Interesting question on COM
The answer given is C) the solution doesn't provide any explanation as to why, can someone suggest to me the thought process of how I should solve this
31 Replies
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.I tried thinking and I think it should be D) as COM of the compartment should not change with the movement of passengers
If I am wrong please correct me 🙏
The track is frictionless.
In the people+train system there is no external force.
So the net center of mass of the system will remain at the same place (which in this case is C2) while the center of mass of train will move about (C1) in direction opposite to center of mass of people combined.
COM for both together will not change
Fnet = 0. Equilibrium is maintained. Therefore momentum can be conserved. From this, we can draw the conclusion that net displacement of c.o.m will be 0, that is, displacement of C2. Since passengers will move along the track, the compartment will move in the opposite direction keeping displacement of c2 zero. But since the compartment has made a displacement, it's centre of mass, I.e. c1, will also move. Hence C is the correct answer.
this is analogous to the man walking on a boat problem, if you've seen those before
that also works on the principle that since there's no external force on the system, the velocity of COM = 0
is the ans A
its (C)
^
because the total COM will be constant and the people are moving so the Trains COm should move too
the train is stationary
velocity of COM = 0 and will remain 0 until an external force is applied
ye
so if people move
train should move
people move = train move correct
but people train combined system will not move since no external force on the system
the combined COM wont move , but i think the train will move
yeah that's (C), not (A)
if C1 moves
the COM wont be constnat
C2 HAS to move
for COM net to be constant
no no
for the net COM to be constant, when the people shift their center of mass (by moving in the compartments)
the train also shifts its center of mass in the opposite direction
Yes thats what am saying
wait nvm
so C2 doesnt move
i read question wrong
i thought C1 is people C2 is train
mb
its C
lmao fair
Ahh so that’s what it is I missed the point of the question lol
It makes sense now yeah I have solved several questions like those
yeap
Oh that explains it perfectly thanks
I suppose that the question confused you with it's wordings, that's natural to happen
Equilibrium is actually established in the net system of people and train so it does make sense for c2 to remain stationary while train will move so that m1 times x1 of train equals m2 times x2 of people similar boat man or sledge problems
Yeah wording is somewhat different than normal
Exactly
@_zbro @Nimboi Thank you all I get this question now
welcome :) you can close the thread by typing
+solved @person1 @person2Okay
+solved @Nimboi @_zbro
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