swansont

You might notice that what I said is not what you had said. You said "If B is moving then the distance is contracted" Not the same thing at all. The length between E and X is at rest in E's frame. ergo, it is not length contracted. What md65536 had said was that the lengths were contracted in that example, because they were indeed moving.

! Moderator Note Personal attacks are not permitted. Attacking what is posted is fair game

Provide a quote where anyone said that,

That’s not a valid conclusion, as your analysis is incomplete. You haven’t accounted for the change in angular speed. So you now have rotation in your rotating frame (have fun analyzing that). There must be a tangential force acting in order for this to happen. Both of these elements have yet to be addressed. (you could, of course, choose a simpler system to analyze)

! Moderator Note crackpot sockpuppet. closed

dock has been banned as a sockpuppet of redstone, Trần Thành, Energizer and the logic00x triplets

I stated that my analysis that I assumed m<<M, so there is no change is the CoM, but that the rotation details would change if that assumption didn’t hold My example with the table is to rebut your claim the if the rod can’t move it can’t pull, which is ludicrous. Motion isn’t required in order to exert a force. The rod exerts a tension, and via friction, this is the source of the centripetal force. Lecturing me on “next time” is out of line. You asked me to analyze this (poorly documented) scenario. I didn’t drop it into the thread unbidden. You have no credibility to make such an assessment. You don’t understand physics - you’ve made that abundantly clear. Most of what you have done is object when physics disagrees with your misconceptions. And yet you insist your misconceptions represent proper physics. What’s “familiar” to everyone else in the thread is physics that works. If you want to show something to be wrong, do an analysis that’s actually compliant with Newton’s laws (nothing made up) or show, experimentally, that Newton’s laws are wrong. Those are your options. Anything else is BS. The burden of proof is yours. Perhaps you an explain what you mean by “the centrifugal is by nature inertial” because it makes no sense to me. What we experience is trying to move in a straight line, and having the car exert a force on us to move us in a circle. There is no real force outward. Illusions are not physical effects. Your senses are tricked by being in a rotating frame but interpreting everything you experience as if you were in an inertial frame.

There is always a reaction force in an inertial frame. The rod exerts a frictional force on the mass, the mass exerts a frictional force on the rod. You keep asking if there’s a reaction force, like you’re hoping that someone will eventually say, “no” A table leg doesn’t move, but exerts a force to keep whatever’s on the table from falling to the ground. Your assessment of what is and is not possible needs to be based in physics, not fantasy. Centripetal is a label, not a kind of force. Tension can be a centripetal force. Friction, gravity, electrostatic, even a normal force can be the centripetal force in a problem. It’s just a label, important because circular motion means certain equations apply to the problem.

It must be accelerating (the direction is changing). Since w and r are both increasing, the tangential speed is increasing (v = wr), and there's a radial component, too. So the speed is increasing.

The mass slides all the way to the stop, as I said. The frictional force can't supply all of the required centripetal acceleration. The (M+m) system would not accelerate, since there is no external force. The axis of rotation would shift by a small amount. But in the limit of M>>m this would vanish.

And I did the inertial frame. There is no centrifugal force in the inertial frame. There are no fictitous forces, because things in inertial frames follow Newton's laws. When you do the analysis in the rotating frame, it will start at rest in that frame.

I am analyzing it in an inertial frame, which is what you indicate in the drawing. The object is rotating, and it clearly labeled as such, so there is no centrifugal force. If you are in the rotating frame, there is no initial rotation of the object. So you have to pick one. Both cannot exist in the same analysis. Either the object is rotating (inertial frame of reference) or the item is fixed and the reference frame is rotating.

But r can increase on its own. You don't need a radial force for this to happen. If there is no radial force, it travels in a straight line, which increases the distance to the center of the circle. I read the part where you said angular velocity was constant. If you increase it, the centripetal force must increase. If friction does not permit this to happen, then the net force is insufficient to keep it moving in a circle. There's no radial force pushing outward on it - just friction. But you had ac1 = v1^2/r1 = w1^2 r1 = Ffr1/m If you increase to w2, the frictional force can't supply the required centripetal force. The mass will slide because of this. Not because there is an outward force. And it won't stop, because the force needed for circular motion is larger than friction can supply. It continues to slide until it hits the stop. There is no centrifugal force. Friction is not directed to the outside. There is an inward radial component. That's all there is.

Then it wouldn't be the twins paradox. Then it's not the twins paradox. That's not the twins paradox. There are myriad examples one can come up with to discuss relativistic effects, but the twins paradox is a particular example — it has the element of the twin returning home to compare their clock with the one which stayed on earth, and thus has acceleration.

