Question

A train of mass $M$ is moving on a circular track of radius $R$ with a constant speed $v$. The length of the train is half of the perimeter of the track. The linear momentum of the train will be

• A. $zero$
• B. $\frac{2Mv}{\pi }$
• C. $MvR$
• D. $Mv$

Answer 1

Answer: (B)

$\frac{2Mv}{\pi }$

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 $\overrightarrow{v}$ 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 $\hat{y}$ and $-\hat{y}$ 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$=L=\pi R$;

Mass per unit length$=\lambda =\frac{M}{\pi R}$;

Take a small element at an angle $\theta$ on the semicircle which subtends angle $d\theta$.

Let $dm$ be the mass of this small element.
Let $dl=$ Length of the small element$=Rd\theta$;

$dm=\lambda dl=\frac{M}{\pi R}Rd\theta$;

$P={\int }_0^{\pi }d{p}_x={\int }_0^{\pi }dm\left(v\sin \theta \right)={\int }_0^{\pi }\frac{M}{\pi R}\ast Rd\theta \left(v\sin \theta \right)=\frac{Mv}{\pi }{\int }_0^{\pi }\sin \theta d\theta$

$=-\frac{Mv}{\pi }{\left[\cos \theta \right]}_0^{\pi }=-\frac{Mv}{\pi }(\cos \pi -\cos 0)=\frac{-Mv}{\pi }(-2)=\frac{2Mv}{\pi }$