# Tag Info

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You are right in saying that $I$ allows one to relate angular velocity and angular momentum in a linear way. It is just not as simple as the momentum and velocity case. An intuition for why things get complicated is that $L = r \times p$ involves a cross product which makes it very sensitive to the choice of a specific set of orthonormal bases(with fixed ...

1

Here is my derivation of this result. I hope you find it helpful: Say we have n different forces $F_1, F_2, F_3... F_n$, applied at n different points. Now we pick two centers $P$ and $Q$, and express the radial vectors (1) from point $P$ to each of the n points (where forces are applied) as $r_1, r_2, ... r_n$ (2) from point $Q$ to each of the n points ...

3

I think that you're making this problem more complicated than it has to be in order to simply determine if the assembly will tip over or not. You don't really need the spatial distribution of the forces being exerted by the table or ground on the assembly. All you need to note is that if the pivot point is at x=D1 then the ground will exert whatever ...

2

If there is no torque, then $$\sum \tau = \mathbf{r_1}\times\mathbf{F_M}+\mathbf{r_2}\times\mathbf{F_m}=0$$ Therefore, $$\mathbf{r_1}\times M\mathbf{g}+\mathbf{r_2}\times m\mathbf{g}=0\tag{1}$$ where $\mathbf{r_i}$ denotes the position of the center of mass of the combined system relative to the force applied. If we give the box dimensions $h$ and $l$, the ...

0

The 2nd solution you wrote down appears to be the correct solution. Offhand, I see two problems in the first solution. First, I think that a problem with the first attempted solution is that you made a subtle mistake in assuming that F=ma means that $F=mr_1α$. That seems like a plausible step at first but if you examine this step more closely you'll realize ...

0

Potential energy like Force occur in pair. If one has some potential energy due to 2nd, 2nd will have the same potential energy as the first. In the Gravitational potential energy equation : $U = \frac{GMm}{r}$ The potential energy is dependent on both the masses. This value is same regardless whether it is for 1st or 2nd. Both the body can do Same amount ...

3

Particles have gravitational potential energy due to its position in the gravitational field. Systems have potential energy. Ascribing the energy to a particle is incorrect. We say the particle has potential energy and not the Earth (the body doing the work). That is incorrect. The potential energy is a function of the system, specifically the ...

0

Potential energy is just energy stored in a static state -without motion. So a spring can have potential energy, and so can a body attached to the spring that's in a gravitational field. So for this type of system (undamped harmonic oscillator in a gravitational field) potential energy is not strictly defined for the spring. If the forces are conservative ...

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the method i use for determining the direction is quite simple. basically when you are to find the direction of a cross product say a x b just put your hand on a such that it is facing the b vector. thus for finding the direction of torque put your hand on the radius ( in the direction perpendicular to the force) such that it faces the force and curl your ...

0

@vishwaas is correct, but here is an explanation without vector math. By convention, $x$-$y$-$z$ ordinates obey their own RHR such that rotating an imaginary right hand from positive $x$-axis to positive $y$-axis around the $z$-axis (counter-clockwise) puts positive $z$ in the towards-the-observer direction (out of the page). The rotation of, and the torque ...

2

You have not considered the direction of torque in you equation. Since the Torque is caused by Frictional Force $F_r$ which is in the direction $-i$. the torque is $\vec{\tau}= \vec{r}\times\vec{F_r}$ and $\vec{r}$ is in $j$ direction So the cross product yields $\vec{\tau}= -|r||F_r|\hat{k}$ which ofcourse is in the negative direction of positive z-axis

1

This isn't too hard. Translations clearly are isometries: if $\vec a' = \vec a + \vec c$ and $\vec b' = \vec b + \vec c$ then $|\vec a' - \vec b'| = |\vec a - \vec b|.$ Consider any isometry $f$; consider the isometry $g(x) = f(x) - f(0)$ which preserves the origin. It's not too hard to see that this has to be linear and is therefore described by a matrix ...

