# Tag Info

6

Your intuition was correct - the shaft will rotate in one direction and the housing/stator will rotate in the other. If you look up "moment of inertia" you will find that it is the rotational equivalent of mass. For almost any reasonable motor the moment of inertia of the shaft/rotor windings will be smaller than the moment of inertia of the housing/stator. ...

4

You usually cannot push your hand through the table, because it's a single solid. The atoms are held together by covalent bonds, which are electromagnetic in nature. Sand on the other hand is grainy - the $SiO_2$ grains do not interact with each other and are only held "in place" because of gravity. You can run your hand through sand similar to driving a ...

3

It is well known that adding a total time derivative to the Lagrangian does not change equations of motion. The Lagrangian above adds a term $$-q\dot q=-\frac{1}{2}\frac{\mathrm{d}q^2}{\mathrm{d}t}$$ (a total time derivative) to the free particle Lagrangian $\dot q^2$. It is thus fully equivalent to to the standard free particle Lagrangian (up to an ...

2

The astronauts working on the Hubble space telescope had to bring special low torque wrenches to counteract the effect of the torque of the motor spinning them around, due to conservation of angular momentum, although this meant far more use of muscular power to hold them in place. And also to avoid damaging the equipment they worked on, such as screws ...

2

@ChrisDrost's answer is correct, but we can actually remove the assumption that the friction is constant by considering conservation of angular momentum instead. If we put our origin at a point along the ground, then there is no net torque on the sphere: The frictional force always points directly towards (or away from) the origin, and the normal force ...

2

Since a system must obey the law of momentum conservation, the center of mass of a system (which can be made of one or many bodies) must have constant velocity if no external force is applied. Hence, a body can rotate around its center of mass, or it can rotate around any other point, but only if under the influence of an external force. Therefore one can ...

2

It's a complicated subject - but very well studied in the context of eddy current brakes, where the retarding force is used to create a braking force without mechanical friction / wear. For me, the starting point for finding out more was this post - in particular the posting by Jim Hardy contained lots of good links. It seems that some of the most ...

1

I guess the Alka Seltzer is an effervescent kind of tablet. When it is immersed in water, gas bubbles form on the tablet surface and surface tension effects prevent the bubbles from separating them from the tablet surface. By this effect the global density of the system tablet plus bubbles is going down overtime, until the global density approaches the ...

1

For this sort of problem it really helps to go back to the Riemann sum. You don't even need to write it down per se; you just need to think about what your $\delta x$ is in the Riemann sum. So, in this case you're trying to find a "sum of forces" or we might say a "net" force. To find this, it helps a little to think about momentum conservation/Newton's ...

1

As noted in some of the comments already, the answer to your question depends on what properties of quantum eigenstates you want your classical analogues to have. If you're thinking of eigenstates of arbitrary observables, quantum eigenstates have the property that the value of that observable is precisely defined. In classical mechanics, individual points ...

1

The general equation is $$x(t) = x_0+ {\bf e}^{-\beta t} \left( A \sin \omega t + B \cos \omega t\right)$$ where the constants $A$ and $B$ depend on the initial conditions, and $\beta$ and $\omega$ on the mechanical properties of the system. Read http://hyperphysics.phy-astr.gsu.edu/hbase/oscda.html for more information For example if the initial ...

1

It sounds like you are interested in symplectic reduction procedures. On of these methods is that of Routh's procedure to eliminate cyclic variables using a Legendre transform to a reduced-variable Hamiltonian called a Routhian. Forming a variational approach may be difficult for some reduction procedures, however we can view conserved quantities as ...

1

While neither the Lagrangian $\mathcal{L}$ nor the action $S$ are invariant under boosts of the form $$\dot{q}(t) \to \dot{q}(t) + v, \quad v \in \mathbb{R},$$ the Euler-Lagrange equations are. The dynamics of the systems are unchanged for any transformation that preserves $\delta S = 0$, i.e. a transformation of the form  \mathcal{L}(q, \dot{q}, t) \to ...

1

I don't have the expertise you're looking for, but here's a crack at it: This is a really hard problem whose answer depends on the material properties. In particular, the answer depends on how big the 'eddies' are: if the current moves in roughly a big circle then the effect is large, but if there are lots of little eddies the effect is small. However, ...

1

I can give a back-of-the-envelope derivation of a drag force that ignores fringe effects and other complications. Say the conductor is a plate of thickness $\Delta z$ traveling with velocity $v$ in the x direction. Take the magnetic field to be constant in a rectangular area, with the the $\Delta y$ side perpendicular to the velocity much longer than $\Delta ... 1 $$\mathcal{L}\left( q,\dot{q},t\right)= \dot{q}^{2} - q\dot{q} \tag{01}$$ $$\dfrac{d}{dt}\left(\dfrac{\partial \mathcal{L}}{\partial \dot{q}}\right)-\dfrac{\partial \mathcal{L}}{\partial q}=0 \tag{02}$$ \dfrac{d}{dt}\left[\dfrac{\partial \left(\dot{q}^{2} - q\dot{q}\right)}{\partial ... 1 Try to think of this problem using a polar coordinate system.$x$is essentially the radius$r$or$\rho$, measured from pivotal.$w$is simply the angular velocity. So the position vector of the object is$x\hat{\vec r}+\theta\hat{\vec \theta}$So the velocity vector is$\dot x\hat{\vec r}+w\hat{\vec \theta}\$ The hatted vectors are unit. So the ...

1

So this is a phenomenon which is known in billiards as "backspin": you hit a ball off-center and it simultaneously has a motion "forwards" but a spin that imparts a force on the ground to send it "backwards". Trick shots where you induce extreme amounts of backspin by hitting the ball almost vertically downwards are known sometimes as "massé shots", if you ...

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