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![[Post New]](/templates/default/images/icon_minipost_new.gif) 8 Feb 2008 02:48:44 IST
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linear density of belt=p
m of belt=(2l+2pier)p
velocity of part DC=0 as belt is in contact with bottom ofroller which is contact with ground having v=0 for no slip
KE of BC=0.5lpvsquare=k1
where v=v+rw=2v
therefore k1=2lpvsquare
KE of APD + BQC=2*0.5Iw square
=2pieRpv square
therefor total ke of belt=(2l+2pieR)pvsquare=mvsquare
now ke of rollers=2*(0.5mvsquare +0.5Iwsquare)
=mvsquare+Iwsqaure
taking roller as hollow cylinder
ke=mvsquare +mvsquare=2mvsquare
therefore total ke=3mvsquare
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Impossible To be Impossible is Impossible |
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