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destinationIIT09

A smooth sphere of radius R is made to translate in a straight line with constant acceleration a= g.A particle kept on top of the sphere is released from there at zero velocity with respect to the sphere.Find the speed of the particle with respect to the sphere as a function of angle @ as it slides down?

plz give full soultion

ans: sqaureroot of 2Rg(1 + sin@ - cos@)

destinationIIT09

somebody dooo help

i am stuck

anandghegde

this will help

http://www.goiit.com/posts/list/mechanics-good-one-49224.htm#247796

ganesha1991

Re:Dcpandey:work,power n energy;

destinationIIT09

@ ganesha1991

could u plz post ur solution 2

the one by anchitsani doesnt seem 2 make me satisfied

anandghegde

Ok lemme post a very easy solution..........

The problem becomes dead easy if you work in the frame of the accelarating sphere.

Apply pseudo force.
Now simple apply work energy theorem.
$m.g.R(1-\cos \theta) + ma.R.\sin \theta = \frac{mv^2}{2}$

here maRsin

is the work done by pseudo force. (work = force X displacement in the direction of force. which is R sin

here.)
Now put a=g

$2g.R(1-\cos \theta+\sin \theta) = v^2$

$v = \sqrt{2g.R(1-\cos \theta+\sin \theta) }$

anandghegde

for your reference

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