Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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.Please give a detailed solution.A) Zero B)2Mv/pi C)MvR D)Mv

290 Points
9 years ago
Hi Siddhant,
Consider the centre of the circle as the origin (for the X-Y Plane)
Now the X-component of the linear momentum will cancel out (in the lower half and the uper half)
Only the Y-component would add up.
So the required momentum would be = 2∫(M/ΠR)vcosx dx {in the limit range from x= 0 to pi/2)
So the Linear Momentum = 2Mv / ΠR.
In Option (B), you should also have "R" in the Dr.
Best Regards,
Rishabh Maggirwar
19 Points
8 years ago

The answer posted by Ashwin (IIT Madras) is absolutely correct but with a silly mistake.

The ''R'' in the denominator gets cancelled out so it remains :

2 x (integral) M/pi x  v cosx  x  dx    {range from 0 to pi/2}

linear momnetum = 2Mv/pi

Shuvrasish Roy
26 Points
4 years ago
For the linear momentum of the system we have to find the centre of mass of the systemAs told in the question the train is half as long the perimeter of the circular track, therefore its centre of mass must lie 2pi/R m away from the centre of the track. And we`d also have to find the angular velocity of the Centre of mass i.e. 2/RHence momentum of the train=mass of the train*location of centre of mass*angular velocity of the centre of massp=pi*2R/pi*2/R