass of the rod is 30 g, what is the value of a?

(b) For the 30g rod, will the center of mass lie at its midpoint, to the left of the midpoint, or to the right of the midpoint? Why?

(c) For the 30g rod, find the center of mass, and compare your prediction in (b).

(d) At what value of x should the 30g rod be cut in order to form two pieces of equal mass?

4) Consider two thin bars of constant cross-sectional area, each of length 10 cm, with respective mass den-

(a) Find the mass of each bar.

(b) Find the center of mass of each bar.

(c) Now consider a new 10 cm bar whose mass density function is f (x) = p(x) + p(x).

i. Explain how you can easily find the mass of this new bar with little to no additional work.

as possible, in light of earlier computations. iii. True or false: the center of mass of this new bar is the average of the centers of mass of the two earlier bars. Write at least one sentence to say why your conclusion makes sense.

5) Consider the curve given by y = f (x) = 2xe —1.25x + (30 — x)e —0.25(30—x) .

(a) Plot this curve in the window x = 0 . . . 30, y = 0. 3 (with constrained scaling so the units on the x and y axis are equal), and use it to generate a solid of revolution about the x-axis. Explain why this curve could generate a reasonable model of a baseball bat.

(b) Let x and y be measured in inches. Find the total volume of the baseball bat generated by revolving the given curve about the x-axis. Include units on your answer

(c) Suppose that the baseball bat has constant weight density, and that the weight density is 0. 6 ounces per cubic inch. Find the total weight of the bat whose volume you found in (b).

(d) Because the baseball bat does not have constant cross-sectional area, we see that the amount of weight concentrated at a location x along the bat is determined by the volume of a slice at location x. Explain why we can think about the function

(where f is the function given

at the start of the problem) as being the weight density function for how the weight of the baseball bat is distributed from x = 0 to x = 30. (e) Compute the center of mass of the baseball bat.


6.3: Density, Mass, and Center of Mass

If the object under consideration is continuous , then

where is the mass density of the object, and is the volume occupied by the th element. Here, it is assumed that this volume is small compared to the total volume of the object. Taking the limit that the number of elements goes to infinity, and the volume of each element goes to zero, Eqs. (325) and (326) yield the following integral formula for the position vector of the centre of mass:

Here, the integral is taken over the whole volume of the object, and is an element of that volume. Incidentally, the triple integral sign indicates a volume integral: i.e. , a simultaneous integral over three independent Cartesian coordinates. Finally, for an object whose mass density is constant --which is the only type of object that we shall be considering in this course--the above expression reduces to

where is the volume of the object. According to Eq. (328), the centre of mass of a body of uniform density is located at the geometric centre of that body.

For many solid objects, the location of the geometric centre follows from symmetry. For instance, the geometric centre of a cube is the point of intersection of the cube's diagonals. See Fig. 72. Likewise, the geometric centre of a right cylinder is located on the axis, half-way up the cylinder. See Fig. 73.

As an illustration of the use of formula (328), let us calculate the geometric centre of a regular square-sided pyramid. Figure 74 shows such a pyramid. Let be the length of each side. It follows, from simple trigonometry, that the height of the pyramid is . Suppose that the base of the pyramid lies on the - plane, and the apex is aligned with the -axis, as shown in the diagram. It follows, from symmetry, that the geometric centre of the pyramid lies on the -axis. It only remains to calculate the perpendicular distance, , between the geometric centre and the base of the pyramid. This quantity is obtained from the -component of Eq. (328):

where the integral is taken over the volume of the pyramid.

In the above integral, the limits of integration for are to , respectively ( i.e. , from the base to the apex of the pyramid). The corresponding limits of integration for and are to , respectively ( i.e. , the limits are at the base of the pyramid, and at the apex). Hence, Eq. (329) can be written more explicitly as

As indicated above, it makes sense to perform the - and - integrals before the -integrals, since the limits of integration for the - and - integrals are -dependent. Performing the -integrals, we obtain

Performing the -integrals, we obtain

Finally, performing the -integrals, we obtain

Thus, the geometric centre of a regular square-sided pyramid is located on the symmetry axis, one quarter of the way from the base to the apex.


