hey, sorry about going off subject, i would like to ask a question on physics. Since there isn't anyone in that section i am helpless. here is my question: A cylinder of radius 20cm and mass 100g rolls down an inclined plane of height 60cm without slipping. the linear velocity of the cylinder at the bottom is?? (g=980cm/s/s)

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hey, sorry about going off subject, i would like to ask a question on physics. Since there isn't anyone in that section i am helpless. here is my question: A cylinder of radius 20cm and mass 100g rolls down an inclined plane of height 60cm without slipping. the linear velocity of the cylinder at the bottom is?? (g=980cm/s/s)

Mathematics
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mgh = KE
1/2 mv^2+1/2 I w^2= Mgh
i did that but i m not gettin the answer.

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Other answers:

KE = KE(r) + KE(t) 1/2(mr^2/2)w^2 + 1/2 x m x v^2
rw=v w=v/r substitute
3/4 x mv^2 = mgh
i went through all that buddy. can you solve ang post your anser.
*and *answer
v=2.8 m/s
ow rats! i made a calculation mistake! neway thnx buddy..
no prob
wat is the radius of curveture if a particle moves in a trajectory given by y=f(x) at any point(x,y)
parameterize the eqn in terms of time....r =v^2/a(radial)
it is given as \[R=[1+(dy/dx)^{2}\left[ \right]^{3/2}\]
divied by d^2y\dx^2
if u know the answer then y do u ask..i dont memorize formulae..its wiser to go through the basic approach...
i want to kno how it came

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