Recent content by Brinx

We did some more experiments here at the department, and we got a couple of interesting results. What we did: we used a ball chain (2.4mm ball diameter) of several meters in length, and piled it into a compact heap onto the middle of a table. We let the end of the chain dangle over the table...

This post refers to simulation software that can be used to study the Ising model on curved surfaces: https://www.physicsforums.com/showthread.php?t=409671 This work treats 'sphere-like' lattices, such as a pillow-shaped one and a cube projected onto a sphere...

In 1 second, you're selling 100 kW * 1 s = 100 'kWs' = 100 kJ. Expressed alternatively, this is about 0.028 kWh because 1 kWh = 1000 W * 1 h = 1000 W * 3600 s = 3.6 * 10^6 J. In the whole hour, you have sold 100 kW * 3600 s = 3.6 * 10^8 J = 100 kWh. Power integrated over time equals energy.

Ah, you're right of course - my apologies. I somehow completely misinterpreted sophiecentaur's post. Dissipation power is proportional to velocity, but dynamic friction force is constant.

If the dynamical friction force on a surface element of the book is linearly dependent on the velocity of that surface element, and if we only consider a purely translational motion for simplicity, then the momentum of the book would decay exponentially it seems: the smaller the momentum, the...

That's what was bothering me a bit too - it is why I considered the whole curved part of the trajectory as a black box in a previous post. Then again, I did assume the tension at both ends of that segment to be constant and I am still not quite convinced one way or the other about that. As...

What is a good way to determine along which part of the chain your derivation holds, in that case? Is there a specific condition by which we may define the start and the end of the arc?

That's a great post JollyOlly, and the line of reasoning is clear. I'm wondering about the statement that the two tension forces (at the start of the arc and the end of the arc) can be unequal though: does this not imply that there is angular acceleration going on in the arc (because of a net...

Right, that makes sense. So running on a treadmill with a given slope will have you expend about as much energy as running up an actual hill with the same slope. Both of those situations require that you expend more energy per unit time than if you were running along flat ground.

A.T., the fact remains that running up a hill takes a lot more energy than running on a flat surface - this is something that is easily experimentally verified. If I'm reading your words right, you seem to think that elevating your center of mass does not require energy. In that case, you'll...

See: http://en.wikipedia.org/wiki/Mach's_principle

Hmm, I'm not sure about this. A simplified case: sliding a mass across a frictionless flat surface at constant velocity requires only the initial impulse and no subsequent work, but sliding the same mass up a frictionless slope (i.e. partly against the force of gravity) does require a steady...

My apologies for the giant wall of text that this post has turned into. I hope people find the patience to read it all! DrClaude: the minimum radius of curvature of the chain is a lot smaller than the radius of curvature that is observed in the 'loop' that forms. I would not expect it to be a...

So, would it be possible to measure the local charge density at each endpoint of a non-closed length of wire moving through the uniform magnetic field? This should be done in such a way that no closed circuit is formed which the wire would be a part of. Think of a dumbbell with the wire being...

Yes, if you have applied the force for the same duration to both objects the magnitudes of p1 and p2 will be the same. The velocities of the objects, however, will depend on their respective masses: v1 = p1 / m1 and v2 = p2 / m2.