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1

There are already many good answers. Besides the fact that the standard definition of work directly relates to the work-energy theorem and the notion of potential energy, here is a geometric argument. I) The force $F_i(x,v,t)$, $i\in\{1,2,3\},$ transforms as $(0,1)$ co-vector $$\tag{1} F_i ~=~\sum_{j=1}^3F^{\prime}_j \frac{\partial x^{\prime j}}{\partial ...


2

For starters, these are not the same thing. The integration by parts rule makes this fairly obvious: $$\int_i^f y\,\mathrm{d}x = y_f x_f - y_i x_i - \int_i^f x\,\mathrm{d}y$$ But then you might be wondering what makes $\int \vec{F}\cdot\mathrm{d}\vec{s}$ the "right" definition for work while $\int \vec{s}\cdot\mathrm{d}\vec{F}$ is the "wrong" one. In a ...


5

The reason the relationship $$ W=\int\mathbf s\cdot d\mathbf F $$ doesn't work is because Work is defined as the result of a force $\mathbf F$ on a point that moves along a distance. The point follows a curve $\mathbf s$ with a velocity $\mathbf v$. The small amount of work, $\delta W$, that occurs of the instant of time $dt$ is $$ \delta W=\mathbf F(\mathbf ...


3

Because, according to your definitions, if I strain a rubber bar with constant force until it rips apart, I haven't done one joule of work to it.


0

Since you are interested in showing that the velocity must be equal to something perhaps you should start by considering the kinetic energy the block has before going up the ramp. That energy will be taken out by the work done by friction and gravity.


0

The box initially has some kinetic energy $KE = \frac{m}{2}v^2$. As the box slides up the ramp, $KE$ is lost due to friction, and $KE$ is also converted to gravitational potential energy as the box rises in elevation. The initial velocity is important because it allows you to calculate the initial $KE$ in the system.


-2

Let us take all the three bodies,together with string as our system. Yes,tension in both the strings will be equal as both masses are tied through string symmetrically about B. So they both experience equal tension. Yes,acceleration of center of mass remains constant as no external force acting on system(tension is an internal force). Although position of ...


1

Let's assume that the balls are identical, and the strings are identical, and are elastic (obey Hooke's Law) and massless. You have stated that you've somehow arranged for B to have a constant acceleration. This means that the net force on B is constant. There are three forces on B: one due to the string connected to A, one due to the string connected to ...


1

As mass B accelerates, the other two "stay behind". The string will no longer be at 90 degrees to the motion, and tension in the strings will pull back on B. So in order to maintain constant acceleration on B, the force needs to increase. The tension in the two strings is the same, by symmetry. As the force on B (and thus, the system) changes, the center of ...


2

1st question: Yes all systems have areas of periodic (or almost-periodic, or isolated periodic) motion. For integrable systems this is the expected. But in fact the existence of isolated periodic solutions is a form of non-integrability a well (related to 2nd question as well). For the exact meaning of isolated periodic solution check the link on ...


3

Let us start from Minkowski spacetime $M$ and construct the trivial bundle $\Phi=\mathbb R \times M \to M$ whose sections $\phi : M \ni p \mapsto (p,\phi(p))$ are the scalar fields you want to discuss their dynamics. Since you correctly wish to see the partial derivatives of $\phi$ as variables independent from $\phi$ itself (this is your second raised ...


0

Pressure is, by dimensional analysis, just a form of energy density or momentum flux density. In the most general terms, it is defined as a second rank tensor given by: $$ \mathbb{P}_{s} = m_{s} \int d^{3}v \ \left( \textbf{v} - \textbf{U}_{s} \right) \left( \textbf{v} - \textbf{U}_{s} \right) \ f_{s}\left( \textbf{x}, \textbf{v}, t \right) $$ where ...


0

Your conclusion is wrong because the expression you use for the kinetic energy of the system is wrong. The easiest way to calculate the total kinetic energy (KE) is to add the KE of the center of mass plus the KE relative to the center of mass. The rotational KE can be expressed as $E=\frac{1}{2}I\omega^2$ where $I$ is the moment of inertia. For a disk ...


