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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
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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
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OpenStudy (anonymous):
exactly.. now calculate v..
OpenStudy (anonymous):
vdisk=9.89
OpenStudy (anonymous):
vsphere=9.89
OpenStudy (anonymous):
is the v correct?
OpenStudy (anonymous):
\[v = \sqrt{2gh/(i/MR ^{2}+1)}\]
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OpenStudy (anonymous):
i for disk = \[MR ^{2}/2\]
and i for sphere is \[2MR ^{2}/5\]
so substitue the values