IIT JEE , AIEEE Forum - Physics , Mechanics

joyfrancis

q) A particle kept at the bottommost point inside a fixed smooth spherical shell of radius R=1m is givena velocity u such that it is just able to complete one full circle. What is the acceleration of the particle when it's velocity is vertical.

a)g

10

b)g

c)g

2

d)3g

karthik2007

See, here we need the acceleration at the right point, bisecting the circumference on the right hand side. We know u, so use newton's laws and get the ans

akhil_o

When the velocity is vertical,

the height is R

TE=5/2Rg

at side position equating energies we hav

v=rt(3Rg)

so centrifugal force=

m(3g) towards right

also

downward force=mg

so net force

=^{[ ]}

(mg)²+3²(mg)²

mg

10

so acceleration=

g

10

^{[ ]}^{[ ]}

anchitsaini

height difference between 2 pts=r
v at bottommost pt=root5gr
v at req pt=root(5gr-2gr)=root3gr where r=1
a=vsquare/r=3g
also tangential accn=g
therefore a=root(9gsquare+gsqaure) as both are perpendicular
=root(10)g

sidsgr88

feel the ans is a
when v is vertical the string is horizontal
g acts downwards
and the v is sqrt(3rg) by work energy theo
so a centripetal acc v(square)/r is applied ==3g
so by pythagoras net is g(sqrt(10))

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