# Tag Info

## New answers tagged classical-mechanics

2

The $\vec\omega\times \vec r_b$ term appear in the rotating (non-inertial) frame. It is there because the motion of a particle in a rotating frame expressed in the lab contains a "pure" motion on the rotating (non-inertial) frame, plus a term to account for the rotation of the non-inertial frame. In equations, $\vec r'(t)=U(t) \vec r$ where $\vec r'$ is ...

0

The vector sum of the external torques determines the rate of change of the angular momentum vector. For an object on an axle, the vector representing the torque (relative to the axle) is defined as being along the axle. Starting from zero, the vector representing the increase in the angular momentum would be in that same direction along the axle. For an ...

0

The bead is constrained to move along a line, so, as you thought, there is only 1 degree of freedom. The problem with your polar coordinates choice is that you did not choose the right coordinates system. The coordinate is "position along the rod", not "distance from center". One way of thinking about this is to presume you want to use Cartesian coordinates ...

1

1) The Lagrangian is simply a function of generalised coordinates and velocities(see also @sanaris above), which when put into an action and extremised gives you the time evolution of the coordinates(and thus, the velocities via differentiation). You do not know the trajectory, and thus the velocity, and thus the kinetic energy, before you have actually ...

0

Lagrangian is defined as the kinetic energy minus the potential energy You are wrong. Lagrangian has nothing to do with kinetic and potential energy. For example Einstein-Hilbert action is $S=\frac{1}{2k}\int R\sqrt{-g}\,d^4x$. There is no fast-hand way to see in there something like "potential" and "kinetic" energy. There is only one true definition of ...

0

In Lagrangian mechanics, we take the kinetic & potential terms as axiomatic, i.e. we don't use Newton's second law to justify $K=\frac12m\dot{x}^2$, we just claim $K=\frac12m\dot{x}^2$. Newtonian, Lagrangian and Hamiltonian mechanics (and a few other options) are equivalent, but they assume different things.

0

You are right. The Newtonian and Lagrangian formulations are equivalent. Either can be used as a starting point for mechanics. Which you think more fundamental is largely a matter of personal choice. My own preference is to formulate Newton's laws from conservation of momentum (the third law contains the physical content, the second just a definition of ...

2

Suppose a rocket always flying in a vertical position at the equator region, and its engine can't move it sideways. In the first kilometers the air density is enough to keep it in the same frame of the base, or rotating with the earth. But after 100 km, that is conventionally taken as the thickness of the atmosphere, there is no more air drag. So its ...

3

It isn't a question of distance from the Earth. It is just a question of relative velocity. For example if you were next to a geostationary satellite in a geostationary orbit then you would not observe the rotation of the Earth; the Earth would still appear to be stationary to you. On the flip side, if you were to (somehow) hover $1\,\rm m$ above the ...

0

An answer from an undergraduate. A potential is defined for a conservative force. For a conservative force, a line integral done on a closed loop has to be zero.Now in the case of dissipative forces like friction the closed line integral will be non zero. So dissipative forces are non conservative hence they admit no potentials.

0

From a purely classical point of view: "Does this [classical mass invariance] stem from other basic principles of classical mechanics or is this an independent experimental fact?" No, this fact doesn't stem from other classical principles. Classically, this is an independent experimental observation. It boils down to Newton's 2nd Law for a ...

3

The idea from early treatments of special relativity that mass increases with velocity was superseded in general relativity and is better not used. It is a fundamental principle that the laws of physics are covariant - they are formulated using tensor (& vector & scalar invariant) quantities so as to be the same for all observers. Proper mass, or ...

-2

This is a known misconception . There are 2 kinds of masses 1) inertial mass : Inertial mass is found in Newton's law of motion F=ma. This m expresses the resistance of an object to change its velocity. 2)gravitational mass: Gravitational mass is found in Newton's law of gravity. g = G*m/r^2.This m expresses the result of an object to gravity. At low ...