Mass,Weight and, Density

I Words: Most people hardly think that there is a difference between "weight" and "mass" and it wasn't until we started our exploration of space that is was possible for the average person to experience, even indirectly, what it must mean to be "weightless". Everyone has been confused over the difference between "weight" and "density". Which of us hasn't fallen for the old riddle: What weighs more, a pound of lead or a pound of feathers? Now that Astronauts regularly seem to be demonstrating "weightlessness", ( yet we know they still have all of the matter or mass they had when they were on earth,) more of us are beginning appreciate the weight - mass confusion. Like many confusing concepts, after they are finally understood, they are still difficult to explain to others who don't understand. We hope we can explain the difference between mass, weight and density so clearly that you will have no trouble explaining the difference to your students.

Mass: This concept is so basic that, like length and time, it is really impossible to define. Isaac Newton called mass the quantity of matter . We can talk all around it but we will finally have to admit that our words fail. Some say mass is the amount of matter in something (and hope that no one asks: What is matter?). Others say mass is the measure of an object's inertia (which assumes we understand the elusive property of inertia). To add to the confusion, mass is related to an object's inertia but it also is related to how hard objects are attracted to the earth. Better minds than ours have been confused over the meaning of the concept "mass" and even today, better minds than ours contemplate what mass really means. Our way of giving up on the impossible task of defining mass is to say: mass is the measure of the amount of "stuff" in something. This definition is properly confusing and you can work on the meaning of "stuff"! In the metric system mass is measured in kilograms and grams and these will be the units we will most often use. (In the United States today, almost no one knows what the unit of mass is called--it's not the pound. The pound is a unit of weight--more about weight in the next paragraph. The more mass something has, the harder it is to move or, the more sluggish it is. The correct US unit of mass is called the "slug"--short for sluggishness-- but, as we said, almost no one uses this unit today.)

Weight: If you can finally accept the concept mass even if we have been unable to define it, weight is easy: The weight of a mass is the force that the earth pulls on the mass. We hope you have a feeling for what force means (and we will discuss it later). The entire idea of weight can be understood as the force of gravity on something. Usually we spend most of our time on Earth so our weight is the force that the earth pulls on us. If we get further away from the earth, the force the earth pulls on us is less and we weigh less. If you lived on Mars, the above definition would probably change to: "The weight of a mass is the force that Mars pulls on the mass." The whole idea of weight is related to the force of gravity (and we hate just to use the word "gravity" since it can bring up even more confusion). It would be correct to say, no matter where you might be in the universe that "the weight of a mass is the force of gravity on the mass." In the metric system force is measured in newtons hence weight is also measured in newtons. You will learn later that on the surface of the earth, a mass of 1 kilogram weighs 9.8 newtons. (You will probably never learn anywhere that on the surface of the earth, one slug weighs 32.2 pounds--don't worry about it, very few people know this!) The pound is the US unit of force hence the US unit of weight is also the pound. We will use newtons for the unit of force (and weight) almost always in the discussions that follow.

Density: There are two kinds of density, "weight density" and "mass density". We will only use mass density and when we say: "density", we will mean "mass density". Density is mass per volume. Lead is dense, Styrofoam is not. The metric system was designed so that water will have a density of one gram per cubic centimeter or 1000 kilograms per cubic meter. Lead is about 10 times as dense as water and Styrofoam is about one tenth as dense as water.

II Purpose of the Activity:

The purpose of this activity is to investigate the meaning of mass, weight and density by looking at how each might be measured.

III Materials required for the Activity:

At least one box of #1 (small) paper clips, 20 (or more) long thin rubber bands (#19 will work--they are 1/16" thick and 3+" long), drinking straws, a fine tipped marking pen (Sharpie), scotch tape, 40 (or more) 1oz or 2oz plastic portion cups (Dixie sells them in boxes of 800 for less than $10--see if your school cafeteria has them), lots of pennies (to use as "weights"), light string, 20 (or more) specially drilled wooden rulers or cut sections of wooden molding, about a pound or two of each of the following as available: sand, rice, sawdust, fine crushed Styrofoam (place Styrofoam packing or cups in a plastic bag and pound--yes, this can make a mess), lead shot (can this be used with children? Lead shot can be obtained at a gun and ammo shop), any other finely ground solid material which can be used to illustrate a variety of different densities. (We have avoided liquids since they seem to make a bigger mess.)

