IIT JEE , AIEEE Forum - Physics , Mechanics

asish

a smooth sphere of radius 'R' is made to translate in a st.line with a const. accln 'a'
a particle kept on top of sphere is released frm there at zero velocity wrt sphere.find speed of particle wrt sphere as function of the angle '

' it slides
pls reply with explanation

tarun_bits

apply the conservation of energy to get the expression for velocity at angle @ ...then draw the FBD

apply the pseudo forces also...and the centrifugal force mv2/r..

balance the forces out and get expression for v as terms of @/..

hope it helps...ask for further doubts

akshay.khare91

At which instant we have to find the speed of particle????

asish

wen it makes an angle

anchitsaini

from the FBD --

m= mass of particle

m dv/dt = tangential acceleration = ma cos @ + mg sin @

multiplying by v on lhs and Rd@/dt on rhs as v = R d@/dt

mv dv/dt = ma cos @R d@/dt + mg sin @R d@/dt

v dv = a R cos @d@ + gR sin @ d@

integrating --

v² / 2 = aR sin@ - gR cos@ + c

where c is constant of integration

at @ = 0 , v is also 0

putting this

c = gR

hence

v² / 2 = aR sin@ - gR cos@ + gR

v =

[2R( asin@ - g cos@ + g) ]

destinationIIT09

can someone plz help me with the FBD

i dont seem 2 get the first step

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