Relative motion,work and energy

[vissh]

A smooth sphere of radius R is made to translate in a straight line with a constant acceleration a . A particle kept on the 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 the angle 'A' it slides.

The Attempt at a Solution

hmmm. can you pls tell me how the particle will move after being released . And hints if you could :D

Thanks in advance :)

 

[Delphi51]

As the sphere accelerates, the particle will be left behind. Given that it slides along the surface, there must be gravity pulling it down perpendicular to the motion of the sphere. In the absence of given information, I would use g = 9.81 and zero friction. The vertical and horizontal motions are not independent; they are constrained by the particle being on the surface of the sphere. Looks like a tough little problem.

 

[vissh]

Thanks Delphi :D Got the answer now :)
Here , the approx diagram i used http://s1102.photobucket.com/albums/g448/vissh/?action=view&current=temp2.jpg"
If the sphere accelerate right,the particle will move left and slide on the sphere due to the downward pull of gravity .
Then,the change in K.E. = 1/2mv² - 0
Next,only two forces will do work gravity and the pseudo force as seen from Sphere.
...W_pseudo :-
For any small displacement ds, dW = ma(ds)cosA = ma(RdA)cosA
And on integrating left from 0 to W and right from 0 to A -- W_pseudo =maRsinA
...W_g = mg[R(1-cosA)]
And thus ,By work energy theorem,
1/2mv² = maRsinA + mg[R(1-cosA)]

And goal achieved for finding v :) Small question indeed :P