Formulario fisica basica
t – t0 ∆t V0x= V0 cos θ x=V0x t x=(V0 cos θ)t t= V0 sin θ g y=V02 sin2 θ 2g yg= xgtan θ –(1/2)gt yg= xgtan θ –
2
V0y= V0 sin θ y= V0yt-(1/2)gt . vy = v0y -gt tT =2t xT=V0 cos θ tT xT= V02 sin 2θ g
2
w = w0 + αt θ = θ0 + w0t + (½)αt2 w2 =w02 + 2α(θ – θ0)
Inercias especificas esfera: (2/3)mr2 e. Hueca: (2/3)mr2 vara centro:(1/12)ml2 vara lado:(1/3)ml2 lamina:(1/12)m(l2+a2)cilindro:(1/2)mr2 c. Hueco:mr2
ENERGIA ROTACIONAL K = ½ mV2 V = rw
KR = ∑ ki = ∑i ½miVi2 KR = ½ ∑i miri2w2
g x 2V02 cos2 θ
2 g
I = ∑ miri2
KR = ½Iw2MOVIMIENTO CIRCULAR UNIFORME ∆V = V – V0 a = ∆V ∆t a→ = ar→ + at→ V = wr a = V∆r r∆t ac = ar =V2 r
MOMENTUM DE INERCIA I= Lim ∑ ∆miri2 = ∫r2dm ∆m→0
ϱ =m/v ϱ = dm/dv I = ϱ ∫r2dv MOMENTO DE TORSION τ = F sen θ r = Fd d = r sen θ
a = √ar2 + at2 w = θ/t at= αr
... Y LA SEGUNDA LEY DE NEWTON ar = V2 r Fr= mV r
2
∑ τ = τ1 + τ2 + ... + τn τ = Ftr at = αr Ft = mat τ = mαr2 τ = rmat τ = Iα
∑ F = ma Fr = T
Fr= mar T = mV r
2
MAS Fϛ = - kx - kx =max
PENDULO CONICO ∑ Fy = may = 0 ∑ Fr = mar Tsen θ = mar Tcos θ mg Tcos θ = mg Tsen θ = mar V = √Lg sen θ tan θ
ax = - kx /m x(t) = Acos(wt + ϕ) T =2π/w = 2π√ m/k f = 1/T = √ k/m| / 2π w = 2πf = 2π/T tan ϕ = - vi /w xi xi = Acos ϕ vi= - wAsen ϕ A =√(xi + (vi/w)2)
ROTACION DE UN OBJETO S = rθ P (r , θ)v = dx/dt = -wAsen(wt + ϕ) a = dv/dt -Aw2cos(wt + ϕ) = -w2x vmax = wA = A √ k/m amax = w2A = Ak /m
w = θ – θ0 = ∆θ t – t0 ∆t w = dθ dt ∆θ = θ – θ0
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