Several mE 3ts students hole the system of rigid 40 dies in

Several mE 3ts students hole the system of rigid 40 dies in Tae position shown in plane. At Professo Huituer\'s signal tical ney release the system from rest Determine ne Mital angula accederetion 0,8m. 7m. \"AB is liguti 0.8 m lon ec is a form plate l o m x 0,7m, Ma ss 2 is a u form disk diameter 0,3 m mass l.7s kg CD DE is an ezulaterne triangular plate O.Sm on eau side, m ass I.D ka mass Center at Cen r E centroidel radius of oils m Connections A, D are Su oof pins. Skow all of your work

Solution

solution:

1)M.I. of each body about there C.G> is as follows

i)for rod=Ir=mL^2/12=.8^2/12=.0533 kg m2

ii)for rectangle=Irt=m/12(b^2+h^2)=.3104 kg m2

iii)for disc=Id=mr^2/2=.14 kg m2

iv)for triangle=It=mk^2=.0225 kg m2

2)here weight of rectangle and disc will act through node point and balance at end support but end reaction will not cause torque on system,hence weight at node point are

Mb=2.5/2=1.25 kg

Md=1.75/2=.875 kg

mc=.875+1.25=2.125 kg

hence torque of each element are about C.G. are as follows

i)for rod,Tr=-1.25*9.81*.4=-4.905 kg m2

ii)for rectangle=Trt=1.25*9.81*.5-2.125*.5*9.81=-4.29185 kg m2

iii)for disc=Td=2.125*.4-.875*.4=4.905 kg m2

iv)for triangle=Tt=.875*.375*9.81=3.220 kg m2

for triangle distnace of point D from E=2*h/3=.3752 m

h=.65sin60=.5629 m

3)hence angular accelaration of each memeber are

i)for rod=Tr=Ir*ar

ar=-4.905/.0533=-91.96 rad/s2

ii) for rectangle=Trt=Irt*art

art=-4.29185/.3104=-13.826 rad/s2

iii)for disc=Td=Id*ad

ad=4.905/.14=35.035 rad/s2

iv)for triangle=Tt=It*at

at=3.220/.0225=143.13 rad/s2