Applying an impulse on a dumbbell from one end

[teoferrazzi]

Homework Statement

So you have this dumbbell, that is lying on a two-dimensional xy plane, and is made of two masses - m1 and m2, with m2 to the ''right'' on the x-axis [the text doesn't specify, but it's probably a good idea to place m1 in the origin]. suddenly you apply a force F in the negative y direction on m1 for dt. Calculate the rotation rate of the dumbbell.

The Attempt at a Solution

an impulse Fdt causes a change in momentum mv. I presume the other mass m2 experiences the same change in momentum, so it will experience the same change in momentum; the change in linear velocity will be inversely proportional to the mass, but the angular velocity will be the same because the distance to the centre of mass is also inversely proportional to the mass. so the dumbbell will rotate uniformly [which is obvious]. the problem is that a hint given in the question says i should use conservation of several momenta. but this force is obviously external. how does that work with conservation? and does the whole impulse translate into angular velocity?

 

[Quinzio]

Rather than applying thoughtlessly some formulas, I think it's better to reason about it. You can solve it without writing down any equation.

First question is:
you apply a force on m2 for an istant dt. Which directions must have the force not to have any influence on m1 (for the duration of dt) ?

 

[teoferrazzi]

for the record the force is on m1 [it doesn't matter, but to avoid further confusion].
that is a hypothetical question. in this problem the force is in the y direction, perpendicular to the distance to the center of mass, so the whole force causes the change in angular momentum [for it is a cross product mv X r]. the dumbbell will not move downwards, it will only rotate, that is my intuition, but how do i prove it?

 

[Quinzio]

for the record the force is on m1 [it doesn't matter, but to avoid further confusion].

Ok, who cares ? The object is simmetrical.
Change my post swapping m1/m2.

that is a hypothetical question. in this problem the force is in the y direction, perpendicular to the distance to the center of mass, so the whole force causes the change in angular momentum [for it is a cross product mv X r]. the dumbbell will not move downwards, it will only rotate, that is my intuition, but how do i prove it?

Your intuition may be not correct, this is what we are proving.

that is a hypothetical question.

I know.
But thinking a bit more broadly will allow you learning new useful things.

Answer my question: you apply a force on m1. If you choose a specific direction, m2 will not move for the duration of the impulse dt (after it will move). which is this direction ?

 

[teoferrazzi]

i can think of two main direction: one is the y direction, which will cause some degree of rotation, and the x direction. if you apply a force in the positive x direction, you will move the dumbbell forwards - the opposite if you move it in the y direction. in every case it seems that m2 will suffer to some extent. all the other directions are combinations of x and y so.. I'm out of ideas, it seems no matter what I do m2 will be influenced!

 

[teoferrazzi]

im sorry, i meant ''the opposite if you move it in the negative x direction''

 

[Quinzio]

Ok, m1 and m2 are connected by a massless rod.
Applying a force to m1 which is ALWAYS perpendicular to the rod, WILL NOT MOVE m2.
Can you see that ?

Imagine you have a space ship, which is made by two big masses (m1 and m2) connected by a long thin rigid tube (rod) (something like this was shown in "2001 Space Odissey", really).
An engine is mounted on m1, and it pushes only perpendicular to the rod.
The engine gives a big pulse for a very short period (dt).
Is m2 inluenced during the dt ?

Pick up a tube in your home (eg. a sweep) hold it with both hands as distant as possible. Move one hand back and forth perpendicular to the rod.
Do you experience a force on the other hand ?

 

[teoferrazzi]

but like this I'm holding the other end steady. here I'm just applying a force and the other end is perfectly free.
the question has it as a given that the dumbbell will rotate. so if you say that m2 will not move, are you saying that m2 will be the center of said rotation?
[this homework question is for my 3rd year advanced mechanics course- I'm doing it here but I'm an exchange student and I am as now lagging behind the other students because I don't do a proper physics program back home -. so forgive me if I put it in the advanced physics section!]

 

[teoferrazzi]

the hint given in the question goes on about how I should apply conservation of linear AND angular momentum. which is my biggest puzzler: this system doesn't seem isolated so I wouldn't assume conservation!

 

[Quinzio]

but like this I'm holding the other end steady. here I'm just applying a force and the other end is perfectly free.
the question has it as a given that the dumbbell will rotate. so if you say that m2 will not move, are you saying that m2 will be the center of said rotation?
[this homework question is for my 3rd year advanced mechanics course- I'm doing it here but I'm an exchange student and I am as now lagging behind the other students because I don't do a proper physics program back home -. so forgive me if I put it in the advanced physics section!]

Wait a moment and follow me.

You agreed that during dt, m2 doesn't move.

Now answer:
immediately after dt, not a moment before, not a moment after
m2 is still, agreed.
m1 moves at speed v.
Where is the center of mass ?
What's its speed ?

 

[teoferrazzi]

the position of the center of mass is Dm2/(m1+m2) since i assumed m1 to be in the origin. I would say that its speed is directly proportional to the distance from m2: so it is something like v*(Dm1/(m1+m2))/D

 

[Quinzio]

the position of the center of mass is Dm2/(m1+m2) since i assumed m1 to be in the origin. I would say that its speed is directly proportional to the distance from m2: so it is something like v*(Dm1/(m1+m2))/D

Great.
So, after the instant dt, no force will be ever applied to the dumbbell.
So, if no force is applied after can the center of mass modify its velocity or direction after dt ?

