Recent content by jbriggs444

Right. You are asked for velocity. You have determined the [coordinate] acceleration. That is all you need. The various forces required to produce that acceleration do not enter in.

The professor is still wrong. Let us grant that the intended threshold for pilot unconsciousness is 6g of coordinate acceleration. This is then 7g of experienced proper acceleration at the bottom of the loop. But that 7g of experienced proper acceleration is not equal to the ##\frac{v^2}{r}##...

Loss of consciousness results from proper acceleration, not coordinate acceleration. Proper acceleration is how much the plane's acceleration deviates from free fall. Coordinate acceleration is the plane's acceleration in a chosen reference frame. Probably the rest frame of the ground. The...

Yes.

When the question asks for mass, do you understand it to mean relativistic mass or invariant mass? Obviously one can discard the idea that invariant mass is meant immediately. The invariant masses are always in the ratio of one to four regardless of velocity. Never one to three. You understand...

Let me be sure I understand. For this calculation we are abstracting away the interaction zone between the falling portion and the static portion and making blind assertions about the behavior there. Your particular blind assertions are that energy is not conserved and that the tensions from...

The static portion has velocity zero. The falling part has velocity ##v##.

A good way to approach infinite quantities in physics is to model them as the limit of finite cases as some parameter is increased without bound (or decreased toward zero). So we take the limit of a flexible ribbon and a tight curve as both flexibility and tightness increase without bound. That...

In line ##\LaTeX##:: ##a_1 = \frac{4 m_2 m_1 + m_0 ( m_1 - m_2 )} {4 m_2 m_1 + m_0 (m_1 + m_2)} g## Which should render as ##a_1 = \frac{4 m_2 m_1 + m_0 ( m_1 - m_2 )} {4 m_2 m_1 + m_0 (m_1 + m_2)} g## I tossed in some white space in the unrendered LaTeX for what I think is better readability...

If you are using an accelerating frame and pseudo-forces then those pseudo-forces should appear in your force balance equations. If, for instance, the pulley is accelerating downward with some acceleration (##a_k##) then the pseudo-force will act upward -- opposite to gravity. The net apparent...

This is fine in isolation. One often picks a sign convention to try to keep all quantities positive. If one guesses wrong, the worst thing that happens is that upon solving the algebra some quantity turns out to be negative instead. This is a common occurrence in circuit analysis. We may not...

Agreed. I like to point to Newton's original text from time to time. The second part is the one that really resonates with me. In modern words: "the force of A on B is equal and opposite to the corresponding force of B on A". The notions of "action" and "reaction" are often mistakenly...

No, that is not the frame that I implicitly chose. The table does move with respect to the table+book COM frame. The frame that I had in mind is the lab frame where the table is continuously at rest. One can tell that this is the choice I made based on the fact that I spoke of the table having...

Newton's third law is all about momentum. It amounts to a statement that momentum is conserved. Newton's third law has less to do with energy. However, let us proceed to your next paragraphs... Let us chase this down. The force of your hand on the book is equal and opposite to the force done...

I had to zoom way in to read the handwriting. Your first assertion is that ##a_1 + a_2 = 2a_k##. This is obviously correct. The acceleration of the upper block is the average of the accelerations of the lower two blocks. Your next two assertions looked strange to me. You have ##T - m_1g =...