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Post Closed as "Duplicate" by Dale, Jon Custer, John Rennie newtonian-mechanics
Changed the question from involving incorrect collision calculations to just force
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Tushar Hegde
Tushar Hegde
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InIf there is a rod on a smooth surface, and a force F acts on it in 2 cases:

enter image description here

1>
F acts on the first casecentre of mass, thereand it is no torquethe only force acting and thereforeon the rod.
Since the change in momentum of the system would be equal to the Force acting on it wont rotate,
MV-1(of the whole system) = F

2> F is the only force acting on the system, but it doesnt act on the centre of mass.
ButHere also, the change in momentum of the system would be equal to the Force acting on it, so
MV-2 = F
And MV-1 = MV-2
Even though the linear momentum (and mass, and therefore translational KE) of the system in the two cases are the same, as in the second case, the Force acts on one side of the rod, won't it should rotate as there isalso have angular momentum due to the torque. that acted on it

Somewhere inThe assumptions of this question :
Net Force on the picture below I feel like I have madesystem as a mistake, but I am unable to figure out where.whole = (Vector) sum of all forces acting on it regardless of position
Where isChange in linear momentum of the mistake that leads me to believesystem as a whole = Net Force on the system as a whole

Now, when there would be more energyis equal force being exerted in both cases, does the second case have more total energy than the first ?
  And if it isn't a mistakedoes, wherewhy does the extra energy come fromimparted depend on the position of where you apply the force?

enter image description here enter image description here

In the first case, there is no torque acting and therefore it wont rotate.
But in the second case it should rotate as there is torque.

Somewhere in the picture below I feel like I have made a mistake, but I am unable to figure out where.
Where is the mistake that leads me to believe there would be more energy in the second case?
  And if it isn't a mistake, where does the extra energy come from?

enter image description here enter image description here

If there is a rod on a smooth surface, and a force F acts on it in 2 cases:

enter image description here

1>
F acts on the centre of mass, and it is the only force acting on the rod.
Since the change in momentum of the system would be equal to the Force acting on it,
MV-1(of the whole system) = F

2> F is the only force acting on the system, but it doesnt act on the centre of mass.
Here also, the change in momentum of the system would be equal to the Force acting on it, so
MV-2 = F
And MV-1 = MV-2
Even though the linear momentum (and mass, and therefore translational KE) of the system in the two cases are the same, as in the second case, the Force acts on one side of the rod, won't it also have angular momentum due to the torque that acted on it

The assumptions of this question :
Net Force on the system as a whole = (Vector) sum of all forces acting on it regardless of position
Change in linear momentum of the system as a whole = Net Force on the system as a whole

Now, when there is equal force being exerted in both cases, does the second case have more total energy than the first ? And if it does, why does the energy imparted depend on the position of where you apply the force?

Source Link
Tushar Hegde
Tushar Hegde
  • 11
  • 2

Is there extra energy in this system and where does it come from?

In the first case, there is no torque acting and therefore it wont rotate.
But in the second case it should rotate as there is torque.

Somewhere in the picture below I feel like I have made a mistake, but I am unable to figure out where.
Where is the mistake that leads me to believe there would be more energy in the second case?
And if it isn't a mistake, where does the extra energy come from?

enter image description here enter image description here