IIT JEE , AIEEE Forum - Physics , Mechanics

karthik2007

Let the sphere be given an initial velocity v.

Now, at the topmost part of the cylinder (see figure) , the sphere goes in a circle of radius of R-r.
Let the sphere have a velocity of v' at the top of the cylinder.
So, we have mv'²/(R-r) = mg + N

As we need the minimum speed, N = 0

So mv^'2/(R-r) - mg

or v'² = g(R-r) - (1)

Now, conserving energies, we have:

1/2mv² + 1/2Iw² = mg(2R-2r) + 1/2mv'² + 1/2Iw'²

As it undergoes pure rolling, w = v/r, and I = 2/5mr².

Putting these in the energy equation, we get

7/10mv² = 2g(R-r) + 7/10g(R-r)

or 7/10v² = g(27R-27r)/10

Which gives v =

27g(R-r)/7

Do tell me if I have gone wrong somewhere, as the final answer seems to be a little scary :)
Note: Click the image for better clarity.
Cheers!

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