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Jahrdenas_GEAS > Mechanics > Flashcards

Mechanics Flashcards

1
Q

Rotational Inertia of a rotating particle

A

J = m . r^2

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2
Q

Rotational Inertia of a rotating Thin Hoop

A

J = m . r^2

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3
Q

Rotational Inertia of a rotating Disk/Cylinder

A

J = 0.5m . r^2

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4
Q

Rotational Inertia of a rotating Solid Sphere

A

J = (2/5)m . r^2

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5
Q

Rotational Inertia of a rotating Hollow Sphere

A

J = (2/3)m . r^2

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6
Q

Rotational Inertia of a rotating Thin Rod

A

J = (1/12)m . r^2

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7
Q

Formula for Steiner’s/Parallel Axis Theorem

A

J = Jg + md^2

Jg - inertia about the figure’s centroid
d - distance from centroid of figure to point of interest where inertia is to be examined from
m - mass of object

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8
Q

Inertia of a rectangle Rotated at the centroid

A

Ix = (b h^3)/12

note, axis that figure rotates must be parallel to the base(b)

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9
Q

Inertia of a Triangle Rotated at the centroid

A

Ix = (b h^3)/36

note, axis that figure rotates must be parallel to the base(b)

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10
Q

Inertia of a Rectangle/Triangle Rotated at the base

A

Multiply Ix(@centroid) by 3

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11
Q

Inertia of a circle Rotated at the centroid

A

Ix = π(r^4) / 4

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12
Q

Inertia of an ellipse Rotated at the centroid

A

Ix = π(a.b^3) / 4

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13
Q

Inertia of a composite figure

A

I = ∑(Ig + Ad^2)

Ig - Inertia about the figure’s centroid
d - distance from centroid of figure to point of interest where inertia is to be examined from
A - Area of the one of the figures that comprise the composite figure

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14
Q

Formulas used in evaluating banked curves

A

tan θ = V^2 / g.r
tan θ = μ

θ - Banking angle
μ - Coefficient of Friction

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15
Q

Evaluation of a situation where an object slides from the top of an inclined plane to the bottom, when the plane introduces friction

A

When Friction is involved, Kinetic energy is not equal to potential energy:

μ = (tan θ) . (%PE dissipated by Friction)

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16
Q

Formula for Torque(τ) (Involving Linear terms)

A

τ = F . r

F - Force
r - distance from force to point where moment is taken

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17
Q

Derivation of Torque(τ) formula involving rotational terms

A 
τ = F . r ;   F = ma
τ = (m . a . r) ; a = r . α
τ = (m . r . α . r)
τ = (m . r ^2. α ) ; I = m . r^2

Final Eqtn:
τ = I . α

α - angular acceleration

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18
Q

Formula involving belt friction in a pulley

A

Fmax / Fmin = e^(μθrad)

θrad - angle that subtends the belt that is in contact with the pulley
μ - coefficient of friction

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19
Q

Torque involving Belt Friction

A

τ = (Fmax - Fmin) . r

r - radius of pulley

20
Q

Power involving Belt Friction

A

P = (Fmax - Fmin) . Velocity

21
Q

A cable designed to carry a load, and sags due to the load’s weight

A

Parabolic cable

22
Q

Distance of the load to the lowest point of the cable

Found in the middle of the cable’s span

A

Sag (expressed as the variable ‘d’)

23
Q

Variables that express the tension at the Sag point and the tension at the support

A

H - Tension @ Lowest point (Sag point, located @ L/2) also the minimum tension

T - Tension @ Highest point (@ the supports)
also the maximum tension

24
Q

The distance between two supports

A

Span (expressed as the variable ‘L’)

25
Q

Formula for the Length of the parabolic cable(S)

A

S = L + (8.d^2)/(3L) - (32d^4)/(5L^3)

26
Q

Formula for the Tension at the lowest point(H) of a parabolic cable

A

H = (ωL^2)/(8d)

ω - Load distribution of the load(Weight per meter)
L - Span of cable
d - Sag of cable

27
Q

Formula for Tension at the supports(T) of a parabolic cable

A

T = sqrt(H^2 + W^2)

W - the concentrated weight located @ L/4, and comprises half of the total load
W = ω . (L/2)

28
Q

CALTECH: Length of a parabolic cable

A

use STAT a + bx +cx^2, and interpret the situation as a coordinate system

get 3 coordinates regarding the parabolic cable (maybe two points at the supports, and one at the sag point)