0

The $\mathbb{R}^{3}$ and the $SO(3)$ respectively parametrize the center-of-mass trajectory and the relative motion (i.e., rotation) about the center of mass. To see why the latter is $SO(3)$, imagine three mutually orthogonal unit vectors that are fixed to the rigid body and whose base points all coincide with the center of mass. One can completely ...

2

Current trend today in MBD is towards writing code, doing simulations for some practical problems. That is not entirely true and it mostly depends on the actual areas and topics you are dealing with, as for all the other subjects. Of course, due to the industrial applications, the practical side always has more money and more academic positions, but ...

0

This is an excellent problem. It is counter intuitive, so the following drawing might help: Assuming the volume of the sphere is small, it will not displace a lot of water when at the bottom of the pool. However, if it's placed on top of the boat, the boat will now have to displace enough water to support the extra weight of the lead sphere. The volume of ...

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You can certainly do this "the hard way", and the potential may be so straightforward that this is easiest: $V(x)$ is a harmonic oscillator potential $\frac12 m \omega^2 x^2$ yielding the equations of motion $\ddot x = -\omega^2 x.$ Probably the best way to do this is to write $x = A \cos(\omega t)$ from $t=-\tau$ to $+\tau$; a derivative gives an equation ...

1

As you may know, You can equate a damper with a capacitor and a spring with an inductor. You'd need to transfer the energy from your wave somehow to those mechanics in order to use the damper and the spring.

2

There is already a very useful mechanical equivalent called the hydraulic analogy. You'll find lots of related posts already on this site. All analogies have their limits, but the hydraulic analogy is remarkably good. You can even represent components like capacitors, inductors and even semiconductor junctions.

1

Yes, in electrical circuits of only passive elements (Resistors, Capacitors, and Inductors) only the Resistors dissipate power (as heat). Active elements like transistors can also dissipate power, and if the currents in the circuit are changing with time, then power can be radiated away in electromagnetic waves. Therefore, the electrical resistance of the ...

1

Yes it is true. The only DC magnets that use "no" power are superconducting magnets (like in MRI systems). Of course for those, there is significant power needed to keep the windings at superconducting temperature... and the cooling system will typically use several kW. "How much power does a junkyard magnet use" is not an easy thing to answer: but ...

0

To the extent that all your matrix elements are <<1, these rotation and strain parameters are doing the matrix transformation (think of the cube being centered on the origin, and M being applied to vectors going to the corners of the cube): $$M=I+ \epsilon + \omega$$ This matrix would simultaneously Strain (stretch) the box fractionally by .2 in ...

3

On a well designed balance scale, the center of gravity of the beam or lever arm will be just slightly below the center pivot point. If the beam is not level, the center of gravity will be to one side or the other of the pivot, and will thus create torque as it tries to move directly below the pivot point. The distance between the pivot point and the ...

1

Your expression for the acceleration due to the kinetic friction is incorrect. Remember, $$f_k = \mu_kn,$$ where $n$ is the normal force. To find the normal force, you have to use what you know about the centripetal acceleration. Draw a free-body diagram, and label the weight of the car and the normal force, and then you know that ...

0

I can use the coordinates $x=R\cos\omega t+l\sin \theta$ and $y=R\sin\omega t-l\cos \theta$. I think these are all I need. Can you leave a comment if you agree or not?

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Both answers are incorrect. The correct answer taking into account both angles as mentioned by others, has R-r in the numerator. The limiting case, in which the ball gets really small, produces a period that cannot go to zero, and this consideration alone can be used to eliminate both of your answers.

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Here is how I understand it: let's say you have some coordinates $p_i$ and some momenta $q_i$. You want to find transformations $$Q_i\equiv Q_i(q_i,\,...,\,q_n,\,p_i,...,\,p_n,\,t)\qquad P_i\equiv P_i(q_i,\,...,\,q_n,\,p_i,...,\,p_n,\,t)$$ These variables $p_i$, $q_i$, $Q_i$, and $P_i$ must satisfy the Hamilton's equations of motion: ...

