Mechanics Flashcards by Mapua Jahrdenas

Rotational Inertia of a rotating particle

J = m . r^2

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Rotational Inertia of a rotating Thin Hoop

J = m . r^2

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Rotational Inertia of a rotating Disk/Cylinder

J = 0.5m . r^2

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Rotational Inertia of a rotating Solid Sphere

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

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Rotational Inertia of a rotating Hollow Sphere

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

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Rotational Inertia of a rotating Thin Rod

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

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Formula for Steiner’s/Parallel Axis Theorem

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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Inertia of a rectangle Rotated at the centroid

Ix = (b h^3)/12

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

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Inertia of a Triangle Rotated at the centroid

Ix = (b h^3)/36

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

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Inertia of a Rectangle/Triangle Rotated at the base

Multiply Ix(@centroid) by 3

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Inertia of a circle Rotated at the centroid

Ix = π(r^4) / 4

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Inertia of an ellipse Rotated at the centroid

Ix = π(a.b^3) / 4

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Inertia of a composite figure

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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Formulas used in evaluating banked curves

tan θ = V^2 / g.r
tan θ = μ

θ - Banking angle
μ - Coefficient of Friction

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

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

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

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Formula for Torque(τ) (Involving Linear terms)

τ = F . r

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

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Derivation of Torque(τ) formula involving rotational terms

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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Formula involving belt friction in a pulley

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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Torque involving Belt Friction

τ = (Fmax - Fmin) . r

r - radius of pulley

Power involving Belt Friction

P = (Fmax - Fmin) . Velocity

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

Parabolic cable

Distance of the load to the lowest point of the cable

Found in the middle of the cable’s span

Sag (expressed as the variable ‘d’)

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

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

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

The distance between two supports

Span (expressed as the variable ‘L’)

Formula for the Length of the parabolic cable(S)

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

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

H = (ωL^2)/(8d) ω - Load distribution of the load(Weight per meter) L - Span of cable d - Sag of cable

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

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

CALTECH: Length of a parabolic cable

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)

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

Catenary cable

Formula for the Length of a catenary cable

S = L + (8.d^2)/(3L) - (32d^4)/(5L^3) Answer: Total Catenary Length = 2S

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

(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

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

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)

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

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

Formula for the Span(L) of a catenary cable

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)

Maximum Height of a Projectile (Ymax)

Ymax = (Vo sinθ)^2 / (2g)

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

R = (Vo^2) (sin2θ) / g

Time Elapsed to complete Projectile Trajectory

T = 2Vo sinθ / g "TWO VEEE, SINE THETA OVER GEEE"....."ALL YOU NEED IS TIME"

Alternate Range Formula (Using time elapsed)

R = (Vo cosθ ) . t

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

``` 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)

the Formula for Lami's Theorem

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

Varignon's Theorem

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

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

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

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

60 degrees

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

56.5 degrees

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

V = sqrt(2g(hinit)) hinit - initial height before sliding

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

@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

Formula for the velocity of the conical vector

V = sqrt(gr tanθ) r - radius of cone θ - angle formed by the string with respect to the normal of the surface