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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...

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...

Grade:11

2 Answers

Vikas TU
14149 Points
7 years ago
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
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
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 v
into 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=πRM​Rdθ;
P=∫0π​dpx​=∫0π​dm(vsinθ)=∫0π​πRM​∗Rdθ(vsinθ)=πMv​∫0π​sinθdθ
=−πMv​[cosθ]0π​=−πMv​(cosπ−cos0)=π−Mv​(−2)=π2Mv​

Thanks and Regards

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