I thought the center of mass equation was derived for general forces, $$\sum{\vec{F_{ext}}}=M\vec{a_{CM}}$$  Then suddenly when the external force on the $i$ particle is of the form $m_ig_i$, where $g_i$ varies throughout the body, we have to use this equation:  $$\vec{W}=M\vec{a_{CG}}$$  where $CG$ is a different point than $CM$. So, what makes gravity special?  EDIT: I'm not asking the difference between them. They have different formulas, so obviously they have different values when $g$ is not constant.  I'm asking why doesn't the resultant gravitational force or $W$ can't be thought of as acting on the center of mass when the center of mass equation is derived for any general external force.
7037
My book says:     For most of the small objects, both are same. But for mammoth ones, they are really different ones. And in a gravity-less environment, COG is absent; COM still exists.   Ok, what's the big deal when things are small and big? How these two :  &amp; ?
116246
So this is the question that's been bothering me:  Say you have a simple rigid body in space that is at rest or traveling with only translational motion at a constant speed. Say that the body is something like a rod and it's not rotating. So, at some point, an external force is acted on the rod, just for an instant and it is not acted along some axis that goes through the center of mass. What will happen??  My guess is that it will start rotating about the center of mass (because of torque) + it will get some translational motion towards the direction of the force. Is that correct? and if so, how much of the force becomes rotation and how much translational movement and why??  Please forgive my English and thnx in advance!!
103419
I got an assignment to prove that in a straight homogeneous rod, you can always choose a coordinate system in such a way that  $$\int_S x_1 \, dx_1 \, dx_2=0 $$  $$\int_S x_2 \, dx_1 \, dx_2=0 $$  $$\int_S x_1x_2 \, dx_1 \, dx_2=0$$  where $S$ is the cross section of the rod  Now, intuitively, I can understand that the first two expressions basically state that if you find the mass center of the rod, then all you have to do is to put the origin of the coordinate system in the mass center, and the integrals turn out to be zero. What I'm curious about is whether or not this is a good enough of an explanation, and also, a hint on how to obtain the third equality.
99658
I remember the first time I saw this formula, $F = ma$, and it seemed perfectly correct and natural to me (in the sense that it is very straight-forward and logical, not that it is easy by any means). But then, I wondered, what if an object is moving with a constant speed? Maybe a car. If it travels with constant speed and hits me, it still hurts. So it's force must not be $0$.  One more thing related to this problem. If the same car is pulling another car behind it, if the car goes with constant speed, it implies that the first car doesn't exert any force to the second car at that moment, since it is not technically pulling it anymore (espacially if we don't take into consideration the friction).  So my question would be the following: How do these two phenomenons describe the same law of motion?   PS: If gravity wasn't behaving as some kind of acceleration pulling me constantly to the Earth's core, then if I jumped upwards, would I start slowly getting away from Earth?
71566
Is there a difference between Weight and Force of Gravity or they just mean the same thing? My textbook does not clarifies this point.
38928
Suppose I have an inertial frame with coordinate $\{q\}$. Now I define another reference frame with coordinate $\{q'(q,\dot q,t)\}$. I obtain the equation of motion in $\{q'\}$ in two different ways:   First obtain the equation of motion in $\{q\}$ by the Euler Lagrange equation $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q}\right)-\frac{\partial L}{\partial q}=0$$ and then rewrite the equation in terms of $\{q'\}$. First transform $L(q,t)$ to $L'(q',t)=L(q(q',t),t)$ and then obtain the equation of motion $$\frac{d}{dt}\left(\frac{\partial L'}{\partial \dot q'}\right)-\frac{\partial L'}{\partial q'}=0.$$   Are the two answers just the same?
73118
I tried doing  just for fun. It is reproduced as follows:          (I'm not allowed to post the entire problem here since I'm a new user.)  Here are my equations:  $$T - mg = ma$$ (equation 1)  $$Mg - T = Ma$$  (equation 2)  (Upper case "M"  used for infinite series of masses on the right arm of the first pulley. Lower case "m" for the small, single mass on the left arm.)  algebra is as follows:  $$Mg - (ma + mg) = Ma$$ $$g (M - m) = a (M + m)$$ $$a = g (M - m) / (M + m)$$   Here is my reasoning to get final answer:   The limit of $a$ for large $M$ goes to $g$. Therefore, acceleration of the top mass is $g$. However, the correct answer is $g/2$. Why does my reasoning get the wrong answer? Any ideas?    (Please excuse the formatting, I'm a newbie here).
104656
So, according to Einstein's theory of general relativity, gravity is not a force instead it is a consequence of objects with mass deforming spacetime, right? And so, according to him, there is no difference between standing on a platform accelerating upwards at 9.81 m/s^2 and standing on Earth, right? But, there is a difference. At least, I think it's a valid difference. If you drop two apples from a high altitude on Earth, not only will they accelerate downwards at 9.81m/s^2 but they will also slightly move towards each other because the Earth is a sphere and gravity acts radially inwards pulling the two apples closer to each other. But if you drop the apples from the same height above the accelerating platform, the apples will only appear to accelerate downwards at 9.81 m/s^2 but will not move towards each other like they would on Earth. Why is this true? Is this even a valid difference? If it isn't, why isn't it?
19257
I have learned that $$\mathbf{F} = m \mathbf{a}$$ where both the force $\mathbf{F}$ and acceleration $\mathbf{a}$ are vectors. This makes sense, since both force and acceleration have a direction.  On the other hand, the kinetic energy $$K = \frac12 mv^2$$ looks completely different. It doesn't seem to depend on the direction. How are these two concepts related?