3

The $L\cdot\Omega$ term comes directly from the change of frame of reference, especially from the transformation from the static frame of reference to the rotating frame of reference. Let $\mathcal{R}\equiv(x,y,z)$ the initial static frame, and $\widetilde{\mathcal{R}}\equiv(x',y',z')$ the rotating frame at a constant velocity ...


-1

Yes this setup violates the second law of thermodynamics .The law states that - "It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work." Now you know the answer


0

Check paragraphs 5-7 for a possible answer...to the compression option. http://www.popularmechanics.com/technology/aviation/airships/airship-of-dreams-lighter-than-air-travel-is-back-16292687


2

As you turn the gears, the point where the outer gear touches the inner gear will travel around the circumference of the inner gear, making one full rotation. But the centre of the outer gear is twice as far (because the gears are of equal size), so it will travel twice as far too. (That is, twice the circumference of the gears.) And because there's no ...


0

I was also involved in this problem for the past few weeks.You can write newtons law of motion for small segment of string and obtain a differential equation.From that equation you can find the normal modes of the string and the general motion of the string is given by a superposition of the normal modes.But i ignored the longitudinal oscillations and ...


3

Speaking as an (ex) experimental scientist, graphing the raw data straight from your experiment is an excellent idea because it gives you a basic view of how large the experimental errors are, i.e. how scattered the data is, and sometimes gives you a clue to the function behind the data. In this case you can see at a glance that the errors in your ...


0

I would attack this problem this way: Locate the center of gravity of the upper hemisphere; specifically determine how far up the central radius the CofG lies Assume the upper sphere rotates ("wobbles") through an angle $\theta$; find the new point of contact between the upper and lower hemispheres in terms of $\theta$ From these new points of contacts, ...


0

It will turn out that there is no gravitational field inside a hollow sphere. I'm not going to do the integral to prove this for you, but here are some tips. $dV = r^2 sin \theta dr d\theta d\phi$, not with an $R$. You are integrating over the entire volume of the sphere. You are approaching the integral in the wrong way by converting to (x,y,z) space, ...


0

There are two forces acting on the mass (omitting any frictional forces). The first one is the centrifugal force. The second one is the gravitational force. At the moment the string slackens, the sum of the forces in the direction of the string will be zero. In a formula: m*v^2/a - m * g / cos(pi/3) = 0 From here you can work out the relation between v^2 ...


0

The key is the statement that the string slackens. This means that the entirety of the centripetal force needed to make the bend of radius $a$ comes from the component of the weight which is along $a$. From kinematics, you know that the centripetal acceleration at the top point is $$a_c=\frac{v^2}{a},$$ and you know that the component of gravity which is ...


0

I think this problem is trying to get you to use generalized coordinates. If you make your coordinates the center of mass and the seperation, you will get decoupled equations for the free particle and the harmonic oscillator respectively. Being agnostic about the coordinates is the secret superpower of the Lagrangean and Hamiltonian approaches.


0

Firstly we divide the work up into internal and external components: \begin{equation} \sum _i\int ^{r_2}_{r_1}\vec{F_i}^{(e)}\cdot d\vec s _i+\sum _{i,j}\int ^{r_2}_{r_1}\vec {F_{ij}}\cdot d\vec s _i \end{equation} The factor of a half comes in since we are summing over both $i$ and $j$, and (I think) we can assume that $V_{ij}=V_{ji}$ (why?). They ...


5

There's a sign error in your equations of motion. The Lagrangian of the system will be $$L=T-U= \frac{m}{2} \left( \dot{x_1}^2 + \dot{x_2}^2 \right)-\frac{k}{2} \left( L + x_1 - x_2 \right)^2$$ So the equation of motion for $x_1$ is: $$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}_1}-\frac{\partial L}{\partial x_1}=0 \\ m\ddot{x}_1+k( L + x_1 - x_2 )=0 \\ ...


