# Tag Info

17

Calculating the power emitted as gravitational waves is relatively straightforward, and you'll find it described in any advanced work on GR. I found a nice description in Gravitational Waves: Sources, Detectors and Searches. To summarise an awful lot of algebra, the power emitted as gravitational waves by a rotating object is approximately: $$P = ... 14 In this case, gravity is still an external force. In a zero-g environment, the mouse would also begin to move around the inside of the wheel, opposite the rotation it causes in the wheel, which would keep the angular momentum at zero. This would happen because the only way for the mouse to exert a force on the wheel and rotate it is for it to push itself in ... 6 Your intuition is correct. For the ball to change its angular momentum (to go from "backspin" to "forward spin"), there needs to be a net torque acting. There are two forces on the ball: gravity, and the normal force of the slope. Both these forces act through the center of mass - so neither force adds torque. Without torque, there is no change in angular ... 6 Electrons in a conducting disk in order to maintain equilibrium will have to have a centripetal force on them equal to the local change in potential energy with respect to a change in radius, that is$$ m_e\omega^2 r = -e{d\phi\over dr} $$After integrating, we get a potential difference between the center and a point R out$$ \Delta\phi = -{m_e\omega^2 ...

4

Suppose you pedal at about 100 rpm (I don't know what a typical rate of pedalling is, but this seems a plausible order of magnitude). To make a fair comparison with your motorbike you need to gear the moorbike engine down from 7,000 rpm to the same 100 rpm that you pedal at, i.e. a factor of 70, and this will multiply the torque by a factor of 70. So at 100 ...

3

You're using a vector formula with cross product. Let's switch to this: t = r * F * sin theta, where theta is the angle between your leg and the lever on the sprocket wheel. Theta will change depending on where you are in the cycle of pedaling your bike. Greatest value for sin theta will be when the sprocket crank is horizontal and your leg is ...

3

Your first quote is correct for an idealised model. There is no rolling friction then. Both wheel and surface are considered completely rigid. Ideal model - no rolling friction Non-ideal/more realistic model - rolling friction comes into the picture These pictures are from this link that gives a very good graphic view on this. Going away from an ideal ...

2

Lots of aspects to this question. Mass per se does not increase drag - volume of the bob of the pendulum might. The increase in mass (inertia) makes the stored energy of the bob larger (thus - longer time for motion to decay); but the projected area (in the direction of the motion of the bob) will presumably also increase, which result in greater drag. ...

2

If you have a good pendulum clock, you may be able to extract a minute amount of energy from it - but usually any additional friction will cause the clock to stop ticking. This is of course a function of the amount of energy you try to extract. Here is the experiment to do: Tape a strong disk magnet to the back of the pendulum. Make a coil of magnet wire ...

2

You correctly identified there are two angles of interest, labeled $\theta$ and $\phi$. I actually want to pick two different angles for my analysis - see this diagram: First thing to note is that if the cylinder rolls without sliding, the length of the green arc and the red arc must be the same. Length of green arc: $(\phi + \theta) a$ Length of red ...

2

The direction of the motion at any time $t$ is the direction of the velocity vector $\textbf{v}(t)$ as derived by solving the equations of motion; likewise $\omega(t)$ gives you back the direction of rotation according to the right hand rule. friction is the force that causes rotation is not entirely correct. Any force with non-zero torque generates ...

2

I suspect this is an example of the spinning-egg problem, in which a prolate spheroid (such as an egg) spun on a table about one of its "short" axes will tend to "stand up" so that it's spinning about its long axis. A few explanations have been proposed for this phenomenon, most notably: H. K. Moffatt & Y. Shimomura, "Spinning eggs — a paradox ...

1

The angular momentum of the earth && mouse wheel system does not change. When the earth pulls on the mouse, the mouse pulls on the earth, so no net moment is seen any arbitrary point in the universe.

1

What data do you have for linear motion? Your equations are correct if you have the acceleration as a function of time and the orientation is constant. The angular accelerometer can give you the angles as a function of time with integration. Unfortunately, drift can be a problem. The received wisdom is to use an accelerometer (linear or angle), integrate ...

1

The rotation of the Earth's dipolar magnetic field produces an electric field in space. Because the electric field is zero in the rotating frame, it is equal to $$\mathbf E=-(\omega\times \mathbf r)\times \mathbf B$$ in a fixed frame, where $\omega$ is the angular velocity of the Earth, $\mathbf r$ the radial distance and $\mathbf B$ the magnetic field. ...

1

Never seen that before, so I just tried it. Cool. I believe that the membrane between the yolk and the white is elastic, so when you first, gently, give the egg a little angular momentum, you are only spinning the white. As the yolk catches up the effective moment of inertia drops, and conservation of momentum therefor implies a higher angular velocity.

1

The hammer can be thought as a mean to deliver enough energy to the nail to deform the underlying material and let it penetrate deeper. Ideally, all the energy the hammer gets from your arm as kinetic energy is transferred to the material and results in its (hopefully) permanent deformation, but as always, that's not the case in practice. What should be ...

1

You are right. In an optimal system, there isn't any friction. But in real life there is, because there aren't any balls or surfaces that are so perfect that they touch just in one point. Also the surface of the ball is never perfectly flat, so there will always be friction between the surface of the ball and the surrounding air. So for calculating the ...

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