39829
I feel like this question has been asked before but I can't find it. would the Euler Lagrange equation for, say, the standard model Lagrangian be $$\frac{\partial L}{\partial \phi}=\partial_\mu \frac{\partial L}{\partial (\partial_\mu \phi)}$$ Where $\phi$ is whatever field is an question and $\mu$ is (I believe) being summed from 0 to 3. Or, ist the correct equation $$\frac{\partial L}{\partial \phi}=D_\mu \frac{\partial L}{\partial (D_\mu \phi)}$$ Where $D$ is the covariant derivative of the theory. My intuition tells me its the second eqn but i just wanted to be sure, and I think I once saw someone say that the two were equivalent.
3958
Take two different reference frames. The first one - say $\lbrace{W}\rbrace$ - is inertial and it's fixed to the Earth surface. The second one $\lbrace{B}\rbrace$ is moving and in general is rotating and accelerated, so $\lbrace{B}\rbrace$ it is not inertial.  The equations of motion for the origin of the frame $\lbrace{B}\rbrace$ can be formulated on both frames as:  $$ \sum_{i}\,  ^wF_{ext} = m\, ^w\dot{v}$$ in the inertial frame and $$ \sum_{i}\,  ^bF_{ext} = m\, ^b\dot{v} - ^b\omega \times m ^bv$$ on the non-inertial one, where $^f x$ represents the vector $x$ expressed on the frame $f$.  Let an accelerometer to be installed over $\lbrace{B}\rbrace$ which by design subtracts gravity from its output.  I'm not sure if it is measuring the term $^b\omega \times m ^bv$ or not so, what is the accelerometer exactly giving as output?
15849
This is a problem that has bothered me for a couple of weeks now, and I can't seem to wrap my head around it and understand it.  Let's say we have a planet with a mass of m. We also have an object of relatively small mass (so small that its gravitational field would not affect the planet), and we know that at time 0 it is at position h.  If we know the acceleration due to gravity (g), we can calculate the position y  at time t:  $y = h - { g \; t^2 \over 2 } $  And we know that the acceleration due to gravity could be calculated:  $g = G { m \over y^2 }$  But since the acceleration depends on the position of the object relative to the planet, and the position of the object is dependent on the acceleration, this would obviously mean that the first formula won't work when we are working with displacement so large that it would change the acceleration significantly.  So, in essence, my question is, what formula can we use to calculate the position of an object, taking into consideration the change in acceleration due to gravity?
75939
A mass of point performs an effectively 1-dimensional motion in the radial coordinate. If we use the conservation of angular momentum, the centrifugal potential should be added to the original one.     The equation of motion can be obtained also from the Lagrangian. if we substitute, however, the conserved angular momentum herein then the centrifugal potential arises with the opposite sign. So if we naively apply the Euler-Lagrange equation then the centrifugal force appears with the wrong sign in the equations of motion.   I don't know how to resolve this "paradox".
110123
This is my first year as a physics student, and I've never learned about vectors past a basic level, so this is confusing me. When we have gravity on an inclined plane, we separate it into two components, which I understand.  However, consider the image below, and there's a box at point A. When separating the gravity components, you draw a triangle AGC (sorry, the G and D are on top of each other and difficult to distinguish). AG becomes the force of gravity in the y-direction, and GC becomes the force of gravity in the x-direction. Then you do the trig functions from there. However, when I tried this myself, I drew triangle ADF instead and tried the trig functions from there. It didn't work. I'm having trouble understanding why you can't compute the trig functions from AGF. The only partial solution I came up with was that force of gravity in the y-direction can't be the hypotenuse, as the force of gravity in the y-direction is always less than the force of gravity. But I think I'm missing something more.
70536
A pulley fixed to a rigid support carries a rope whose one end is tied to a ladder and a man and the other end to the counterweight of mass $M$. The man of mass $m$ climbs up a distance of $h$ with respect to the ladder and then stops. If the mass of the rope and the friction in the pulley axle are negligible find the displacement of the center of mass of the system.   My question is that the net external force on the system is 0 then why is there any displacement of centre of mass?
106534
So I would have thought that this would be how you derive the  on a spring: basically the same way you do with gravity and other contexts, use $$W=\vec{F}\cdot \vec{x}.$$ If you displace a spring by $x$, then it exerts a force $-k x$, so $F=-kx$, since the displacement is $x$.   So $$W=-kx^2.\qquad \leftarrow\text{ (however, apparently wrong!)}$$  I've seen the correct derivation of work in a spring (with an extra half) and don't doubt that it's correct, but also don't see where my logic fails in this alternate derivation.
96518
I want to outline a solid argument (or bulletpoints) to show how weak is the idea of diff(M) being the gauge group of general relativity.  basically i have these points that in my view are very solid but i want to understand if there are misconceptions on my part that i'm simply not getting and if its so, i ask for help to make the case more solid, or understanding why it doesn't apply (to gravity):    (thanks to Raymond Streater for making that point completely clear) gauge invariance in electrodynamics is an observation that physical observables are unchanged after a gauge transformation without changing coordinate frame ( we are ask to believe that in gravity someone did the same? that is, someone made the observation that physical observables are unchanged after a diffeomorphism-gauge-transformation, only to later argue that because of this, that there are no physical observables to begin with, that doesn't make a lot of sense, to not say that its just plain stupid circular argument ) . The invariance is of course broken when the theory is quantized and $\hbar$ makes an appearance, because it assumes a preferred scale for certain energies. The key point here is: classical gravity is not special regarding having diff(M) as a symmetry group from bulletpoints 2 and 3, if i cannot infer that Diff(M) is a gauge-invariance of electrodynamics, the same should apply to gravity   For this question, i would say that a valid answer would either disprove any of the arguments as fallacies themselves (hence showing a solid argument why gravity is special and diff(M) is without a doubt its gauge group), or improve the argument for making it bullet-proof (sorry for the pun)