4

I would say pressure is better defined by $$ \vec{F} = P \vec{A}. $$ Yes, we are defining a quantity without having it all alone on the left-hand side. And yes, area is a vector. And as you guessed trying to divide one vector by another leads to trouble, so we won't do it. Let me explain where this comes from and what it is shorthand for. In continuum ...


0

There are different mathematical way to define pressure (all equivalent), but perhaps the most common one is using the component of the force normal to the surface. That is why in the definition you are using you are actually dividing scalars. For this way to define pressure, you see http://en.wikipedia.org/wiki/Pressure. You can also consider area as a ...


1

Yes, area is a vector, which is the normal to the surface. ($\vec{A}=A\vec{n}$) $\vec{F} = -P\vec{A}$ In this case P is simply the proportionality constant to the vectors F and A, which also mean that F and A has to be in the same direction (F is the normal force and not shear forces). The negative sign accounts for the fact that the force and normal ...


1

REMARK. Perhaps I wrongly interpreted the question. I interpreted it as if were referred to the total volume of phase space. The answer is negative if the question regards general changes in time of topology of the total space of phases and if you do not impose any generic restriction on the topology of the spaces, like compactness (see the final ...


5

Here's something I believe is a simple proof. Unfortunately it uses a little bit of cohomology. Consider the canonical 2-form in extended phase space $T^*M \times \mathbb{R}$ $$\omega = \sum_{i=1}^N dq_i \wedge dp_i - dH(\vec{q},\vec{p},t) \wedge dt ,$$ where $N = dim(M)$. A function $f: M \to M$ is said to be a canonical transformation iff $f^* \omega = ...


2

By the main theorem of connectedness in general topology, continuous maps preserve connectedness. Time evolution of Hamiltonian systems preserves connectedness because it is continuous. I think it is independent of from Liouville's theorem, it just requires the proving Hamiltonian time evolution is continuous. This is just a formal way of restating ...


0

The problem is that your analysis is all done from the perspective of the frame that measures the box to be moving at 0.1c -- in this frame, it's true that the time for the light to get from the source to the wall is different from the time for the light to get from the wall back to the source. But if these same events are measured by someone inside the box ...


0

From physics we know that the net force on a mass moving in a circular path at constant speed always points towards the center and has magnitude $$|F_{net}| = m*a_{c}=\frac{mv^2}{r}$$ For an object moving in a vertical circle, when the object reaches the side the net force must be pointing towards the center (west). This implies that the force of weight is ...


1

The reasoning is the same as the two-particle system: $$ E_i = E_{1i} + \frac{p_0^2}{2m_1} + E_i' + \frac{p_0^2}{2(M-m_1)}, $$ so that $$ E_i - E_{1i} - E_i' = \frac{p_0^2}{2}\left(\frac{1}{m_1} + \frac{1}{M-m_1}\right) = \frac{p_0^2}{2}\frac{M}{m_1(M-m_1)}. $$ Therefore $$ \frac{p_0^2}{2m_1} = \frac{M-m_1}{M}(E_i - E_{1i} - E_i'). $$


0

I'm not sure what you're referring to by 'straight' acceleration. If that angle is the angle at which a force, $F$, is being applied, then the horizontal force is: $$ F_x=F\cdot cos(\theta)\\ F_y=F\cdot sin(\theta) $$ These will get the forces along the X and Y axis respectively. Assuming width, $w$, length, $l$, height, $h$, mass, $m$ and $\text{'some ...


0

Here is one take on how to understand the relation between force and potential energy, which I think is the closest modern version of how it would have been seen originally. Let's take as the conditions for a force to be conservative $$\nabla \times \mathbf{F} = 0$$ and $\mathbf{F}$ is a function of position only (this leaves out the magnetic force on a ...


1

We can start at the relationship: $W=-\Delta U$, which is work done by a conservative force. The math A (conservative) force $F$ will do this work on an object when doing a displacement $\Delta x$, and $W=F \Delta x$. In the general case, the force might be different at different points as the object is moved (the force of gravity is not constant along ...