IV What the teacher must do in advance of the activity:

We feel the most difficult preparation item for this activity will be making the drilled ruler or preparing sections of wood properly drilled. (Is it possible to assign tasks like this to parents who have better shop facilities at home?) These will become the balance beams for our mass balance. The actual dimensions of the balance beam is not too critical and a wooden ruler is about right--the best are old ones that have lost the metal strip they often have along one edge. What follows is a description and illustration of how the balance beam should be constructed:

The center hole should be exactly in the center so that when the beam is supported by a nail through this center hole, it will spin around freely. The two end holes should be in line with the center hole and equidistant on each side. Two other holes should be drilled on a line directly above the center hole. The hole size is not critical, about 1/8 of an inch is fine. (We sincerely hope it will not be too difficult for you to accomplish this. We fooled around with metal hangers, etc. but nothing simple seemed to work as well as a carefully prepared stick.)

It will be necessary to poke holes in the portion cups. Should you do this in advance or can your students do it? A small nail works well for this purpose. Probably time could be saved in class if the necessary strings were cut to length in advance (and perhaps even tied to the cups).

Building the "Weight Scale" requires some careful cutting of a straw that can be done with a good pair of scissors or a sharp knife. We think kids can do all of it but it will take time. You should build one prototype Weight Scale in advance so you can work out the details of construction and decide how much of the cutting should be done in advance.

Teaching outline and Presentation suggestions:

(Once again we remind you that we really don't know the best way to teach these concepts to young students. The following are suggestions for construction and use of the equipment and how we envisioned one might use this stuff but only you can know the best way to present this material to your students.)

Measuring mass: Mass is usually measured with a balance. The idea is to compare the unknown object with the mass of a known amount. Illustrated below is the device we will use to measure mass and we will call it "the Mass Balance".

Since everyone seems to have lots of pennies and all pennies are about the same mass, we will use the penny as our standard of mass. (It turns out that the average penny has a mass of about 2.6 grams and you can convert to grams if you wish but for now, we will simply determine mass in "pennies".) The mass measurement is accomplished simply by placing the unknown object in one cup of the Mass Balance and finding out how many pennies placed on the other side it takes to achieve balance. You should first check the Mass Balance with nothing in either cup to see if it is properly "zeroed". You should notice that the balance is most sensitive when the upper paper clip is in the center hole (in fact it is really too sensitive here) and it will be less sensitive when you use the higher holes. Slight errors in the zero reading can be corrected by using shorter or longer string sections on the appropriate side. Make sure all paper clips rotate freely in the drilled holes. The balance will not work properly if the paper clips hang up. The Mass Balance can be loaded with the cups on the table and pulling upward slightly on the support paper clip will test the balance condition. We suggest that students begin by matching pennies on the left with pennies on the right (and they should discover that all pennies aren't really the same--this is real!) After the students become familiar with the use of the balance, we suggest that nearly equal volumes of the assorted materials (sand, rice, metal shot, Styrofoam) be measured. If you are using the 1 oz Dixie portion cups, it is possible to draw a line on the cup 1.4 cm above the bottom and it will represent 10 cubic centimeters and a line 2.3 cm above the bottom of the cup will represent 20 cubic centimeters (or milliliters).

A very important question to consider now is: If you used this Mass Balance on the moon or on Mars, would the same amount of material on one side require the same number of pennies on the other side to balance it as it did on Earth? Naturally there is no easy way for us to perform such an experiment but, having your students think about this should help them to start understanding the difference between weight and mass. Mass or as Newton would say, the quantity of matter in an object, does not change when you change your location in space but, as we will see shortly, weight does change.

Measuring weight: Using a carefully segmented straw, a bent paper clip, a rubber band, some string, a small cup, a 3X5 card and some scotch tape we will construct the "Weight Scale" shown below:

( Since we had difficulty in joining the segment of rubber band to the string, we decided to show you a "carrick bend" which works quite well for this situation. A simple slip knot works well on the bottom where the rubber band attaches to the bent paper clip bail. A later construction detail diagram will give you a better illustration of the Weight Scale.)

The students will calibrate this Weight Scale with pennies and mark the 3 X 5 card with a marking pen during the calibration exercise. Attaching the rubber band to the bail of the cup is easily accomplished with a slip knot but attaching the string to the rubber band is a slight problem--a suggested knot is shown with the illustration. The whole idea is to have the zero of the scale at the bottom of the card using the string-rubber band junction as the pointer. With about 25 pennies in the cup, the rubber band will stretch to about the top of the card. (You hold the scale with the string which has been passed through a small piece of straw taped to the card.) The students will carefully load the cup with pennies and mark the card at about 5 penny intervals.

