Recent content by jbriggs444

If the motion is like a jump rope then the ##x## axis (the line between the anchored end points) would remain in place. One then has a choice to make. One could use an inertial coordinates where points on the rope are rotating in the ##y## and ##z## directions. With this choice, the bits of...

You've mentioned Feynman's analogy twice now. Let me link it for you. I love that lecture also. https://www.feynmanlectures.caltech.edu/I_04.html

In a typical bicycle wheel, the spokes are in radial tension while the rim is under circumferential compression. The spoke tension is significantly larger than the centripetal force that would be required to keep the bits of rim in uniform circular motion at the rotation rates encountered in...

Then stop yapping? Yes. You can apply any force you like. But if you apply a centripetal force, it will supply zero torque. The centripetal force required to keep an object in uniform circular motion is ##F_c=\frac{mv^2}{R}##, yes. Presumably any such force is supplied by the spokes of the...

Computer people and mathematical people tend to think about modulus differently. For computer folks, we think about modulus arithmetic as a function. You put a dividend and a divisor in. You get a remainder out. So "3 mod 2" is a number: The remainder of three upon division by two. Just as you...

I do not see anything missed, although it seems like more work than is required. I would have liked to see the point more clearly and directly made that $$|a_A^x| = \frac{T_1}{m} > \frac{T_1 \sin \theta}{m} = |a_B^x|$$ Personally, I would have taken a different approach. Reason first that the...

Per the problem description you have four forces.

It is not clearly specified, but I believe that Alice is only told which row she is to fill in and Bob is only told which column he will fill in. Alice reads from the agreed-upon 3 x 3 table by looking at the selected row and making the set of three choices listed in that row. Bob reads from...

That is the formula for the magnitude of the Coriolis force. Not for its horizontal component.

When I want to reconcile this simplifying assumption with the real world, I tend to think about things relaxing toward an equilibrium. This will make more sense once one has an intuition for spring-like behavior under one's belt. As one pushes harder and harder on an object, it deflects more...

You do not need or want a time standard based on time passage at the center of the Earth. Better to use a standard at the geoid. Which is what is actually done. https://en.wikipedia.org/wiki/International_Atomic_Time

As @jack action has indicated, centripetal acceleration has essentially nothing to do with friction -- except to the extent that it could conceivably be used as a surrogate for a speed measurement. Halve the wheel diameter and you've doubled centripetal acceleration without affecting friction...

[Human] power-limited acceleration and friction-limited deceleration should be very different.

So not invariant. And also not necessarily constant. So speaking about "the distance" is already erroneous. The definite article is not warranted. We are only talking about "a distance".

You are going to need to be much more clear on what test you are running. Are you measuring the drag force (rolling resistance) during acceleration? Are you measuring the kinetic friction during decelleration during a skid to a stop? The coefficient ##\frac{a}{g}## is applicable to a skidding...