IIT JEE , AIEEE Forum - Physics , Mechanics - a uniform solid cylinder of radius r=15 cm rolls over a horizontal plane passing into an i

anchitsaini

considering pure rotational motion about instantaneous axis of rotation --

initially energy = mgR + 1/2 * I * w²

                     = mgR + 1/2 [ mR²/2 + mR²] (v / R)²

                     = mgR + 1/2 [ 3mR²/2 ] (v / R)²

final energy = mgh + 1/2 [ 3mR²/2 ] (v₁ / R)²

h = R cos @

hence

conserving energy we get --

3mv² / 4 + mgR = 3mv₁² / 4 + mgR cos @      ---------1

also as N = 0

mg cos @ provides necessary centripetal acceleration

we get

mg cos @ = mv₁² / R

v₁² = gR cos @

putting this value in 1

we get

v =

[gR(7 cos @ - 4)] / 3 = 1m/s on putting values

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