3

As the first question has received sufficient exposition, I would like to make a point with regard to the second one. First thing to understand is that integrability and non-linearity of a system are two different concepts. It is true though that all linear systems in classical mechanics (i.e those that are described by systems of linear equations, be them ...


7

I'll write here a list of my personal favorites plus some commonly used books. I wouldn't be surprised if your teacher chose either one of the books below as a textbook: i) Mechanics, the first volume of the Landau course on Theoretical Physics; ii) Goldstein's book "Classical Mechanics"; iii) Taylor's book "Classical Mechanics"; iv) Marion's book ...


3

The essential idea of a Poincaré map is to boil down the way you represent a dynamical system. For this, the system has to have certain properties, namely to return to some region in it’s state space from time to time. This is fulfilled if the dynamics is periodic, but it also works with chaotic dynamics. To give a simple example, instead of analysing the ...


4

(1) In general, what is meant by non-linear system in classical mechanics? A linear system is described by a set of differential equations that are a linear combination of the dependent variable and its derivatives. Some examples of linear systems in classical mechanics: A damped harmonic oscillator, $$m \frac{d^2 x(t)}{dt^2} + c \frac{d x(t)}{dt} + k ...


3

As requested by the OP, I gather my points in an answer. Linear systems Linear systems are systems which are linear with respect to a physical quantity. Mathematically, their evolution can be written as a (possibly differential) equation. Examples: A linear spring is linear in the sense that is produces a force proportional to the displacement it ...


4

A linear system is one whose dynamics obeys linear differential equations, in contrast with those that are non-linear whose dynamics obeys non-linear differential equations. So if the dyanmics of the variable $x(t)$ obeys a a differential equation $$f\left(x(t),\frac{d}{dt}x(t),\dots,\frac{d^n}{dt^n}x(t),t\right)=0,$$ if $x_1(t)$ and $x_2(t)$ are differente ...


0

This is a collection of resources related to complexity. Many physical systems can be represented with graphs: Quantum graphs Mechanical systems Complexity of a graph or a weighted graph is a notion established by algebraic geometry. http://arxiv.org/abs/0705.2284 (The weighted complexity and the determinant functions of graphs) Audrey Terras. Zeta ...


0

I understand that you're not talking about entropy (which is not the same as complexity or disorder), but $\text{complexity}$ is a rather subjective term. I would assume that the relative complexity of a system (which is broad in and of itself, by the way) would have to do with statistical analysis of its parts, the energy required to achieve a certain ...


0

Direct integration give you the time it takes to go between $v_1$ and $v_2$. $$ t = \int \limits_{v_1}^{v_2} \frac{1}{ \dot{v}(t)} \, {\rm d}v $$ So if $\dot{v}(t) = a_0 - \beta v^2$ then $$ t = \frac{1}{2 \sqrt{a_0 \beta}} \ln \left( \frac{a_0+\sqrt{a_0 \beta}(v_2-v_1)-\beta v_1 v_2 }{a_0-\sqrt{a_0 \beta}(v_2-v_1)-\beta v_1 v_2} \right) $$


2

I do not know if it is "plausible" (I do not think so), however a trivial model can be constructed for the one-dimensional case with continuous forces depending on velocities, for $c>0$ constant: $$F_{12}(v_1,v_2) = c\sqrt{|v_1-v_2|} \quad\mbox{and}\quad F_{21}(v_1,v_2) = -c\sqrt{|v_1-v_2|}$$ The system of these two particles does not admit a unique ...


0

The concept of an isolated system is an approximation. Like all approximations, it applies to some systems better than others. A billiard ball feels only a weak frictional force in the form of rolling resistance, whereas a wall typically has foundations buried in the ground which can provide a strong resistive force. Thus, the former system is closer to the ...


0

It all depends on what you want to study. The billiard balls are generally viewed as an isolated system for the purposes of explaining elastic collisions, but you could as well introduce friction with the pool table, and the consider the system balls+table as the isolated one. This just means you have to consider the friction. In the case of the car hitting ...



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