Sabemos que:

∴∝=a10,75

=a21,5

∴ I D=M k02

∴T−Fr=mB x a1

T=mB∝×0,75+Fr

∴Por τ2 τ=∝ I D=P×1,5−T ×0,75

∴∝ I D=P ×1,5−(mB∝×0,75+mB g μk )0,75

∝(I D+mB0,752¿=P ×1,5−mB g μk

a) w (t )=∫( 12365

t 3+ 121825

t2−294365

)

ω ( t )= 3365

t 4+ 41825

t 3−294365

t+ω0

ω0=20rad

s

Para t=5

ω (5 )= 3365

×54+ 41825

×53−294365

×5+20

ω (5)=21,37249 rads

b)

V B=ω ( s )×0,75

V B=21,37249×0,75

V B=16,02937ms

c)

Calculamos P(s)

P (s )=5×53+52

P (s )=650N

R0=√T 2+P2

R0=√375,068492+6502

R0=750,45077N

d)

T=mB∝×0,75+mB g × μk

T=50×( 12365

×53+ 121825

×52−294365

)+50×9,8×0,5

T=375,06849N

k 0=150×10−3m

M=20kg

I=Mk 02

Tenemos:

A y B son iguales por ser simétricos

F=F A=FB=kdx

dx=300×10−3×θ

∴ Iα=−F A ×θ−FB ×300×10−3

θ=−2 F ×300×10−3

I=−2k ×300×10−3×θ ×300×10−3

M ×(150×10−3)2

θ+ 2kθM

×( 300×10−3

150×10−3 )2

=0

θ+ 8kθM

=0

a)

M eq=M=20kg

b)

k eq=8 k=8×500=4000 Nm

c)

w=√ 400020 =14,1421 rads

V 1=0 T 2=0

T 1=12

I θm2+ 12

m(0,75)2 θ2mV 2=12

k1 x12+12

k2 x22−mg0,75 senθ

T 1=[ 12 .112

m .1,52+ 12

m.0,752] θ2m V 2=12.330 .0,92+1

2.167.1,52−mg .0,75 .θ2m

T 1=0,5739 θ2m V 2=[ 12.330 .0,92+1

2.167.1,52−15 .0,75]θ2m

V 2=310,275θ2m

Sabemosθ2m=ω2nθ2m

T 1+V 1=T2+V 2

a)

0,5739 θ2m+0=0+310,275θ2m ≫0,5739θ2m−310,275θ2m=0

b)

ω2n=310,2750,5739

ωn=23,2517rad

s

c)

V máx C=θm×1,5 θm=23,2517×θm1,5θm=125×10−3

θm=23,2517×0,12515

θm=0,1251 ,5

V máx C=23,2517×

0,12515

×1,5≫2,90646ms

Se desplaza 0,2 m, t=0

c=20 N .sm

k=32 Nm

8 x+20 x+32 x=0

ε= 202√8×32

<1 Subamortiguado

X ( t )=A e−ε ωnsin (ωd t +φ)

ωd=ωn √1−ε2

ωd=frecuencia natural devibración

ωn=frecuencia naturaldel sistema

a)

ωn=√ 328ωn=2

rads

b)

ωd=2√1−2021

ωd=2√ 38ωd=1,224

rads

c)

X (0 )=0 ,2

A=0,2

X (0 )=sinφ

0,20,2

=sinφ

φ=90 °= π2

rad

d)

X (t )=A e−ε ωnsin (1,224 t+90)

Cuando : t=2 s

X (2 )=0,2e−58

× 2× 2sin (1,224×2+ π

2)

X (2 )=−0,0126m