0

Also, an alternative to the accepted answer. By the work-energy theorem, $W_{int} + W_{ext} = \Delta T$ And by the general statement of conservation of energy, in the absence of any heat transfers, if $U$ is the total potential energy of all of the pairs of particles in the system, $W_{ext} = \Delta T + \Delta U$ Subtracting the equations yields, $W_{... 3 This is an interesting question. You are right in how you think about potential energy. There are more bodies involved than just the one we are trying to get the trajectorie from. Saying a particle moves in a static external potential is always an approximation. I want to motivate this: In classical mechanics you often deal with closed systems: You have a ... 2 The fundamental forces are conservative, so any non-conservative forces are due to neglected degrees of freedom or degrees of freedom we've averaged over. The energy stored in those hidden degrees of freedom is the internal energy, kinda by definition. So internal non-conservative forces must conserve the total energy, internal + mechanical. This is the "... 1 An infinitesimal off-shell transformation that preserves the action up to boundary terms is a so-called quasisymmetry. The main reason to consider quasisymmetries is that Noether's first theorem also works for them: They also produce on-shell conservation laws. In fact they are the natural class to study in order to have a bijective correspondence with ... -1$a_g$is the acceleration at a given time, so it makes no sense to differentiate against time, since you want the maximum depending on theta you differentiate just for theta. No chain rule involved, at a different time you will have a different$v$, so the max will be at a different point. -2 The string has to hold against the buoyant force wich is the weight of the water, the volume of the ball takes, the end of the string tears with this fore on the bottom or the balance, when the ball is released and goes to the top it is only much less immersed if you hold some He balloons and stand on balance your weight is less than without, this is the ... 1 The axiomatic system you are looking for needs the mathematical axioms and theorems used for establishing differential equations and calculus, to be complete, Newton's law's are not enough. Euclidean geometry is simple, as it needs only algebraic relations as a mathematical framework. Classical mechanics needs a more sophisticated mathematical framework, ... 2 The point of the first paper is that$F = ma$doesn't actually indicate how matter moves, at least not without specifying what$F$is, which is correct. You can't really get much of use out of Newtonian mechanics, outside of general theorems, if you don't specify the forces. I don't think this is quite the takedown it is, because an axiom system can get by ... 0 For each object for energy conservation you will have$\Delta\text{KE}=-\Delta\text{PE}$. Since both of these terms are directly proportional to the mass, the mass variables on each side of the equation will cancel out. Therefore, you do not need to know the mass of each object to solve this problem. Also, in this case speed means the translational speed of ... 1 I think Introduction to Electrodynamics by Grittiths would be perfect for you! It has a chapter that is solely dedicated to multivariable calculus, which the rest of the book then uses for the majority of the problems. It also has a more "chill" tone which makes it easier to read, if you're just starting to learn physics. 0 An object travelling in a circle of radius$r$with angular velocity$\omega$(expressed in radians per second) is under a constant acceleration$a$towards the centre equal to$\omega^2 r$(the acceleration can also be expressed as$v^2/r$where$v$is the speed of the rim). According to Newton's$F=ma$, this means there must be an inward force, which we ... 2 If a particle is constrained in circular motion by a circular wall then there is a normal force accelerating it inwards. This is called a centripetal force. The equal and opposite force is the force which it exerts on the wall. 0 The Lagrangian is motivated by the following observation. The Lagrangian of a QM particle moving in n-dimension with a coordinate$ \vec x(t)=(x_1(t),x_2(t),\dots,x_n(t))$is given by $$\frac{1}{2}m \dot {\vec x}^2\;,$$ Which is just the kinetic energy of the particle. I will set$m=1$below. Now if we want to have a Lagrangian for a particle moving on ... 3 The mechanical energy needed in both cases to lift the body will stay the same. You say "Arguments for reducing: When you raised your legs, your center of gravity moves a shorter distance, thus you use less energy to hoist your body up". This is not true. Once you have lifted your legs, your center of gravity will be higher to start with, but the movements ... 1 Your Lagrangian doesn't accurately describe the energy of the system. You have a term in your kinetic energy $$\frac{\mu(t)}{2}(\dot{\xi(t)} - u)^2$$ which is intended (I assume) to describe the kinetic energy$T_\text{fuel}(t)$of all of the fuel that has been thrown out of the rocket up to time$t$. But the fuel that's thrown out at a time$t_e < t$... 0 I just realized of a way to answer this question which has a different flavor from the previous answers. I hope this answer complements them. Consider a configuration space$Q$, with coordinates$q$, and its corresponding velocity space$TQ$, with coordinates$(q,\dot{q})$. Given a Lagrangian$L:TQ\rightarrow\mathbb{R}$, we have its fibre-wise derivative$...

1

1) If we can visualize angular momentum in 3D as a vector, how should we think about its counterpart in 4D, a tensor? And what about the law of conservation of angular momentum. In general, angular momentum is a geometric object called a rank $2$ differential form. To oversimplify a bit, you can think of it as a plane, along with a signed magnitude. In ...

2

Let's say you start with two identical disks, 1 and 2. You start by slightly tilting disk 1 and grinding it with its center along the edge of disk 2. In this way, the edges of disk 2 as well as the center of disk 1 are ground down. In other words, disk 1 starts becoming slightly concave, while disk 2 starts becoming slightly convex. What happens when we ...

1

In many-particle problems one has to distinguish the constants characterizing the movement of the system as a whole (it energy, the three components of the center-of-mass momentum, and the three components of the angular momentum), and the relative movement of the parts in the system. So, indeed, the relative movement of the particles will eventually cease ...

0

The solution for this kind of differential equation $${\frac {d^{2}}{d{t}^{2}}}q(t) +{\omega}^{2} q(t) =0$$ is: $$q(t)=A\,\sin(\omega\,t+\varphi)$$ where: $\omega$ eigen angular velocity $\varphi$ phase $A$ Amplitude the frequency of the sine wave is $f=\frac{\omega}{2\pi}\quad [\text{hz}]$ and the period of the wave is $T=\frac{2\pi}{\omega}\quad [\... 5 Consider a coordinate system with the$x$-axis parallel to the initial position of the rod and let$y(x)$describe the shape of the spinning rod. The potential energy of a small piece of the spinning rod is the sum of the elastic energy,$\frac{\kappa y''^2}{2} dx$(here we assume that$y' \approx 0$) and the energy due to centripetal force$\frac{-\rho \...

