Pencil movement on a no-friction table

[DimosNat]

Homework Statement

[/B]
So here is the problem:
We have a pencil , horizontally towards us, sitting on a table(see the picture). There are no external forces applying on the system table-pencil and friction of air doesn't exist. There is also no friction between the pencil and the table.
So, we apply the same force F two times. The first time we apply the force on the point M (see the picture) which is the center mass of the pencil. The second time we apply the same force on the point A which is the edge of the pencil.The force is applied vertical on the pencil in both cases (see the picture).
Question:
The center mass of the pencil will reach the end of the table:
a) faster when F is applied on the center mass of the pencil , point M.
b) faster when F is applied on the edge of the pencil, point A.
c) at the same time in both cases.

Homework Equations

The Attempt at a Solution

So, I have argued with two of my high school physicists about this problem and both gave me a different explanation, but I disagree with both of them, so here are the two opinions of my teachers and my opinion.

1st teacher:
The first teacher answers c .
His explanation is that in both cases the pencil will only have linear kinetic energy so it will reach the end of the table at the same time in both cases.

2nd teacher:
The second teacher answers c.
His explanation is that in the first case,when F is applied on M (center mass), the pencil will reach the end of the table in t seconds and F will produce work W which will be transformend into linear kinetic energy E₁.
In the second case though, the pencil will acquire both linear and rotational kinetic energy and F will produce a bigger amount of Work. A part of that Work will be transformed into linear kinetic energy E₂ and the rest into rotational kinetic energy. He also said that E₁=E₂ and that's why the pencil will reach the end of the table at the same time t in either case.

He also said that ΣF=ma_cm and because F is the same in both cases the
acceleration of the center mass will be the same.

My opinion:
I disagree with them both and i believe that correct answer is a.
I believe that in both cases F will produce the same amount of work W.
In the first case all of this amount will be transformed in linear kinetic energy.
In the second case a part of Work will be transformed into linear kinetic energy and the rest will become rotational kinetic energy, and this is why the pencil will reach the end of the table faster when F is applied on M.

P.S: My answer to the first teacher who supported that the pencil will only do linear movement and no rotational movement when F is applied on A:
If the pencil doesn't do rotational movement then it should be ΣT=0 (t=torque) as to any point of the pencil.
So, if we consider that the rotation axis is on point M and that the length of the pencil is x then we have:
ΣΤ_Μ=0 <=> F*(x/2)=0 then F=0 or x=0 which is wrong.

I would really appreciate some help on this point.
Thank you!

 

[haruspex]

The force is applied vertical on the pencil in both cases

I assume you mean horizontally, the picture being the view from above.

in both cases the pencil will only have linear kinetic energy

Clearly not. In the second case it will also have rotational energy.

ΣF=ma_cm and because F is the same in both cases the
acceleration of the center mass will be the same.

Correct explanation. The preceding discussion of energies is true but does not constitute a proof.

in both cases F will produce the same amount of work W.

Only if F advances the same distance. In the second case the rotation of the pencil means the force will get to advance further.

 

[DimosNat]

Only if F advances the same distance. In the second case the rotation of the pencil means the force will get to advance further.

Namely, in the second case F will produce more work in comparison with the first case ,because the pencil in the same amount of time t will advance the same distance d and will rotate θ rad.
So W₁=E₁=F*d
while W₂=E₂=F*d + F*(x/2)*θ
W₁<W₂
Is that right?

 

[haruspex]

Namely, in the second case F will produce more work in comparison with the first case ,because the pencil in the same amount of time t will advance the same distance d and will rotate θ rad.
So W₁=E₁=F*d
while W₂=E₂=F*d + F*(x/2)*θ
W₁<W₂
Is that right?

Yes.