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GRUPO
5
INTEGRANTES
APARCO RAMOS WILLIAM OSWALDO20080056B
AUQUI TOVAR ALEXANDER JOSE
20101141CGOICOCHEA CRUZADO EDIXON
20101211AERIA MORENO EDSON A!
20102558EEXAMEN PARCIAL 201"#III MC#""8S$%%&'() A B*+,-$)
C
TEMA) MOVIMIENTO RELATIVO EN 2D
BLOQUE C .PARCIAL 201"#"
En el mecanismo mostrado ⃗ω2 = -10 K (rad/s) constante. En el
instante
mostrado determine:
a) La magnitud de la velocidad angular absoluta del disco 3.
(rad/s).
b) La magnitud de la velocidad angular absoluta de la barra
4.(rad/s).
c) La aceleracin angular absoluta del disco 3.(rad/
s2
)
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d) La aceleracin angular absoluta de la barra 4.(rad/
s2
)
SOLUCION:
!"LL"#$% 'L%:
$*+% 3,""
4.33
,
%4
1.
2.
$*+% 2 3.210"
2m4
%2
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%5"#$% 6EL%+*$"$ , E7E+% $E %4:
⃗V B=⃗V O4
+⃗ω4 x⃗ R B /O
4
+⃗V B /O4
⃗V O4
=0 ;⃗ V B /O4
=0
V B=⃗ω4 x RB /O4
⃗V B=ω4 k̂ x (−7.9768 î−0.6087 ̂j)
⃗V B=−7.9768ω4 ̂j+0.6087ω4 î
$E *'"L 8%5" %5"#$% 6EL%+*$"$ $E " E7E+% $E %2:
V A=V O2
+⃗ω2 x R A /O
2
+V A /O2
V O
2
=0 ; V A /O2
=0
V A=⃗ω2 x R A /O2
⃗V A=−10 k̂ x (−1.4142 î+1.4142 ̂j)
⃗V A=+14.142 î+14.142 ̂j
6EL%+*$"$ $E 72 E7E+% $E ":
V P2
=V A+⃗ω2 x R P2/ A+V P
2/ A
V P2/ A=0
⃗V P2=14.142 î+14.142 ̂j±10 k̂ x (0.2734
î+2.4850 ̂j)
⃗V P2
=14.142 î+14.142 ̂j−2.734 ̂j+24.850 î
⃗V P2
=38.9920 î+11.4080 ̂j
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6EL%+*$"$ $E 73 E7E+% $E 72:
⃗V P3
=⃗V P2
+⃗ω2 x⃗ R P
3/ P
2
+⃗V P3/ P
2
R P3/ P
2
=0; V P3/ P
2
=0
⃗V P3
=⃗V P2
=38.9920 î+11.4080 ̂j
6EL%+*$"$ $E , E7E+% $E 73:
V B=V P
3
+⃗ω3 x R
B / P3
+V B/ P3
V B / P3
=0
⃗V B=38.9920 î+11.4080 ̂j+ω3 k̂ x (0.1640
î+1.4910 ̂j)
⃗V B=38.9920 î+11.4080 ̂j+0.1640ω3
̂j−1.4910ω3 î ¿
⃗V B=(38.9920−1.4910ω3)î+(8.0550+0.1640ω3) ̂j
*'"L"#$% L" 6EL%+*$"$E $E , E7E+% $E %4 9 %2:
⃗V B=(38.9920−1.4910ω3)î+(8.0550+0.1640ω3) ̂j
⃗V B=0.6087ω4 î−7.9768ω4 ̂j
⃗ω3=26.7886 rad /s↺
⃗ω4=−1.5605 rad /s↺
+"L+L"#$% "+ELE"+*# $E , E7E+% $E %4:
⃗ω
⃗ω4 x (¿¿4 x⃗ R B /O4)+2.⃗ω4 x⃗ V B
/O4
+⃗aB/O4
⃗aB=⃗aO4
+⃗α 4 x⃗ R B/O
4
+¿
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⃗aO4
=0 ; V B /O4
=0 ;⃗ aB /O4
=0
⃗ω
⃗ω4 x (¿¿4 x⃗ R B /O4)⃗
