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Physics 17 Online
OpenStudy (anonymous):

f

OpenStudy (dls):

|dw:1354088903047:dw| is this the diagram for disc?

OpenStudy (anonymous):

Yes,th disk and the sphere are on the same ramp

OpenStudy (dls):

yeah i cant show it in the same diagram

OpenStudy (dls):

now @Mashy will help you :p

OpenStudy (anonymous):

hmmm use energy conservation and derive the expression for the speed at the bottom of the ramp.. initial energy = pure potential energy final energy = pure kinetic energy of rotation + translation i am assuming its rolling without slip/skid

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

ok so do it step wise.. expression for potential energy and ke of rotation and translation i hope you know!

OpenStudy (anonymous):

mgh=1/2mv^2+1/2iw^2

OpenStudy (anonymous):

for the disk mgh=1/4mr^4v^2+1/2mv^2

OpenStudy (anonymous):

for the sphere mgh+1/5mr^4v^2+1/2mv^2

OpenStudy (anonymous):

exactly.. now calculate v..

OpenStudy (anonymous):

vdisk=9.89

OpenStudy (anonymous):

vsphere=9.89

OpenStudy (anonymous):

is the v correct?

OpenStudy (anonymous):

\[v = \sqrt{2gh/(i/MR ^{2}+1)}\]

OpenStudy (anonymous):

i for disk = \[MR ^{2}/2\] and i for sphere is \[2MR ^{2}/5\] so substitue the values

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