3

The question duplicates $\hbar \rightarrow 0$ in QM where it has been soundly answered. The most direct bridge is through deformation quantization, the phase-space formulation of QM, where operator observables are mapped injectively onto their Wigner transforms, c-number phase-space functions, just like their classical counterparts. It is then evident that ...

0

I believe although there were no large physics hurtles to overcome in developing the space pen, all of the pens that were currently on the market at the time were highly unreliable without gravity assisting them, and were thus unacceptable for use in space. Additionally, pencils produced graphite dust which in space will float around until it lands ...

3

I'll start by setting the tangential part of the question aside: In quantum field theory there are currents and charges which are conserved classically, but not quantum mechanically, that is, there are violations of these conservation laws to order $\hbar^2$, and hence the symmetries hold classically but not quantum mechanically. (See below for what that ...

2

Well.. 1) If you were running at the speed of sound, you probably wouldn't be for long. The human body isn't designed to handle those kinds of stresses. 2) Assuming you're listening to the iPod using ear buds (in your ear) You can probably think of the air between the seal on the ear bud and your ear drum as isolated from the air you're running through, ...

2

how do you find potential in a place where we have no intuition of force and are not allow to find it. Well I think this might be your problem; I've certainly never heard it said that you are not allowed to find forces. The Euler-Lagrange equations are simply another tool to finding the dynamics, but that doesn't mean you have to start from scratch and ...

0

As illustrated HERE on Astronomy SE, since the velocity of the meteor or comet nucleus is going to be about 40 km/s (essentially the escape velocity from the sun at at 1 AU) and the velocity of Earth is about 30 km/s, the relative velocity of the two can be anywhere from 10 km/s to 70 km/s (varying by a factor of 7) depending on the direction the object is ...

0

In a general Lagrangian formalism, $L$ doesn't equal $T - V$. Rather it is a function of some field (be it scalar field, vector field, or whatever other field that is useful...) $\phi$, derivatives of $\phi$ and spacetime (x,y,z,t). This function is chosen so that the equations of motion produce the correct physical phenomena. In general, canonical ...

0

So you started with $L' = L(|v|^2 + 2v\bullet \epsilon + \epsilon^2)$ If you treat $2v\bullet \epsilon + \epsilon^2$ as a variation of $|v|^2$, you may use the taylor's theorem treating $|v|^2$ as a variable, and you get the following expression: $L(|v'|^2) = L(|v|^2) + \frac{\partial L}{\partial |v|^2} (2v \bullet\epsilon + \epsilon^2)$ + higher order ...

4

Yes, it does, A dynamo diriven in reverse simply changes the sign of the generated voltage. As a dynamo produces AC and changes the sign of voltage a few time each second, you wont notice that. So even alterating the rotation every minute has close to no effect (except from having to slow down and reaccelerate a potentialy heavy axis)

1

The experiment you propose seems fairly sound, but you would need to test it many times to check whether a consistent co-efficient of friction can describe a beetle's stickyness well. With small creatures, friction is often way too blunt a concept to measure or describe a their ability to cling to a surface. Some creatures can switch their clinging ability ...

1

Given $$\int_{t_1}^{t_2}\textrm{d}t\,\left(\frac{\partial L}{\partial q}\,\delta q + \frac{\partial L}{\partial v}\,\delta v \right)= \int_{t_1}^{t_2}\textrm{d}t\,\left(\frac{\partial L}{\partial q}\,\delta q\right) + \int_{t_1}^{t_2}\textrm{d}t\,\left(\frac{\partial L}{\partial v}\,\frac{d}{dt}\delta q \right)$$ then the second contribution on the right ...