2

...you need $\omega = \sqrt{\frac{k}{m}}$ to even show (which the book doesn't) that $x(t) = A\sin(\omega t)$ is a solution. Actually, you don't. You can go through the whole process of deriving the differential equation that governs a simple (or physical) pendulum, approximating it, and solving it without ever using the symbol $\omega$ or talking about a ...

3

It is indeed an oddity, but it is quite consistent. In the first reference frame work has been done moving the box. In the second, the box ends up where it started, no net displacement, no net work has been done, and no change to kinetic energy (although there has been a change to direction of motion). As anna v said in her comment '"conservation laws hold ...

4

The key to understanding how this works is to include the conservation of momentum and the mass of the earth as well. Let $m=10 \ \text{kg}$ and let $\Delta v = 10 \ \text{m/s}$. Then the change in KE of the mass $m$ is $$\Delta KE_m = \frac{1}{2}m (v_0+\Delta v)^2-\frac{1}{2}m v_0^2 = \frac{1}{2}m(\Delta v^2+ 2 v_0 \Delta v)$$ and the change in momentum is $... 0 From the point of the guy who fuels the truck the reference frame that matters is the one where the road is stationary. because it is the road that the wheels work against to propel the truck. A scooter riding thief traveling at a steady 5m/s relative to the trucks starting speed waits for the truck to match her speed before stealing the box that way she ... 1 My guess is that at high rotation speeds any small defect of excentricity of your refill will generate a rotating force due to rotation unbalance. This force may in turn excite a fundamental flexural mode of your refill with a clamped and a free edge condition. The mode shape (in particular, the position of the nodes) should then depend on the rotation speed.... -1 First of all, you need to know that any external force can change the mechanical energy of a system. The proof of : $$\Delta E_m=W=-Q$$ Maybe you can use the first law of thermodynamics, it can be helpful !! It's given by the following relation : $$\Delta U=W+Q$$ Note that :$\Delta U= U_{final}-U_{initial} $In this case$\Delta U=0$, "Cyclical ... 2 Short Answer: It is zero. Long Answer: Assume you are on a frame of reference of road (i.e you are standing on road) then the system is truck + block. You observe both of them moving now looking closely friction is holding block on truck by directing itself in forward direction, using Newton's third law equal and opposite forces are acting on the truck. ... 1 Per the work energy theorem the net work done on an object equals its change in kinetic energy. Since the block begins and ends at rest its change in kinetic energy is zero. Therefore the net work done on the block by the static friction force is zero When the truck accelerated the work done by friction was positive giving it kinetic energy since the force ... 1 You can use Newtown's equations of motion with constant acceleration (Here, the acceleration is only due to gravity neglecting the air resistance). Try to solve for the the displacement (height) for each ball separately as a function in time. Once you get both displacements as functions in time, you can equate them and solve for the time and consequently, ... 2 As mentioned in the answer by sammy gerbil, there are two conservation laws (momentum and angular momentum) that are true, whatever the coordinate system, under specific conditions (no external force or torque). However, you are on to something with your intuitive view. If you study more advanced mechanics, you will learn about formulations of mechanics ... 0 There are only 2 laws of conservation here : angular momentum and linear momentum. Both are vectors. Vectors are expressed differently in Cartesian and cylindrical co-ordinate systems. In a Cartesian system the components of linear momentum are constant in the absence of external forces; in a cylindrical system the components can change (unless the object ... 1 If you change the generalized coordinate the eigenvalues of your new equation of motion must be the same otherwise the dynamic of your new system is changed. to obtain the eigenvalues you must linearize your equation of motions. Example: $$\underbrace{\begin{bmatrix} \ddot{x} \\ \ddot{y} \\ \end{bmatrix}}_{\vec{\ddot{q}}}+\underbrace{\left[ \begin {... 0 The first constraint tells you that distances between particles constituting the rigid body are fixed - which is definition of rigid body. In this case, the distance between particle i and particle j is \Delta r=\left|r_j-r_i\right| and this must be a constant to be called c_{ij}. But because we don't want to deal with absolute values, we write it as:... 2 From what you suggest, the book does not claim that every quadratic Hamiltonian has an exact solution. Moreover, it’s not clear what is mean by “exact”. If you consider the Hamiltonian with arbitrary quadratics$$ \hat H=\sum_{ij} c_{ij} \hat L_i\hat L_j$$you could find an arbitrarily accurate numerical solution by diagonalizing within each$(2\ell+1)\...

2

Forces often come from force fields like gravitational or electromagnetic. These differ in different parts of space, so they may cancel in one point but not in another. Now, suppose a particle is at a point where all forces cancel (and there's a non-empty set of forces being in superposition). Let's now perturb its position. In general, the forces now won't ...

0

The following answer assumes we are dealing with an inertial frame of reference. It really depends on what is meant by a "free" particle. The dictionary definition of "free" as an adjective is "not under the control or in the power of another". If a particle is not subjected to a net force then what we do know is that the particle either remains at rest or ...

Top 50 recent answers are included