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
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?
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...
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.
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...
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...
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?
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?
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...
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
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!
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...
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...