A more detailed construction of the Weight Scale is shown "below." (Again, we suggest that you construct one in advance so you can evaluate how difficult it is to build.) Note that in the construction diagram "below", we show how the straw should be sectioned so that it can be attached to the 3 X 5 card. (In the final scale, naturally, the string and rubber band fit inside of the straw sections.) It is important that the lower length of the straw be made just long enough to extend from the top of the paper clip "bail" to the bottom of the card with the rubber band sticking out the top. You must be able to tie the rubber band to the string and have the junction of the two be on the lower end of the 3 X 5 with nothing in the cup.

This is the "below" referred to in the above paragraph. We decided it would take a large image to show the necessary details so, if you have the time, click here for Weight Scale Construction Details.

After the students have calibrated the Weight Scale, it might be fun to have them see if they can guess how many pennies have been loaded into their cup by another student. From this they will learn how to read between the marks they have placed on their cards (this is called interpolation) and they will also learn that the scale is really not too accurate. However, all instruments are less than perfect at some level and this crude scale should help them to realize this fact. (We think it is nice that the scale is quite inexpensive and students who wish can construct one at home.)

This Weight Scale can also be used to measure some of the other materials that were measured with the Mass Balance. Hopefully they will find that they will get pretty close to the same answer in "pennies" for the mass as measured on the Mass Balance and the weight as measured on the Weight Scale. (So you ask--what is the difference between weight and mass?) Now comes the key question to ask the class: If you took the Mass Balance and your calibrated Weight Scale to the Moon, do you think they would give the same measurement as on Earth? Remember, you always balance the unknown object against several pennies with the Mass Balance but you just let the unknown object pull down against the calibrated rubber band on the Weight Scale. We hope that this thought experiment will help the students see that the Mass Balance will measure the same no matter where you locate it in space but the Weight Scale, which measures how hard gravity pulls down on the object, will give a smaller reading on the moon. (This is confusing stuff and most college students will have difficulty understanding it. Perhaps if your kids start thinking about it early enough, they may come to a better understanding of the difference between weight and mass when they are older.)

Measuring Density: Since density is mass per volume, the most straight forward way of measuring the density of something is to measure its mass, then measure its volume and divide the mass by the volume. We could do exactly that in this activity but at this point we have no good way to measure volume. If you have a graduated cylinder (they aren't expensive but most elementary schools don't have them) you could use it with some water to mark the small "portion cups" at specific volumes. (We have already suggested that the small 1 oz cups will hold 10 cubic centimeters when filled to a point 1.4 cm above the bottom and it will hold 20 cubic centimeters when filled to a point 2.3 centimeters above the bottom.) Rather than actually measuring the density, we feel it will be sufficient for the students to appreciate that the same volume can be a large mass or a small mass depending upon the material involved. Our plan is to have the same volume of several different materials and measure their mass with the Mass Balance. Hopefully this exercise will help the students to begin to see the relationship between mass, volume and density.

(The next page begins the "Student Activity Sheet". We suggest that these be reproduced in sufficient numbers for the entire class. Whether the students work individually or in groups is best decided by you, however, some of the exercises need at least two people to hold and use the apparatus.)

Mass, Weight and Density--how matter is measured, how it interacts with other matter and how it fills space.

Which is heavier, a pound of feathers or a pound of lead? If you have never heard this old trick question before--think about it. Now try this one: which takes up more space, a pound of feathers or a pound of lead? Finally, think about this one: which weighs more 100 pennies on the earth or 100 pennies on the moon? The answer to each of these questions requires that you understand the difference between mass, weight and density.

You will measure the mass of objects by comparing them to the mass of pennies with a thing we will call a "Mass Balance". Although mass is usually measured in kilograms or grams, we will measure mass in "pennies". The Mass Balance is shown below.

This balance measures mass in "penny" units.

First test the Mass Balance to see if it is "zeroed". When you lift it by the center paper clip, it should stay fairly level. When the clip is in the top hole, it will balance easily. If you put the clip in the bottom hole, it probably will be too sensitive to balance at all. Test the Mass Balance by placing 5 pennies in both cups and gently lift it off the table--it should balance. Have one student secretly place a number of pennies in one cup and see if you can figure out how many pennies there are in the cup by matching them with pennies in the other cup. (You will find that not all pennies are exactly the same.) After you learn how to use the Mass Balance, you will be given several different materials to measure. Always measure the same volume of the given materials, that is, always fill the material to the same level on the cup on the left and find its mass by placing pennies in the cup on the right. Record your data in a table like the one below:

Name of material being measured

Mass of material in "pennies"

(name first materal meaured here)

(record its mass in pennies here)

Here is an important question to think about: If you took your Mass Balance to the Moon and repeated this experiment, would you get the same result? (Have your teacher discuss with you what mass means--this is quite confusing to many people.)