How does the traveling twin return without accelerating (which changes their frame)?

You said you didn't follow my explanation. Does this explain the situation adequately, or not? (You showed an arrow on your diagram. No mention of a motor, or that it would have its own mass and angular velocity)

No. Your mouse maps onto your computer screen, which represents a coordinate system. You pick an origin when you start moving the mouse.

You asked me for analysis, and that's my analysis. There is no centrifugal force, no matter how much you want there to be one. We could just as easily done this with a rope, where there physically could not be a centrifugal force. Using a rod doesn't sneak one into play. Your analysis is inconsistent with Newton's laws. No force is necessary to move the mass to r2. It would move there on its own. Look at Ghideon's animations. That shows the path the mass would take if it were able to move freely. See how r increases? No forces are acting on it once it's released. You need a motor to keep this thing rotating. You could spin it up and turn the motor off, but the system wouldn't maintain constant w when the mass moved. Also, the motor would be spinning in the opposite direction to conserve angular momentum.

That was the graph that clearly shows temperature increasing over the last ~140 years. So no, your point doesn't stand. The animation shows that looking at only a few locations wouldn't give us an accurate read on what is happening, since there are fluctuations (i.e. weather happens) Water and land each have some specific heat capacity, so their temperature will rise or fall if the absorb or release energy. Q = mc∆T If I measure in enough places to be representative of the whole, I can sum up the Q for all those areas and figure out if heat was absorbed or emitted overall. The worldwide average ∆T is representative of that value, which (as Area54 pointed out) is easier for non-experts to grasp. Saying the global average increased by 1ºC is saying we absorbed enough energy for the whole surface to increase by 1º even though some areas saw a larger increase and some saw a smaller increase, or possibly a decrease, because this is not a system in steady-state

You need coordinate systems to say something is relative to something else. Empty space is not a coordinate system.

Yes It's better to no use numbers, but if your speed changes from 1 m/s to 2 m/s in 1 second, you have a ∆v of 1 m/s and an acceleration if 1 m/s^2. (if a is constant, a = ∆v/∆t) But, if you're moving at 1 m/s in a circle with a radius of 1m, you also have an acceleration of 1 m/s^2, since your direction of motion changes. Acceleration and velocity are vectors. Changing your direction of motion requires an acceleration. (this is pointed out in Newton's first law) No, velocity is velocity But a constant velocity implies straight-line motion

First critique, of course, is that there is no centrifugal force, and no reaction force should appear in a free-body diagram. The acceleration (i.e. from the net force) is centripetal, and exerted by the rod. ac = v^2/r = w^2 r. (v = wr) You want to increase r (from, say, r1 to r2) while keeping w constant. We are assuming M>>m so we don't have to worry about the axis of rotation moving (which it will) That means the new centripetal force is larger, and also means v has to increase, which represents an increase in kinetic energy. Work has to be done, and a centripetal force doesn't do work. A tangential force must be applied by the rod. The final state of the system the rod exerts a larger tension such that ac = w^2 r2 The only way for the mass to move would be to release it from the connection to the rod, reducing the effective tension it feels. If it were free to move, it would travel a straight line tangent to its original path, which would increase r. Once you reconnect it to the rod the rod would be doing work on it to get it up to speed. But it's not free to move if there is still some attachment to the rod. The difference in the KE tells you how much work the rod must do, and that tells you the force it must exert over the path. From that you might be able to solve for the path of the mass, but it will be a messy calculation. Another thing we know is that if there are no external torques, angular momentum will be conserved. But we know the larger mass is rotating at a constant w and also transferring angular momentum and doing work on the small mass — the only way for that to happen is if it's being driven by an external torque, so angular momentum is not conserved. If you're sitting on a carousel that's rotating counter-clockwise, the world looks like it's rotating clockwise. IOW, if we look at the top of the circle, the carousel is moving left, so it looks like the outside world is moving to the right.

a = dv/dt Any change in velocity is an acceleration

How do you figure that? “This animation shows monthly temperatures for January–December 2019 compared to each month's 1981-2010 average.” 1981-2010 is not considered pre-industrial by most historians

Accelerations are not relative. The people in the boat know they are moving. (You tend not to get seasick if you’re stationary) Newton’s first law and all. And yet we can tell we’re moving. Not that this is a relevant example. I think you overestimate your mastery of physics. It was not at all clear to me you meant mathematical derivative. Derivative of what variable, with respect to what? Was it so hard to use “force” when you meant “force”? Something bobbing up and down - let’s assume a sinusoid motion in time. The second derivative is the acceleration. So maximum at the max displacement