aB=⃗α 4 x⃗
R B /O 4+¿
⃗aB=α 4 k̂ x (−7.9768 î−0.6087
̂j)−1.5605 k̂ x (−1.5605 k̂ x (−7.9768
î−0.6087 ̂j))
⃗aB=(−7.9768α 4 ̂j+0.6087α 4 î)+19.4247
î+1.4822 ̂j
⃗aB=(0.6087α 4+19.4247 ) î+(1.4822−7.9768α 4)
̂j
+"L+L"#$% L" "+ELE"+*# $E " E7E+% $E %2:⃗ω
⃗ω2 x (¿¿2 x⃗ R A /O 2)+2.⃗ω2 x⃗
V A /O2
+⃗a A /O2
⃗a A=⃗aO2
+⃗α 2 x⃗ R A /O
2
+¿
⃗aO2
=0 ;⃗ α 2=0 ; V A /O
2
=0 ;⃗ a A /O2
=0
⃗ω
⃗ω2 x (¿
¿2 x⃗ R A /O 2)⃗a A=¿
⃗a A=−10 k̂ x(−10 k̂ x (−1.4142 î+1.4142 ̂j ))
⃗a A=(+141.42 î−141.42 ̂j )
+"L+L"#$% L" "+ELE"+*# $E 72 E7E+% $E ":
⃗ω
⃗ω2 x (¿¿2 x⃗ R P 2/ A )+2.⃗ω2 x⃗
V P2/ A+⃗a P
2/ A
⃗a P2
=⃗a A+⃗α 2 x⃗ R P2/ A+¿
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⃗ω
⃗ω2 x (¿¿2 x⃗ R P 2/ A )+2.⃗ω2 x⃗
V P2/ A+⃗a P
2/ A
⃗a P2
=⃗a A+⃗α 2 x⃗ R P2/ A+¿
⃗α 2=0
; V P2 / A=0
;⃗
a P2 / A=0
⃗a P2
=141.42 î−141.42 ̂j±10 k̂ x (−10 k̂ x (0.2734
î+2.4850 ̂j ))
⃗a P2
=141.42 î−141.42 ̂j+ (−27.34 î−248.50 ̂j )
¿
⃗a P2
=141.42 î−141.42 ̂j+ (−27.34 î−248.50 ̂j )
¿
⃗a P2=114.08 î−389.92
^ j
+"L+L"#$% L" " "+ELE"+*# $E 73 E7E+% $E 72:
⃗ω
⃗ω2 x (¿¿2 x⃗ R P 3/ P2)+2.⃗ω2 x⃗
V P3/ P
2
+⃗a P3/ P
2
⃗a P3
=⃗a P2
+⃗α 2 x⃗ R P
3/ P
2
+¿
⃗α 2=0 ; R P
3/ P
2
=0; V P3/ P
2
=0
⃗a P3/ P
2
=ω
2
3/2. ρ1 . ρ2
ρ1+ ρ
2
.(^ μ)
ω3=ω
2+ω
3/2
ω3 /2=36.7886 rad /s
⃗a P3/ P
2
=(36.7886)2
(3.75 )4
.(0.994 ̂j+0.1093 î)
⃗a P3/ P
2
=138.6813 î+1261.2006 ̂j
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⃗a P3
=114.08 î−389.92 ̂j+138.6813 î+1261.2006 ̂j
⃗a P3
=252.7613 î+871.2806 ̂j
!"LL"#$% L" "+ELE"+*# $E , E7E+% $E 73:
⃗ω
⃗ω3 x (¿¿3 x⃗ RB/ P3)+2.⃗ω3 x⃗ V B
/ P3
+⃗aB / P3
⃗aB=⃗a P3
+⃗α 3 x⃗ R B / P
3
+¿
V B / P
3
=0 ;⃗ aB / P3
=0
⃗ω
⃗ω3 x(¿¿3 x⃗
RB / P3)⃗aB=⃗a P
3
+⃗α 3 x⃗ RB / P3
+¿
⃗aB=252.7613 î+871.2806 ̂j+α 3 k̂ x
(0.1640 î+1.4910 ̂j )+26.7886 k̂ x (26.7886
k̂ (0.1640 î+1.4910 ̂j))
⃗aB=252.7613 î+871.2806 ̂j+(0.1640α 3
̂j−1.4910α 3 î )+(−117.6911 î−1069.9849
̂j)
⃗aB=(252.7613−1.4910α 3−117.6911)
î+(871.2806+0.1640α 3−1069.9849) ̂j
*'"L"#$% L" "+ELE"+*%#E $E , E7E+% $E %4 9 %2:
⃗aB=(252.7613−1.4910α 3−117.6911)
î+(871.2806+0.1640α 3−1069.9849) ̂j
⃗aB=(0.6087α 4+19.4247 ) î+(1.4822−7.9768α 4)
̂j
α 3=67.8899 rad /s2
↺
α 4=23.7043 rad / s2↺
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EL"$% $EL ,L%;E +
