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  • A train of mass M is moving on a circular track of radius ' R ' with constant speed V. The length of the train is half of the perimeter of the track. The linear momentum of the tr

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

Option: 1

0


Option: 2

\frac{2MV}{\pi}


Option: 3

MVR


Option: 4

MV


Answers (1)

best_answer

 

Centre of Mass of semicircular ring -

It lies at a distance of \frac{2R}{\pi} from centre of the ring along its axis.

-

 

 

 If we treat the train as a ring of mass 'M' then its COM will be at a distance \frac{2R}{\pi}  from the centre of the circle. Velocity of centre of mass is :

V_{CM}= R_{CM}.\omega = \frac{2R}{\pi }.\omega = \frac{2R}{\pi}\left ( \frac{V}{R} \right )         \left ( \because \omega = \frac{V}{R} \right )

\Rightarrow V_{CM}= \frac{2V}{\pi} \Rightarrow MV_{CM} = \frac{2MV}{\pi}

As the linear momentum of any system = MVCM

 \therefore The linear momentum of the train   = \frac{2MV}{\pi}

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