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5

See Anand's answer: I'm not sure whether this one is simply misguided or instead subtle fraud (as calls for money to fund research are involved). Actually the claims made in the article are true in one sense, which gives the idea the whiff of sophisticated fraud. In the linked article, it is claimed that the device is powered by a 1kW source and then ...


5

When you compute the energy of a rotating object you either consider the tangential velocity or the angular velocity. The two expressions are two different ways to look at the same observable, they both lead exactly to the same result and summing them is a mistake. For instance for a point like mass on a straight line we have: $$E_l = \frac{1}{2}m v^2$$ ...


4

No, it does not gain energy. The confusion arises because there's a force that does no work. If the car moves a distance $d \vec l = \vec v dt$, then the work done during that time $dt$ is $dW_{rope} = \vec F_{rope}\cdot d \vec l= 0$. This follows because $ \vec F_{rope}$ and $d \vec l$ are always perpendicular to each other (draw and check this!) which ...


3

Your expression for the velocity looks right; but we have to get a few other things taken care of. First - the center of the marble doesn't move from 0 to 2R, it moves from r to 2R-r - so the potential energy due to this is smaller than mg(2R) which is what you had in your expression. On the other hand, you need to take account of the energy of the sphere ...


2

An energy conservation law only arises when the system studied has a Lagrangian which is invariant under time translations up to total derivatives, due to Noether's theorem. More generally, a quasi-symmetry under spacetime translations gives rise to an entire host of conserved quantities, encoded in the stress-energy tensor. Consider a scalar field; under ...


2

Matter possesses energy from Einstein's equation $E = mc^2$. This equation describes how matter is a form of energy as well, and can be converted from one form to the other. This is exactly how the sun is power, in a process called Nuclear Fusion. With this idea in mind, check this hypothesis out, called Zero-Energy Hypothesis which simply says that matter ...


1

When you have a sphere, let's say, rolling down a ramp, the gravitational potential energy will be converted into two energies: Rotational kinetic energy and translational kinetic energy. You use energy to keep it spinning (AKA moving angularly) in addition to keeping it moving translationally. Therefore, the sphere's total energy at the bottom of the ramp ...


1

This would be space elevator type of setup but it is too short to stay in orbit. It should extend beyond the geostationary orbit (~22,000 miles). International space station is a an altitude of ~200 miles and goes around the planet every 90 mins. Even if you build a proper space elevator there is no way to generate more energy than you already use to build ...


1

First of all, I would like to say that the answer Floris gave is the correct way to do the problem you've set forward, but I thought it worthwhile to note that the result ${5 \over 2}(R-r)$ is the answer if, rather than rolling a marble down the track, you are sliding a cube. If this problem comes from a textbook, there were some unstated assumptions in ...


1

Something must be proven or understood. So would beware from such answers like @Anand did. As example Nuclear reactor could be accepted like Perpetuum Mobile device, would it be invented some time ago. But definitely chances, that someone just invents something from future, without proper knowledge's , equipment, theory etc, are low (as my opinion I ...


1

According to the 1st law of thermodynamics, energy cannot be created or destroyed, it can only change form and is thus conserved. If it did work you would be going against 200 year of scientific consensus. Good luck!


1

Most, if not all, scientific analysis of real situations involves approximations. If some of the kinetic energy gained from falling was converted to kinetic energy of downstream flow (like a more sliding board shaped waterfall) it could affect the calculation. The environment could affect the temperature of the pool of water at the bottom of the falls, ...


1

There follows my try to decompose the solution into a minimal amount of calculation and apart from that only geometrical considerations. The centripetal acceleration is $a_c=\frac{v^2}R$. It is directed towards the center. We define the $z$ coordinate as starting at the top and pointing vertically downwards (see the following Figure). The conservation ...


1

A mass moving in a circle has centripetal acceleration $v^2/r$ directed toward the center of the circle. You can get $v$ from potential energy. The mass here has two forces on it. Gravity is constant and down. The reaction force of the surface (assuming no friction) is normal to the surface. When the sum of these two forces becomes less than centripetal ...


1

The potential energy $ P $ of the car doesn't change (the car stays on the ground the whole time), and because it moves uniformly in a circle, its speed $\left | \mathbf{V} \right |$ doesn't change. But the kinetic energy $ K $ of the car is $\frac {1}{2} m\left | \mathbf{V} \right |^2 $, so it doesn't change aswell. Hence, the total energy $ E= K + P $ ...


1

$\def\om{\omega}\def\vr{{\vec r}}\def\l{\left}\def\r{\right}\def\ve{{\vec e}}\def\vom{{\vec\omega}}\def\ds{\,'}$ Let the car move in the (x,y)-plane, let $m$ be the car's mass and let $J$ be the moment of inertia for the rotation about the axis through the center of mass aligned parallel to the z-direction. If you have a straight line and a circle with ...


1

I will give you a starting point: You could solve this problem by equating the change in potential energy with the change in kinetic energy, $\Delta P.E. = \Delta K.E.$ You could solve the problem with calculus starting from $$ F=ma=mdv/dt=mvdv/dr=F_{gravity}(r)\Rightarrow\\ \int mv dv = \int F_{gravity}(r) dr $$ Even better: solve it both ways and check ...


1

Say the density of the string is $\mu$ and the tension is $T$. It's clear that the kinetic energy of an infinitesimal piece of string is $$dT = \frac{1}{2} (\mu \, dx) u_t(x)^2$$ The length of the infinitesimal piece of string from $(x, u(x))$ to $(x + dx, u(x + dx))$ is \begin{align} d\ell &= \sqrt{dx^2 + (u(x+dx) - u(x))^2} \\ &= \sqrt{dx^2 + ...



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