You will construct and calibrate a Weight Scale. This scale works by measuring how far a certain number of pennies are able to stretch a piece of rubber band. The scale is illustrated below:

Measuring weight in units of pennies.

After you have constructed the Weight Scale (your teacher will give you instructions) you will calibrate it using pennies. While one student carefully holds the Weight Scale by the string the other student will make a line on the card. First mark on the card where the knot touches the card with nothing in the cup (this will be the zero line on the scale). Now add 5 pennies to the cup and carefully mark where the knot touches the card. (Practice with pencil first until you learn how to do this.) Carefully "calibrate" your scale by adding 5 pennies at a time and marking where the knot touches the card. The numbers on your scale will be 0, 5, 10, 15, 20, 25 you probably will not be able to get any more. (Don't expect the numbers to be evenly spaced--rubber bands don't work that way.) If you have calibrated your scale well, you should be able to tell how many pennies someone has placed in the cup without actually counting them. As with the mass balance, you should weigh some of the materials your teacher provides. Always fill the cup with the same volume of material. Make a table like the following:

Name of material being weighed

Weight of material in "pennies"

(give the name of first material weighed here)

(give the weight of the material in pennies here)

Although your measurements are not perfect (no measurements ever are) you can use your calibrated Weight Scale to find the weight of assorted things and you could even use it to count pennies. Now comes a very important question: If you took your calibrated Weight Scale to the Moon, would it work the same way it did on Earth? If you filled it with the same amount of pennies, would the rubber band stretch to the same mark? Have your teacher discuss this with you and see if you can understand the difference between weight and mass.

Density tells us how much stuff has been packed into a certain amount of space. Lead is very dense, Styrofoam is not very dense at all. Density can be measured in grams per cubic centimeter. In our experiments, density could be measured in "pennies per cup". In your experiments with the Mass Balance, you always measured the mass of the same volume of material. Use the data you took to list the materials you measured in order of density with the most dense at the top of the list down to the least dense at the bottom of the list. You don't have to calculate the density in "pennies per cup" since you always measured the same volume of material--just look at your data to make the list.

After you have made your list of the densities of the materials, think about the following important question: How do you think the density of a substance would change if you measured it on the Moon rather than on the Earth?

Additional questions to think about:

1. You have an object and you want to know if it will float in water. To answer the question: "will it float?" do you need to know the objects mass, weight or density?

2. A student's mass on Earth is 50 kilograms. If this student went to the Moon, would her mass be more, less, or the same?

3. A student's weight on Earth is 100 pounds. If this student went to the Moon, would he weigh more, less, or the same?

4. A block of wood easily floats on water when on the Earth. If the same block of wood were taken to the Moon, would it float on water? (Really give this some careful thought, most people can not answer this question.)


Module 5 -- Center of Mass: definition

When describing the translational motion of an extended object, we can represent the object as a point particle, as if all its mass was concentrated at one point. This point, located at the average location of the distributed mass, is called the center of mass. For example, a uniform rod has its center of mass right at its middle. In this module we will learn how to calculate the center of mass of an object or a group of objects.

Moreover, we will show that the overall translational motion of a system can be described in terms of the center of mass of the system. The system under consideration can be a group of particles, such as the molecules of a gas in a container, or a group of extended objects such as the boy and the girl playing on the icy lake from the previous unit.

The athlete shown in the movie below is an example of a non-rigid extended object. The motion of all the different parts of his body can be complicated, but the motion of the athlete's center of mass (red solid path) is the same as the motion of a point particle with a mass equal to the athlete's mass. The two dash lines represent the paths of other parts of the extended object (the head and right foot). While the athlete is in air, the external force acting on the athlete is gravity. As a result, his center of mass follows a projectile motion.

We will see that the location of the center of mass can be either fixed relative to an object, as in the example of the athlete, or it can be a point in space (e.g. a discrete distribution of masses as in examples 1 and 2 below).

The center of mass of a system of 2 point particles

Consider a system formed by two point particles of masses m1 and m2 located at positions and measured with respect to point Q as shown in the figure below. The position of the center of mass with respect to the coordinate system with origin at point Q is obtained as:

Then, the position of the center of mass is:

The center of mass, red cross in the figure, is located along the line joining both masses at a distance d/3 from particle 1, the more massive particle.

b) What will be your answer if the origin of the coordinate system is in particle 2?

