Anexos Lab 5 Fisica 2
-
Upload
andienaves -
Category
Documents
-
view
11 -
download
0
description
Transcript of Anexos Lab 5 Fisica 2
![Page 1: Anexos Lab 5 Fisica 2](https://reader035.fdocuments.es/reader035/viewer/2022072114/563dbb7e550346aa9aada9e2/html5/thumbnails/1.jpg)
Anexo1
Anexo 2
Calculo del momento de inercia respecto al punto P
I p=I cm+md2
Momento de inercia respecto al centro de masa de una tabla rectangular
I cm=112m (a2+b2 )
I cm=112
(0.518 kg ) [ (0.197m )2+ (0.803m )2 ]
I cm=112
(0.518 kg ) [ (0.197m )2+ (0.803m )2 ]
I cm=2.95x 10−2 kg .m2
I p=(2.95 x10−2 kg .m2 )+ (0.518kg ) (0.228m )2
I p=5.64 x 10−2 kg .m2
I p=(2.95 x10−2 kg .m2 )+ (0.518kg ) (0.194m )2
I p=4.90x 10−2 kg .m2
Anexo 3
Determinación del centro de oscilación M
L=I pmd
![Page 2: Anexos Lab 5 Fisica 2](https://reader035.fdocuments.es/reader035/viewer/2022072114/563dbb7e550346aa9aada9e2/html5/thumbnails/2.jpg)
L= 5.64 x10−2kg .m2
(0.518kg)(0.228m)
L=0.478m
Anexo 4
Calculo del momento de inercia respecto al punto M
Del anexo 2 se sabe que:
I cm=2.95x 10−2 kg .m2
Im=I cm+md2
Im=(2.95 x 10−2 kg .m2)+ (0.518kg ) (0.263m )2
Im=5.64 x 10−2 kg .m2
I cm=2.95x 10−2 kg .m2
---------------------------
Im=I cm+md2
Im=(2.95 x 10−2 kg .m2)+ (0.518kg ) (0.297m )2
Im=7.52 x10−2kg .m2
![Page 3: Anexos Lab 5 Fisica 2](https://reader035.fdocuments.es/reader035/viewer/2022072114/563dbb7e550346aa9aada9e2/html5/thumbnails/3.jpg)
Anexo 5
Calculo de la longitud del péndulo simple equivalente
T=2π √ LgL=T
2g4 π2
Con respecto a “P”
L=(1.38 s)2(9.78 m
s2)
4 π2
L=0.472m=47.2cm
Con respecto a “M”
L=(1.39 s)2(9.78 m
s2)
4 π2
L=0.479m=47.9cm