Recent content by brotherbobby

I should reply to you later when time and the advantage of having done some mathematics first permits. For the case is involved. The particle has acquired a speed (and deflection) due east projected up, after having reached its highest point (P in the northern hemisphere). On its way down, it...

Due to the coriolis force, the particle acquires a velocity due east (along X) on its way up. For a place in the northern hemisphere. Likewise, doesn't it acquire an equal and opposite velocity (along -X towards) west on its way down, that should result in no net horizontal displacement?

Yes good question. Sorry I was under the impression that, precisely for my thinking up there, namely that "going up in the northern" is the same as "going down in the southern" and vice versa, that I believed that the coriolis force would disappear in places on the equator. I was also "helped"...

Isn't going "up" in the south hemisphere, far as the coriolis force is concerned, the same as going "down" in the northern? (up and down being relative to the place(s) in question, relative to land)

Thank you all for your comments. I agree with the corrections to my drawing and with the expression of the coriolis acceleration, namely that the mass of the object must be absent : ##\vec a_C = -2(\vec\Omega_0\times\vec v)##. However, crucial thing is that the answer, that the coriolis force...

Yes. The Z plane goes vertically up from the place. Y is along the north direction from the place P. Rotation of the earth is in the Y-Z plane, same as the North-Up plane.

The earth rotates in the Y-Z plane. It's angular velocity vector ##\vec\Omega_0## makes an angle of ##\lambda## with the Y axis and ##\pi/2-\lambda## with the Z axis. It has no component along the X axis, i.e. the east west direction.

Attempt : I start by copying and pasting the diagram for the problem to the right. The rocket marked with R hurls "up" along the z axis with a speed of ##\vec v=v\hat k## from a place P whose latitude is ##\lambda\;\text{(south)}##, or ##-\lambda##. The axes are so chosen that ##x## points to...

I copy and paste the relavant paragraphs and images from the text. I might have to do it several times, each time underlining in red what I failed to follow. Here's the first. Statement : "Here the acceleration is in a plane perpendicular to ##\mathbf\Omega##". Sure, the acceleration itself...

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...