And, since you have m1 moving at v and the center of mass moving at its speed, can you calculate the angular speed ?

 

[teoferrazzi]

the linear velocity of m1 with respect to the center of mass is the difference between their two respective velocities, so it is v*(m2/(m1+m2)), and it is translated into angular velocity..
To get the angular velocity then you divide that velocity by the radius, which is the distance between m1 and the center of mass..
so it's v*(m2/(m1+m2))/(Dm2/(m1+m2))... and this in the end simply gives me v/D.. and seeing that Fdt=m1v, then v=Fdt/m1: in total, the angular velocity is Fdt/m1D.. but here the angular velocity is depending on m1.. i assume that m2 will have the same angular velocity, it will contain m1 also? I could calculate to confirm that, but first.. does all this make sense?

 

[Quinzio]

You did a mistake.
center of mass is at Dm2/(m1+m2)

radius of m1 is Dm1(m1+m2)

w = v m2/m1

v = F / m1

w = F m2 / m1^2

 

[teoferrazzi]

I don't understand.. why when the position of the centre of mass is that, the radius of m1 is different? isn't the radius the distance between m1 and the centre of mass? so shouldn't those two things be the same, since m1 is in the origin?
and where did you take that expression for w from?

 

[Doc Al]

It's true that if the impulse is applied perpendicular to one end of the dumbbell, the instantaneous axis of rotation will be about the other end. But you don't have to assume that if you're unsure about it. Find the angular impulse about the center of mass and set it equal to I_cmω. That's a basic principle that's always true.

 

[teoferrazzi]

ok that's a good point of reference.. another thing is.. the text says i should take into consideration conservation of momenta - both linear and angular, but that puzzles me because it seems like this force is external. what does the text mean?

 

[Doc Al]

the text says i should take into consideration conservation of momenta - both linear and angular, but that puzzles me because it seems like this force is external. what does the text mean?

That seems an odd thing to say. Perhaps they are referring to the impulse-momentum theorem.

(What text are you using?)

 

[teoferrazzi]

the impulse-momentum theorem simply states that the impulse causes a change in momentum? oh well. this text is just a handout from our teacher. now if you take a look at quinzio's solution [he helped me before] he said something that perplexed me: the position of the centre of mass is such, and the radius of m1 is totally different. now isn't that wrong?

 

[Doc Al]

he said something that perplexed me: the position of the centre of mass is such, and the radius of m1 is totally different. now isn't that wrong?

What do you mean by 'the radius of m1'? Used for what?

 

[teoferrazzi]

the rotational radius. we came to the conclusion that m1 had a certain linear velocity, and the centre of mass had another: the difference in linear velocity would translate into rotational velocity [which is something that i need more tangible proof for, by the way] and to get that i would need to divide the difference in linear velocity by the rotational radius, which i thought was the distance between the centre of mass and m1 all along..

 

[Doc Al]

which i thought was the distance between the centre of mass and m1 all along..

Yes, if you're using the difference between the speeds of m1 and the center of mass, then the 'rotational radius' will be the distance between them.

 

[teoferrazzi]

ok. so if you see the formula that I extracted, where I assumed that the two things ought to be the same, does that make sense to you? It's a few posts above this one. I get an angular velocity that is inversely proportional to m1.

 

[Doc Al]

ok. so if you see the formula that I extracted, where I assumed that the two things ought to be the same, does that make sense to you? It's a few posts above this one. I get an angular velocity that is inversely proportional to m1.

Yes, your formula in post #13 is correct: ω = Fdt/m1D.

 

[teoferrazzi]

Ah! Great! I knew something was fishy with Quinzio's contribution [although up to that point he did a great job of letting me see things for myself].
I am, however, a bit shaky on the theoretical background for this.. that is.. -why m2 is utterly unaffected while I apply the impulse? I answered my own question as to why the difference in linear velocities translates to rotational velocities - fairly obvious: the distance between m1 and CM would increase if some of the difference in linear velocity was even partly translated into more linear velocity.. that's absurd. But yeah, could you clear the fog on that doubt of mine?

 

[Doc Al]

I am, however, a bit shaky on the theoretical background for this.. that is.. -why m2 is utterly unaffected while I apply the impulse?

The bottom line is that they are two point masses. But do this: Figure out the speed of m2 by combining the velocity of the center of mass with the rotational speed of m2 about the center of mass.

 

[teoferrazzi]

didnt i get the velocity of the center of mass already by assuming that m2 had no linear velocity due to the impulse?

 

[Doc Al]

didnt i get the velocity of the center of mass already by assuming that m2 had no linear velocity due to the impulse?

You can find the velocity of the center of mass without making any assumptions about the motion of m2. (Just use the impulse-momentum theorem.)

 

[teoferrazzi]

so Fdt= (m1+m2)V.. and then I calculate the difference between the rotational velocity of m2 (multiplied by its radius)and the linear velocity of the centre of mass and ought to find zero?

 

[Doc Al]

so Fdt= (m1+m2)V.. and then I calculate the difference between the rotational velocity of m2 (multiplied by its radius)and the linear velocity of the centre of mass and ought to find zero?

Exactly.

 

[teoferrazzi]

hah! thank you so much. I had gotten to the solution but I needed some justification for it and you gave it to me. No more questions from my part. Cheers