Recent content by brotherbobby

Because the force of the rope is not the only force that acts on them. There's the force of gravity (weight) too, which is different for the two masses. To write it out, for masses labelled by ##i##, the acceleration of each mass ##a=\dfrac{\Sigma F_i}{m_i}##, which can be the same if the forces...

That is what I have a problem with. If it is the same (inextensible) rope, how can different parts of it move with different accelerations? For the two mass Atwood-Machine with a single fixed pulley, our argument was that the two masses go down (and up) with the same acceleration because it is...

[The problem involves the use of Lagrange's equations for conservative systems. I am first trying to solve it using Newton's mechanics, similar to that of a school student] Diagram : The diagram of the problem is shown to the right. Solution : The tension is the same all over the (same) rope -...

Thank you @julian , while I am struggling to understand your method, especially that of the "stars and bars", your results are precious enough to note down and commit to memory. In answer to my question therefore, which you had highlighted, there does exist formulae for the number of independent...

Attempt : [The author uses the term "system" with respect to indexed quantities. He reserves the term "tensor" when the components of those quantities respect certain rules when co-ordinates transform.] Both the "systems" have a total of ##3^3 = 27## components. Of course, they are not all...

That looks extremely good. I am still struggling with the problem and, despite your efforts, I might continue to do so. Still, thank you very much. Not only do I want to understand how vectors (and co-ordinate axes) rotate under arbitrary transformations about an axis, I am also eager to learn...

I could visualise what you mean. So thank you for that. Indeed, the ##x## axis starts with a precise angle to the axis of rotation and, if we take any point on the ##x## axis, the line would continue having a fixed length throughout the rotation. The angle would remain the same too. In my...

Euler's angles are useful in that they help you find the answer. But you don't get a feel of "how" in real time. I'd like a way to visualise directly; what happens to a point (or several points on a line) as one effects a rotation about a given axis.

I have very little clue as to how to imagine (visualise) the rotation and how the axes will look. All I can do is to draw the image of what I mean by the task. Of course, this is before the rotation takes place. In the diagram, ##\mathbf{OP}## is the axis of rotation. For simplicity, we may...

Brilliant argument. Better than using the BAC-CAB rule which you referred to earlier. Still, there's an element of brilliance in that too. Let me use that as a second way out. I don't mean to close the thread, but as the OP, I should solve the problem. Problem statement : Solution : (2)...

Ok you mean the BAC CAB rule in reverse. That will require some thinking. To derive or imagine the double cross product on the RHS looking at the two vectors with their scalar multipliers on the LHS. One of those terms is in the standard form, which helps. But the other is the vector itself...

Yes. I know what's going on. But this is not the time to apply this rule. Because it involves what's in the answer which I am supposed to obtain. Let's imagine that we don't know the answer and proceed from there.

I know what you did. I am asking where did you get the vector ##\hat n\times(\mathbf A\times \hat n)## from?

Where did you obtain the first term from : ##\hat n\times(\mathbf A\times \hat n)##?

Unit vector, you mean. Only one unit vector is given. I agree with you. There are two unit vectors as I show here but the decomposition of the perpendicular component can only be along one unit vector (marked ##\hat n_{\perp}##) or another. In three dimensions, for instance, there will be an...