linear momentum center of mass - de anza...

Linear MomentumCenter of Mass

Lana Sheridan

De Anza College

Nov 14, 2017

Last time

• the ballistic pendulum

• 2D collisions

• center of mass

• finding the center of mass

Overview

• center of mass examples

• center of mass for continuous mass distributions

Center of Mass

Last lecture we introduced the center of mass, the average point ofthe mass for a collection of particles:

rCM =1

M

∑i

mi ri

This is a vector equation, so we can also express it as a set ofcomponent equations, one for each dimension.

Example 9.10 - Three Particles in a plane

A system consists of three particles located as shown. Find thecenter of mass of the system. The masses of the particles arem1 = m2 = 1.0 kg and m3 = 2.0 kg.

9.6 The Center of Mass 269

Example 9.10 The Center of Mass of Three Particles

A system consists of three particles located as shown in Figure 9.18. Find the cen-ter of mass of the system. The masses of the particles are m1 5 m2 5 1.0 kg and m3 5 2.0 kg.

Conceptualize Figure 9.18 shows the three masses. Your intuition should tell you that the center of mass is located somewhere in the region between the blue particle and the pair of tan particles as shown in the figure.

Categorize We categorize this example as a substitution problem because we will be using the equations for the center of mass developed in this section.

S O L U T I O N

We can express the vector position of the center of mass of an extended object in the form

rSCM 51M

3 rS dm (9.34)

which is equivalent to the three expressions given by Equations 9.32 and 9.33. The center of mass of any symmetric object of uniform density lies on an axis of symmetry and on any plane of symmetry. For example, the center of mass of a uni-form rod lies in the rod, midway between its ends. The center of mass of a sphere or a cube lies at its geometric center. Because an extended object is a continuous distribution of mass, each small mass element is acted upon by the gravitational force. The net effect of all these forces is equivalent to the effect of a single force M gS acting through a special point, called the center of gravity. If gS is constant over the mass distribution, the center of grav-ity coincides with the center of mass. If an extended object is pivoted at its center of gravity, it balances in any orientation. The center of gravity of an irregularly shaped object such as a wrench can be determined by suspending the object first from one point and then from another. In Figure 9.16, a wrench is hung from point A and a vertical line AB (which can be established with a plumb bob) is drawn when the wrench has stopped swinging. The wrench is then hung from point C, and a second vertical line CD is drawn. The center of gravity is halfway through the thickness of the wrench, under the intersec-tion of these two lines. In general, if the wrench is hung freely from any point, the vertical line through this point must pass through the center of gravity.

Q uick Quiz 9.7 A baseball bat of uniform density is cut at the location of its cen-ter of mass as shown in Figure 9.17. Which piece has the smaller mass? (a) the piece on the right (b) the piece on the left (c) both pieces have the same mass (d) impossible to determine

Figure 9.17 (Quick Quiz 9.7) A baseball bat cut at the location of its center of mass.

rCMS

2

021

1

3

y (m)

x (m)3

m1 m2

m3

Figure 9.18 (Example 9.10) Two particles are located on the x axis, and a single particle is located on the y axis as shown. The vector indi-cates the location of the system’s center of mass.

continued

y

x

z

riS

rCMS

CM!mi

An extended object can be considered to be a distribution of small elements of mass !mi .

Figure 9.15 The center of mass is located at the vector position rS

CM, which has coordinates xCM, yCM, and zCM.

AB

C

D

The wrench is hung freely first from point A and then from point C.

The intersection of the two lines AB and CD locates the center of gravity.

A

B

Figure 9.16 An experimental technique for determining the center of gravity of a wrench.

1Serway & Jewett, page 269.

Example 9.10

2 dimensions: can find the x and y coordinates of the center ofmass separately.

Total mass, M = 2(1.0 kg) + 2.0 kg = 4 kg.x-direction:

xCM =1

M

∑i

mixi

=1

M(m1x1 +m2x2 +m3x3)

=1

4(1× 1 + 1× 2 + 2× 0)

= 0.75 m

Example 9.10

x-direction: xCM = 0.75 m

y -direction:

yCM =1

M

∑i

miyi

=1

M(m1y1 +m2y2 +m3y3)

=1

4(1× 0 + 1× 0 + 2× 2)

= 1.0 m

Putting both components together:

rCM = (0.75 i+ 1.0 j) m

Continuous mass distribution

On a microscopic scale we don’t really believe mass is continuouslydistributed. (Atoms, etc.!)

However, on a macroscopic scale it sure seems like it is, and its avery good approximation. It is also much easier than trying todescribe every last atom / subatomic particle in a solid.

We need to revise our previous center of mass (CM) definition tosuit this case:

rCM = lim∆mi→0

1

M

∑i

ri ∆mi

rCM =1

M

∫r dm

Continuous mass distribution

rCM =1

M

∫r dm






S O L U T I O N


rSCM 51M

3 rS dm (9.34)




rCMS

2

021

1

3

y (m)

x (m)3

m1 m2

m3


continued

y

x

z

riS

rCMS

CM!mi




AB

C

D



A

B


Discrete vs Continuous mass distribution

Compare a discrete and continuous mass distribution along 1dimension.

