Recent content by brotherbobby

If you put the value of the tension ##2T = \dfrac{4m_1m_2}{m_1+m_2}g## and subtract it from the force due to gravity ##(m_1+m_2)g##, it would come out to be ##\dfrac{(m_2-m_1)^2}{(m_1+m_2)^2}##, which is the total mass ##M(=m_1+m_2)## times the acceleration of the center of mass...

Gravity is an external force. I suppose what you mean is that if ##m_1## were to accelerate "up" with ##g##, we'd need an agent to apply a force of amount ##2m_1g## on it upwards. We need no such agent to let it go "down" with acceleration ##g##. Thank you. I suppose the matter is settled unless...

Oh am sorry; you made the correction. However, I find myself struggling with something else. How can the tension be ##T = 2m_1g##? Imagine the situation where both masses are going "down" with acceleration g. The system would be freely falling. The rope connecting them will be slack, the...

I failed to understand. Mass ##m_2\; (\rightarrow\infty)## is accelerating downward at ##g##. How is mass ##m_1## accelerating upward with acceleration ##2g##? Aren't them both going down and going up with acceleration ##g##?

System : The Atwood's machine is well-known, and drawn alongside, where ##m_2 > m_1##. The (common) acceleration of the masses ##a = \dfrac{m_2-m_1}{m_1+m_2}g## and the tension in the (massless) rope is ##T=\dfrac{2m_1 m_2}{m_1+m_2}g##. Question : Make sense of the above result(s) in the event...

The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But...

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...