1133(D)
Soumyadeep Basu Yeah, I am also getting something like it. Could you post the solution?Upvote·0· Reply ·2014-02-12 22:53:08
A smooth sphere of radius R is made to translate in a straight line with constant acceleration a=g. A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. The speed of the particle with respect to the sphere as a function the angle θ as it slides down is:
(a) √[Rg(sinθ+cosθ)2] (b) √[Rg(1+cosθ+sinθ)] (c) √[4Rgsinθ] (d) √[2Rg(1+sinθ-cosθ)]
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2 Answers
11331133Take the sphere as reference frame and conserve enegry.
mgR + 0∫θ(ma)(Rdθ)cosθ = mgRcosθ + 12mv2
substitue a = g
Akash Anand Why you are not considering the gravitational forces on small particle?- Sourish Ghosh I did. I conserved potential energy. Am I going wrong somewhere?
- Akash Anand No..you are absolutely correct :)
Aniq Ur Rahman Howddugt dz trm 0∫θ(ma)(Rdθ)cosθ ???- Soumyadeep Basu Thanks.
- Sourish Ghosh @aniq that is the work done by the force 'ma'.
Note: 'ma' is the pseudo force we must consider while taking the sphere as the frame of reference.