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Question: A train of mass M is moving on a circular track of radius \(R\) with a constant speed \(v\). The len...

A train of mass M is moving on a circular track of radius RR with a constant speed vv. The length of the train is half of the perimeter of the track. Find the linear momentum of the train.
A) Zero
B) 2Mvπ\dfrac{{2Mv}}{\pi }
C) MvR{\text{MvR}}
D) Mv{\text{Mv}}

Explanation

Solution

Here the length of the train is given to be half of the perimeter of the circular track. The velocity of the train is said to be constant. But only its magnitude remains constant, at different points on the track, the direction of the velocity varies. The linear momentum of the train can thus be obtained by integrating the momentum of a small element on the circular track.

Formula used:
The linear momentum of a body is given by,
p=mvp = mv, where mm is the mass of the body and vv is the velocity of the body.
The linear density of an object is given by,
λ=ml\lambda = \dfrac{m}{l}, where mm is the mass of the object and ll is the length of the object.

Complete step by step answer:
Step 1: Sketch a figure depicting the motion of the train along the circular track and list the parameters given in the question.

The above figure depicts the train along the circular track. As shown in the figure, the train is symmetrical about the Y-axis. A and B represent the two points which are symmetrical about the Y-axis and which correspond to two points in the motion of the train. The velocity vv at these two points (or any point along the circle) will have a vertical component and a horizontal component.
The mass of the train is given to be MM .
The radius of the circular track is given to be RR and its perimeter will be 2πR2\pi R .
Let ll be the length of the train. As mentioned in the question the length of the train can be expressed as l=2πR2=πRl = \dfrac{{2\pi R}}{2} = \pi R .
The mass per unit length or the linear density of the train will be λ=Ml=MπR\lambda = \dfrac{M}{l} = \dfrac{M}{{\pi R}} .
Step 2: Consider a small element of the train to obtain an expression for its linear momentum.
We now consider a small element making an angle θ\theta with the X-axis. This element subtends an angle dθd\theta . This is shown in the figure below.

The length of the small element will be dl=Rdθdl = Rd\theta .
Then the mass of the small element can be expressed as dm=λdldm = \lambda dl -------- (1)
Substituting for λ=MπR\lambda = \dfrac{M}{{\pi R}} and dl=Rdθdl = Rd\theta in equation (1) we get,
dm=MπR×Rdθ=Mπdθ\Rightarrow dm = \dfrac{M}{{\pi R}} \times Rd\theta = \dfrac{M}{\pi }d\theta
Thus the mass of the small element is dm=Mπdθdm = \dfrac{M}{\pi }d\theta .
From the above figure, we see that the vertical component of the velocity vcosθv\cos \theta at A and B is directed along the positive Y direction and negative Y direction respectively and so they will cancel out each other. Thus only the horizontal component vsinθv\sin \theta prevails.
Now we can express the linear momentum of the small element as dp=dm(vsinθ)dp = dm\left( {v\sin \theta } \right) ------- (2)
Substituting dm=Mπdθdm = \dfrac{M}{\pi }d\theta in equation (2) we get,
dp=Mπ(vsinθ)dθ\Rightarrow dp = \dfrac{M}{\pi }\left( {v\sin \theta } \right)d\theta -------- (3)
Equation (3) expresses the linear momentum of the small element.
So integrating equation (3) will provide us with the total linear momentum of the train.
Step 3: Integrate equation (3) to obtain the total linear momentum.
The total linear momentum of the train is given by, P=0πdpx=0πMπ(vsinθ)dθP = \int\limits_0^\pi {d{p_x}} = \int\limits_0^\pi {\dfrac{M}{\pi }\left( {v\sin \theta } \right)d\theta } --------(A)
Integrating the above expression we get,
P=Mvπ0πsinθdθ=Mvπ[cosθ]0π\Rightarrow P = \dfrac{{Mv}}{\pi }\int\limits_0^\pi {\sin \theta d\theta } = \dfrac{{ - Mv}}{\pi }\left[ {\cos \theta } \right]_0^\pi ------- (4)
Applying the limits we get,
P=Mvπ[1(+1)]=2Mvπ\Rightarrow P = \dfrac{{ - Mv}}{\pi }\left[ { - 1 - \left( { + 1} \right)} \right] = \dfrac{{2Mv}}{\pi } ------ (5)
Thus the linear momentum of the train is obtained as P=2MvπP = \dfrac{{2Mv}}{\pi }

Hence the correct option is B.

Note:
Here the start of the train is taken to be from X-axis and so for the length of the train, the angle θ\theta varies from 00^\circ to 180180^\circ and we have the lower limit of the integral given by equation (A) as zero and its upper limit as π\pi . The integral of sinθ\sin \theta is cosθ- \cos \theta, this is substituted in equation (4). We have cosπ=1\cos \pi = - 1 and cos0=+1\cos 0 = + 1, these values are substituted in equation (5). The direction of the linear momentum will be along the horizontal component of the velocity.