IIT JEE , AIEEE Forum - Physics , Mechanics

sachin_gupta1991

Q.A hemispherical bowl of radius r=0.1 m is rotating about its own axis (which is vertical), with an angular velocity

. A particle of mass 10^-2kg on the frictionless inner surface of the bowl is rotating with the same

.The particle is at a height h from the bottom of the bowl.

Obtain the relation between h and

. Also find the minimum value of

needed in order to have a non-zero value of h.

pannaguma

let mass of bowl be m1,
and particle m2.
let radius = r.
take @ with vertical.
so perpendicular dist of particle from axis at any time is R=rsin@
and at any time height from bottom H=r(1-cos@)

thus from above
H=R(1-cos@)/sin@

for equilibrium
w^2Rcos@ = gsin@ [force eqn]

substituting from above
H=g(1-cos@)/w^2 * cos@

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