The position vectors of the two particles measured from the new coordinate system:


Comparing the above result, with the answer obtained in part a):

we conclude that the position of the center of mass depends on the coordinate system used but its distance relative to particle 1 is still d/3

Example 2: Two identical particles

The position of particle 1:

,

then, the position of the center of mass is:

Knowing that M1 = M2, the masses cancel to have:


The center of mass is shown in red in the figure. It is located along the line joining both masses and at the midpoint between the equal masses.

:

1. Select a coordinate system:

In this problem we are asked to use a coordinate system with origin at the vertex of the L-shaped object. We will pick the orientation of the coordinate systeme in such a way that the rods coincides with the x and y axis as shown below:

Now we apply the definition of the center of mass:

.

One way to check if your calculation makes sense is to use the following graphical tool:

As shown in the figure below, the center of mass of the 2 rod system is the blue circle in the intersection of the vertical and horizontal dash lines. The center of mass of this system is at the midpoint because both rods have the same mass.

1. Select a coordinate system:

Let's consider the rod to be along the x-axis. We will choose a coordinate system with origin at the left end of the rod.

2. Divide the extended object in small elements of mass dm.

The rod can be thought to be formed by a continuous distribution of small particles of mass dm. Each of these particles are occupying a given volume inside the object. In the case of this problem, where the rod is assumed to be thin, a particle occupies an element of the rod of length dx.

3. Define the position of the mass element dm.

With respect to our coordinate system, the element of mass dm is located at position:


. We realize that we need to have a relationship between the mass element dm and its position vector . This relationship is obtained from the information about the mass distribution within the object.


In the case of the thin rod, all the mass is distributed along the x-axis, therefore the position of the center of mass is also along the x-axis:


We need to find the relationship between dm and x. In this problem, the mass is distributed uniformly along the rod. This statement implies that the amount of mass dm in each element of rod dx is the same and independent of the position. As a result, the ratio dm/dx = constant.

The ratio dm/dx is called mass density per unit length and is noted with the letter λ:λ = dm/dx..


If the mass is distributed along a line in the -axis then dm = λ dx .

If the mass is distributed along a line uniformly, the λ = constant.


Using this relationship in the integral above we have:


To obtain the value of λ we use the fact that the total mass and the length of the rod are M and L, then:

5. Solve the integral. Replacing λ = M/L in (eq. 1) we obtain:


The mass is not uniformly distributed, because the mass density increases linearly with position. Therefore we write that:

where b is a positive constant.

Using this relationship in the integral above we have:


Note that the value of b is not given but it can be obtained using that the total mass of the rod is M:

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and the position of the center of mass is:

The center of mass is now shifted towards the right end. This is reasonable because the further we go from the left end, the more massive the mass elements.

Center of mass of a symmetric object: If the object is symmetric and the mass is distributed uniformly throughout the object's volume, then the center of mass lies on an axis of symmetry. For example, the center of mass of a uniform sphere or cube lies at its geometric center.


Center of mass of an asymmetric object: If an object has an irregular shape the center of mass can be determined experimentally as shown in the video below by TSG @ MIT Physics

This video is provided by the Technical Services Group of the MIT Department of Physics.
Used with permission.
Video: Finding the Center of Massachusetts.

The center of mass and the center of gravity of a rigid object

The gravitational force acts on each element of mass within a rigid object. Adding the forces on each element, we obtain the total gravitational force on the object of mass M . This force is applied at a given point in the object that is called the center of gravity of the object. If the acceleration of gravity is the same throughout the object, then the center of gravity coincides with the center of mass. When an object is close to the Earth surface this is a very good approximation. If an object is pivoted on its center of gravity it balances in any orientation (no rotation). This is not true when an object is far from the Earth surface, for example an artificial satellite.

Question: Where is the center of gravity of Massachusetts?
The center of gravity is in Worcester

Replace the right piece by a point particle of mass mR and the left piece by a point particle of mass mL. Place each of the masses in their center of mass. Measured with respect to the baseball center of mass, the center of mass of the right mass is at a distance dR and the one of the left mass is at a distance dL. Because of the way the mass is distributed throughout the right and left pieces we can say that dR < dL. Then the center of mass is closer to the right piece. The right mass is the largest.


Watch the video: Mass and center of mass with triple integrals KristaKingMath (September 2021).