0

Partial integration is employed only for the second term in (1): $$\underset{{t}_{1}}{\overset{{t}_{2}}{\int }}\frac{\partial L}{\partial v}\delta v=\underset{{t}_{1}}{\overset{{t}_{2}}{\int }}\frac{d}{dt}\left(\frac{\partial L}{\partial v}\delta q\right)dt-\underset{{t}_{1}}{\overset{{t}_{2}}{\int }}\left(\frac{d}{dt}\frac{\partial L}{\partial ... 2 Well, OK, this is a resolutely vague question, but there is something special, actually, about angular momentum dimensions. In quantum mechanics, the fundamental constant, \hbar has dimensions of angular momentum (and is very small in terms of angular momenta, actions, or phase-space areas of our macroscopic world experience). Classical mechanics results ... 1 So I think this is going on here: The green rope rotates about A. The blue rope rotates about B. The disk rolls off the green rope and thus has a relative center of rotation at D. The disk also rolls off the blue rope and thus has a relative center of rotation at E. The absolute center of rotation of the disk has to be on the line AD as well as on the ... 0 The linear acceleration of B due to the rotational movement:$$ a_r = {L \over 2} {FL/2 \over ML^2/12} = {3F \over M} $$The linear acceleration of B due to the linear acceleration of the center of mass of the rod (which is anti-parallel to the rotational acceleration):$$ a_t = {F \over M} $$Since these two accelerations are on opposite directions to ... 1 The short answer is that the 770 HP is the more powerful. However, I suspect the question is about the difference between power and torque. 'Torque' is the rotational equivalent of force and if an engine is exerting a 'high torque', it's pushing hard. This occurs when the vehicle is accelerating, especially from rest. 'Power' is the rate at which ... -1 The 770 horsepower one. The torque just refers to the ability of the engine to turn the shaft, which is then converted by the transmission. However, the horsepower refers to the actual amount of power or 'acceleration' the engine can dish out. It is the job of the transmission to ensure that the engine has the ability to efficiently supply power to the ... 0 Irrational 3.14 It seems to me that you are using the equation torque = moment of inertia * angular acceleration. This particular equation ONLY holds under three conditions: The rotational center is fixed. The rotational center is the center of mass. The rotational center is moving at a constant velocity. In your scenario, the center of the rod ... 11 Well, \{f, \cdot \}, similarly to \{H,\cdot\}, computes the derivative of the argument \cdot with respect to the action of the one-parameter group of canonical transformations generated by f,$$\phi_a^{(f)} : F \to F\:,\quad a \in \mathbb R\:.\phi_a^{(f)} \circ \phi_b^{(f)}= \phi_{a+b}^{(f)}\:, \quad\phi_{-a}^{(f)} = (\phi_a^{(f)})^{-1} \:, ...

1

Yes it does. In fact, it is one of the( if not the) most important conclusions of Quantum mechanics. If {f,g}= 0 it means that the variables have simultaneous eigenvalues i.e. you can measure both of them on same instant of time. but {f,g} can be non-zero which leads to the theoretical conchusion that the eigenstates of f and g are not simutaneously ...

0

Now, someone tried to mark this question as a duplicate of this other one. From my point of view, having derived the Euler-Lagrange equations before even mentioning the least action principle or the action, it didn't seem too related. The point was, I wanted to have a physical interpretation of the Lagrangian, and leave the action and the principle as ...

3

is there any reason for this quantity to be introduced? It is a quantity for which the actual dynamics makes the integral of the thing be stationary with respect to changes of paths when you consider alternate, but nearby changed paths. Does it have any physical meaning? One problem is that many Lagrangians give the same equations of motion, so ...

0

The kid pulls himself, so there are two tensions 2T to support his body: 2T-mg=(1/5)mg so T=(3/5)mg The force the kid applied equals the normal force n exerted by the seat, therefore T+n-mg=ma n=mg-T+ma=mg-(3/5)mg+(1/5)mg=(3/5)mg

1

As you have already mentioned, L is NOT in general T-V. T-V only holds in classical mechanics. And I will try to motivate the construction of T-V in classical mechanics following "The Variational Principles of Mechanics" by Cornelius Lanczos. To start out, let's talk about statics. It it well known that the condition for a physical system to be at ...

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