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A train of mass M is moving on a circular track or radius R with a constant speed V. The length of the train is half of the perimeter of the track.What will be the linear momentum of the train. Plz tell in detail...
Hello Gunit,I would like to inform U that one of the Askiitians Expert has already answered this question asked very earlier.Check:http://www.askiitians.com/forums/Mechanics/10/41804/centre-of-mass-linear-momentum-collision.htm
Dear Student,Please find below the solution to your problem.The train fits onto the semicircle and has symmetry about the y−axis as shown in the figure below.For the two points shown on the semicircle, we resolve the velocity vinto horizontal and vertical components.Since the two points are symmetrical about the y−axis, the magnitude of the vertical components(y-components) will be equal and point along y^ and −y^ directions respectively.Hence, both will cancel out.The horizontal components(x-components) will point along the x−axis.Mass of the Train=M;Length of the Train=L=πR;Mass per unit length=λ=πRM;Take a small element at an angle θ on the semicircle which subtends angle dθ.Let dm be the mass of this small element.Let dl= Length of the small element=Rdθ;dm=λdl=πRMRdθ;P=∫0πdpx=∫0πdm(vsinθ)=∫0ππRM∗Rdθ(vsinθ)=πMv∫0πsinθdθ=−πMv[cosθ]0π=−πMv(cosπ−cos0)=π−Mv(−2)=π2MvThanks and Regards
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