29
Q

A cable that sags due to its own weight, and has a graph of y = cosh(x)

A

Catenary cable

30
Q

Formula for the Length of a catenary cable

A

S = L + (8.d^2)/(3L) - (32d^4)/(5L^3)

Answer:
Total Catenary Length = 2S

31
Q

Formula for the height of any point (Ypoint)on the catenary cable with respect to the ground

A

(Ypoint)^2 = (Spoint)^2 + C^2

Spoint - Length of the cable from Ypoint in inquiry to the Sag point
C - the height of the sag point with respect to the ground

32
Q

Formula for Tension at the supports(T) of a Catenary cable

A

T = ωy

NOTE: T is the greatest tension in the cable

ω - cable weight distribution (weight per meter)(WITH RESPECT TO SPAN(L), NOT THE LENGTH(S))
y - height of the supports (Highest point on the cable)

33
Q

Formula for Tension at the Sag Point(H) of a Catenary cable

A

H = ωC

NOTE: T is the least tension in the cable

ω - cable weight distribution (weight per meter)(WITH RESPECT TO SPAN(L), NOT THE LENGTH(S))
C - the height of the sag point with respect to the ground

34
Q

Formula for the Span(L) of a catenary cable

A

X = C ln ((S + y)/C)

Answer:
L = 2X

X - distance from one support to the sag point
C - the height of the sag point with respect to the ground
S = L + (8.d^2)/(3L) - (32d^4)/(5L^3)
y - height of the supports (Highest point on the cable)

35
Q

Maximum Height of a Projectile (Ymax)

A

Ymax = (Vo sinθ)^2 / (2g)

36
Q

Range of a projectile (Max. Horizontal distance)(R)

A

R = (Vo^2) (sin2θ) / g

37
Q

Time Elapsed to complete Projectile Trajectory

A

T = 2Vo sinθ / g

“TWO VEEE, SINE THETA OVER GEEE”…..”ALL YOU NEED IS TIME”

38
Q

Alternate Range Formula (Using time elapsed)

A

R = (Vo cosθ ) . t

39
Q

Derivation of the Instantaneous height(Yinst) of the trajectory given the instantaneous horizontal distance travelled(Rinst)

A 
R = (Vo cosθ ) . t
t = R / (Vo cosθ )  ----- EQTN1

y = (Vo sinθ)t +- 0.5gt^2 —EQTN2

EQTN1 IN EQTN2:
y = (Vo sinθ)(R/(Vo cosθ )) +- 0.5g(R/(Vo cosθ ))^2

Final EQTN:
Yinst = Rinst (tanθ) - (g(Rinst)^2)/(2Vo^2 (cosθ)^2)

40
Q

the Formula for Lami’s Theorem

A

NOTE: USED ONLY WHEN THERE ARE 3 FORCES IN EQUILIBRIUM, & ALL 3 FORCES’ LINE OF ACTION INTERSECT

FA / Sin α = FB / Sin β = FC / Sin θ

FA, FB, and FC are the forces, and α β θ are the opposing angles of the forces respectively

41
Q

Varignon’s Theorem

A

Used when obtaining the centroid for irregular shape:

Xbar = (m1x1 + m2x2 + …) / (m1 + m2 + …)

Ybar = (m1y1 + m2y2 + …) / (m1 + m2 + …)

Xbar/Ybar represent the x and y coordinate of the irregularly shaped figure

42
Q

Where should the Moment point be placed when evaluating Summation of Moment problems?

A

a point where the most number of unknown variables are eliminated (forces that intersect Moment point are ignored)

43
Q

The angle of the projectile that makes the area under the curve of the trajectory to be maximum

A

60 degrees

44
Q

The angle of the projectile that makes the length of arc of the trajectory to be maximum

A

56.5 degrees

45
Q

The Formula for the velocity of a material sliding down an inclined plane

A

V = sqrt(2g(hinit))

hinit - initial height before sliding

46
Q

Formulas for KE/PE of a rolling object from an inclined plane

A

@top of ramp:
PE = mgh
KE = 0

@bottom of ramp:
PE = 0
KE = 0.5mv^2 + 0.5Iω^2

therefore:
mgh = 0.5mv^2 + 0.5Iω^2

47
Q

Formula for the velocity of the conical vector

A

V = sqrt(gr tanθ)

r - radius of cone
θ - angle formed by the string with respect to the normal of the surface

Jahrdenas_GEAS (9 decks)

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