Chapter 7-9

Question 1

Center of Mass (CM)

Answer 1

the average position of mass for a system of particles

Equation on equation sheet

Question 2

Conservation of Linear Momentum

Answer 2

if the sum of the external forces acting on a system of particles is zero, then the linear momentum of the system must remain constant

-if sum(Fi) = 0 then p is constant since F = dp/dt = 0

Question 3

Torque

Answer 3

or moment, is the rotational counterpart to force

tau = r cross F

Question 4

Conservation of Angular Momentum

Answer 4

if the sum of the external torques acting on a system of particles is zero, then the angular momentum of the system must remain constant

-if sum(taui) = 0 then L is constant since tau = dL/dt = 0

Question 5

Laboratory Coordinates

Answer 5

reference frame in which the origin is at rest in the laboratory

Question 6

Center of Mass (CM) Coordinates

Answer 6

reference frame in which the origin moves with the center of mass

Question 7

Reduced Mass

Answer 7

mu = m1 x m2 / (m1 + m2)

Question 8

Coefficient of Restitution

Answer 8

epsilon = abs(v2’ - v1’) / abs(v2 - v1) = v’ / v

v’ / v = final relative speed / initial relative speed

Question 9

Elastic Collision

Answer 9

coefficient of restitution (epsilon) is equal to 1, so Q = 0 and total Kinetic Energy (T) is constant

Question 10

Inelastic Collision

Answer 10

coefficient of restitution (epsilon) is greater than or equal to 0 and less than 1, so Q > 1 and Kinetic Energy (T) is not constant

Question 11

Rigid Body

Answer 11

non-deformable object (the distance between any two points is constant)

Question 12

Center of Mass for a Rigid Body

Answer 12

rcm = int(r dm) / int(dm)

for a composite object:
rcm = sum(mi (rcm)i) / m

if a uniform body has a plane of symmetry (a plane that divides it in half), the Center of Mass lies in that plane

Question 13

Moment of Inertia for a Rigid Body

Answer 13

I = int[ (r perpendicular)^2 dm]

I = sum(mi ri^2)

Question 14

Moment of Inertia

Answer 14

(I) rotational counterpart to mass (indicates an object’s inertia to being rotated–called “rotational inertia”)

Question 15

Newton’s 2nd Law for Rotation

Answer 15

tau = (Iz) alpha

Question 16

Rotational Equations generated by substituting analogous quantities

Answer 16

x                     theta
v                     omega
a                     alpha
m                    I
p                     L
F                     tau
p = mv            L = I omega
F = ma            tau = I alpha
T = 1/2 m v^2  Trot = 1/2 I omega^2

Question 17

Moment of Inertia for Specific Rigid Bodies

Answer 17

Uniform thin rod rotated about its center of mass:
I = 1/12 m L^2

Uniform thin ring (or cylindrical shell) rotated about its cm:
I = m R^2

Uniform Disk (or cylinder) rotated about its cm:
I = 1/2 m R^2

Question 18

Parallel-Axis Theorem

Answer 18

a rigid body’s movement of inertia (I) about any axis is equal to “I” about a parallel-axis through its cm plus the product of the body’s mass (m) with the square of the distance (L) between the two axes

I = Icm + m L^2

Question 19

Physical (or Compound) Pendulum

Answer 19

a rigid body that is free to rotate about a horizontal axis under its own weight

T = 1/f = 2 pi sqrt(I/(mgL))

Question 20

Laminar Motion

Answer 20

all particles in a rigid body move parallel to a fixed plane

Question 21

Critical Value

Answer 21

(mu crit) the minimum value of coefficient of static friction (mu s) that allows rolling motion with no slipping

Question 22

Moments of Inertia

Answer 22

Moment of Inertia about x-axis:
Ixx = int[ (y^2 + z^2) dm]

Moment of Inertia about y-axis:
Iyy = int[ (x^2 + z^2) dm]

Moment of Inertia about z-axis:
Izz = int[ (y^2 + x^2) dm]

Question 23

Products of Inertia

Answer 23

xy Product of Inertia
Ixy = Iyx = - int(x y dm)

xz Product of Inertia
Ixz = Izx = - int(x z dm)

zy Product of Inertia
Izy = Iyz = - int(z y dm)

Question 24

Moment of Inertia Tensor

Answer 24

9-component mathematical quantity that contains the moments and products of inertia for a rigid body. It is often written as a 3x3 matrix

I = | Ixx Ixy Ixz |
| Iyx Iyy Iyz |
| Izx Izy Izz |

Question 25

Matrix Multiplication

Answer 25

each matrix entry (Mij) is obtained by multiplying corresponding entries in a Row “i” and Column “j” and adding these products together

Question 26

Moment of Inertia Equation

Answer 26

I = Ixx cos^2 (alpha) + Iyy cos^2 (beta) + Izz cos^2 (gamma) + 2 Iyz cos(beta) cos(gamma) + 2 Izx cos(gamma) cos(alpha) + 2 Ixy cos(alpha) cos(beta)

I = (n with a ~ ) I n –Note that right side are all matrices

Question 27

Angular Momentum Equation

Answer 27

L = I dot omega

L = I omega –Note that right side are both matrices

Question 28

Kinetic Energy Equation

Answer 28

T = 1/2 I omega^2

T = 1/2 (omega with a ~ ) I omega –Note that right side are all matrices

Question 29

Principal Axes of a Rigid Body

Answer 29

a set of 3 coordinate axes for which the “products of inertia” are zero

-found by diagonalizing the moment of inertia tensor
0 = | Ixx - Ii Ixy Ixz |
| Iyx Iyy - Ii Iyz |
| Izx Izy Izz - Ii |