The first sentence of the problem statement says "The figure shows an overhead view".
It is purely a matter of convention. If you are working in two dimensions with one of them vertical (for instance, a child carrying a sled up a hill) then one would normally use x for horizontal and y for...
Nope. I am out as well. I asked for an explanation and got nothing but a regurgitation.
Your response to "how does that work" was:?
That is not an answer. That is an assertion of possibility. But I do not understand what you are even describing. I am asking you to describe it.
Your response...
I will try my hand. But, like @Dale, I do not understand the arrangement.
We seem to have a wheel on a frictionless axle anchored to the ground. We seem to have an array of disks somehow bound to the wheel. Perhaps there are spokes holding them in.
There is no friction between wheel and disks...
Right. Those are a third law pair. Particle on spring and spring on particle.
The point was that the force of particle on spring and floor on spring are not a third law pair. The fact that those are equal and opposite is a consequence of the second law: ##\sum F = ma##. If ##ma## is negligible...
That's a property, not a definition.
Edit: I take it back. It is a definition:
|x| denotes the unique non-negative number that, when squared, yields ##x^2##.
Although, if you have a definition of "non-negative" in hand and are working in a field, it seems simpler to just do the definition by...
But it's not the definition of ##|a|##. It's the definition of ##\sqrt{a^2}##. There is nothing in taking a square root that demands that the result be positive or negative. We have decided by fiat that the ##\sqrt{}## notation denotes the positive square root.
A cyclist adopting a reasonably fixed cadence will be in twice as high a gear at 20 mph as at 10 mph. At a consistent effort, this means half the torque delivered to wheels rotating at twice the speed.
Gaining an incremental 5 miles per hour takes twice as long because the available torque is...
Using radians per second gets rid of a mildly annoying conversion factor if you try to use torque and rotation rate to compute power. Or when you try to use rotation rate and wheel radius to compute linear speed.
Us math types get used to radians and have trouble imagining any other way to do...
In principle, this is a curve-fitting problem. You have three data points that you are using to fit a quadratic equation.
There are three unknowns in a quadratic equation (the "a", "b" and "c" in f(x) = ax^2 + bx + c). There are three data points that you have been asked to use (0.5, 0.1)...
Doing it is fine. Expecting it to be useful in a computation of energy needed to escape is questionable. If you need infinite energy to escape to infinity and you have infinite energy coming from the centrifugal field, how is a little bit of a boost to start with supposed to help?