is it 5:1
or i am going to sleep amd reply tommorow.. :P
portion AB of wedge is rough and BC is smooth.A solid cylinder rolls without slipping from A to B.If AB=BC,then the ratio of translational KE to rotational KE when cylinder reaches point C is??
u have not given values....
but the concept should b that it will accelerate from a to b(both rotationally and linearly)
and after that it will accelerate only linearly.calculate the values using this info.
was this useful.
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at B mgh/2=1/2mv2+(1/2)mr2/2ω2=3/4mr2ω2
v =ωr
h is total ht of wedge...
now from B to C only trans KE will increase by mgh/2 and rot KE will be same so...
at C
Tran KE=1/2m(ωr)2+mgh/2
put mgh/2 from above...
trans KE=5/4mr2ω2
now rot KE is same as that was at B i.e. (1/2)mr2/2ω2
trans KE at C = 5/4mr2ω2
rot KE at C mr2ω2/4
hence 5:1..
[1]
is it alr8...
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