Assume all particles have equal mass and each are separated byone meter. Where is the center of mass of this distribution?

y

x


y

x

The calculation would be:

xCM =1

M

∑i

mixi

=1

6m

5∑i=0

m × i

= 2.5 m

The center of mass is at the midpoint of the line of point masses.


Center of mass of a rod of uniform density.

Density, ρ = MV , is mass per unit volume. Here however, we will

consider a rod that can be treated as 1 dimensional.

We need the mass per unit length: λ = ML .

270 Chapter 9 Linear Momentum and Collisions

Use the defining equations for the coordinates of the center of mass and notice that zCM 5 0:

xCM 51M a

imixi 5

m1x1 1 m2x2 1 m3x3

m1 1 m2 1 m3

511.0 kg 2 11.0 m 2 1 11.0 kg 2 12.0 m 2 1 12.0 kg 2 10 2

1.0 kg 1 1.0 kg 1 2.0 kg5

3.0 kg # m4.0 kg

5 0.75 m

yCM 51M a

imi yi 5

m1y1 1 m2y2 1 m3y3

m1 1 m2 1 m3

511.0 kg 2 10 2 1 11.0 kg 2 10 2 1 12.0 kg 2 12.0 m 2

4.0 kg5

4.0 kg # m4.0 kg

5 1.0 m

Write the position vector of the center of mass:

rSCM ; xCM i 1 yCM j 5 10.75 i 1 1.0 j 2 mExample 9.11 The Center of Mass of a Rod

(A) Show that the center of mass of a rod of mass M and length L lies midway between its ends, assuming the rod has a uniform mass per unit length.

Conceptualize The rod is shown aligned along the x axis in Figure 9.19, so yCM 5 zCM 5 0. What is your prediction of the value of xCM?

Categorize We categorize this example as an analysis problem because we need to divide the rod into small mass elements to perform the integration in Equa-tion 9.32.

Analyze The mass per unit length (this quantity is called the linear mass density) can be written as l 5 M/L for the uni-form rod. If the rod is divided into elements of length dx, the mass of each element is dm 5 l dx.

S O L U T I O N

x

dm = l dxy

dx

x

L

Figure 9.19 (Example 9.11) The geometry used to find the center of mass of a uniform rod.

Use Equation 9.32 to find an expression for xCM: xCM 51M

3 x dm 51M

3L

0 xl dx 5

l

M

x2

2`L0

5lL2

2M

Substitute l 5 M/L: xCM 5L2

2MaM

Lb 5 1

2 L

One can also use symmetry arguments to obtain the same result.

(B) Suppose a rod is nonuniform such that its mass per unit length varies linearly with x according to the expression l 5 ax, where a is a constant. Find the x coordinate of the center of mass as a fraction of L.

Conceptualize Because the mass per unit length is not constant in this case but is proportional to x, elements of the rod to the right are more massive than elements near the left end of the rod.

Categorize This problem is categorized similarly to part (A), with the added twist that the linear mass density is not constant.

Analyze In this case, we replace dm in Equation 9.32 by l dx, where l 5 ax.

S O L U T I O N

Use Equation 9.32 to find an expression for xCM: xCM 51M

3 x dm 51M

3L

0 xl dx 5

1M

3L

0 xax dx

5a

M 3

L

0 x2 dx 5

aL3

3M

▸ 9.10 c o n t i n u e d

Where is the center of mass?


How to do the calculation:

xCM =1

M

∫x dm

=1

M

∫x dm

Observe that dm = λ dx:

xCM =1

M

∫x(λ dx)

=1

Mλ

∫L0x dx


Observe that dm = λ dx:

xCM =1

M

∫x(λ dx)

=1

Mλ

∫L0x dx

=1

Mλ

(L2

2− 0

)=

1

M

(M

L

)L2

2

=L

2

In this case (uniform density) the center of mass is at the center ofthe rod.

Center of Mass Intuition Question

Quick Quiz 9.71 A baseball bat of uniform density is cut at thelocation of its center of mass as shown. Which piece has thesmaller mass?






S O L U T I O N


rSCM 51M

3 rS dm (9.34)




rCMS

2

021

1

3

y (m)

x (m)3

m1 m2

m3


continued

y

x

z

riS

rCMS

CM!mi




AB

C

D



A

B


(A) the piece on the right

(B) the piece on the left

(C) both pieces have the same mass

(D) impossible to determine


Center of Mass Intuition Question

Quick Quiz 9.71 A baseball bat of uniform density is cut at thelocation of its center of mass as shown. Which piece has thesmaller mass?






S O L U T I O N


rSCM 51M

3 rS dm (9.34)




rCMS

2

021

1

3

y (m)

x (m)3

m1 m2

m3


continued

y

x

z

riS

rCMS

CM!mi




AB

C

D



A

B


(A) the piece on the right

(B) the piece on the left ←(C) both pieces have the same mass

(D) impossible to determine


Continuous mass distribution: Another ExampleCenter of mass of a cylinder of uniform density, ρ.

10.6 Calculation of Moments of Inertia 309

Conceptualize To simulate this situation, imagine twirling a can of frozen juice around its central axis. Don’t twirl a nonfrozen can of vegetable soup; it is not a rigid object! The liquid is able to move rela-tive to the metal can.

Categorize This example is a substitution problem, using the defini-tion of moment of inertia. As with Example 10.7, we must reduce the integrand to a single variable. It is convenient to divide the cylinder into many cylindrical shells, each having radius r, thickness dr, and length L as shown in Figure 10.16. The density of the cylinder is r. The volume dV of each shell is its cross-sectional area multiplied by its length: dV 5 L dA 5 L(2pr) dr.

S O L U T I O N

L

dr

z

r

R

Figure 10.16 (Exam-ple 10.8) Calculating I about the z axis for a uniform solid cylinder.

Express dm in terms of dr : dm 5 r dV 5 rL(2pr) dr

Substitute this expression into Equation 10.20: Iz 5 3r 2 dm 5 3r 2 3rL 12pr 2 dr 4 5 2prL 3R

0 r 3 dr 5 1

2 prLR 4

Use the total volume pR2L of the cylinder to express its density:

r 5MV

5M

pR 2L

Substitute this value into the expression for Iz: Iz 5 12pa M

pR 2LbLR 4 5 1

2MR 2

Check this result in Table 10.2.

What if the length of the cylinder in Figure 10.16 is increased to 2L, while the mass M and radius R are held fixed? How does that change the moment of inertia of the cylinder?

Answer Notice that the result for the moment of inertia of a cylinder does not depend on L, the length of the cylinder. It applies equally well to a long cylinder and a flat disk having the same mass M and radius R. Therefore, the moment of inertia of the cylinder is not affected by how the mass is distributed along its length.

WHAT IF ?

The calculation of moments of inertia of an object about an arbitrary axis can be cumbersome, even for a highly symmetric object. Fortunately, use of an important theorem, called the parallel-axis theorem, often simplifies the calculation. To generate the parallel-axis theorem, suppose the object in Figure 10.17a on page 310 rotates about the z axis. The moment of inertia does not depend on how the mass is distributed along the z axis; as we found in Example 10.8, the moment of inertia of a cylinder is independent of its length. Imagine collapsing the three-dimensional object into a planar object as in Figure 10.17b. In this imaginary pro-cess, all mass moves parallel to the z axis until it lies in the xy plane. The coordinates of the object’s center of mass are now xCM, yCM, and zCM 5 0. Let the mass element dm have coordinates (x, y, 0) as shown in the view down the z axis in Figure 10.17c. Because this element is a distance r 5 !x2 1 y2 from the z axis, the moment of inertia of the entire object about the z axis is

I 5 3r 2 dm 5 3 1x2 1 y2 2 dm

We can relate the coordinates x, y of the mass element dm to the coordinates of this same element located in a coordinate system having the object’s center of mass as its origin. If the coordinates of the center of mass are xCM, yCM, and zCM 5 0 in the original coordinate system centered on O, we see from Figure 10.17c that

▸ 10.8 c o n t i n u e d

Continuous mass distribution: Another Example

Let’s choose our axes so that z points along the length of thecylinder, and the origin is right in the center of the cylinder.

Probably, you can easily guess where the CM will be.

At the origin!


Let’s choose our axes so that z points along the length of thecylinder, and the origin is right in the center of the cylinder.

Probably, you can easily guess where the CM will be. At the origin!


How could we prove it?

Along the z-direction this cylinder is very similar to the thin rod wejust considered.

Observe that dm = (πR2)ρ dz:

zCM =1

M

∫z(πR2ρ dz)

=(πR2)ρ

πR2Lρ

∫L/2−L/2

z dz

= 0

Continuous mass distribution: Another ExampleAlong the x-direction this cylinder we are looking down on a circle.Let x = r cos θ, y = r sin θ

Observe that dm = (2yL)ρ dx:

(Looking at just the positive x part of the cylinder, viewed topdown.)

1Or, leave the integral in terms of x and use the substitution u = x2.


Along the x-direction this cylinder we are looking down on a circle.Let x = r cos θ, y = r sin θ

Observe that dm = (2yL)ρ dx:

xCM =1

M

∫x(2yL)ρ dx

=Lρ

πR2L

∫0π

r2(2 sin θ cos θ)(−r sin θ dθ)

=ρ

πR2

[−

2

3r3 sin3 θ

]0π

= 0

1Or, leave the integral in terms of x and use the substitution u = x2.


By symmetry the evaluation for yCM must go exactly the same wayas for xCM, therefore,

yCM = 0 .

This is what we expected all along, but this same technique can beused in cases where we cannot guess the answer.

Summary

• center of mass for continuous mass distributions

3rd Collected Homework! due Monday, Nov 20.

Next Test Monday, Nov 27.

(Uncollected) Homework Serway & Jewett,

• Ch 9, onward from page 288. Probs: 45, 47, 48, 49, 50

• Look at example 9.12 and make sure you understand it.

• Read Chapter 9, if you haven’t already.

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