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0

You can definitely have more generalised coordinates than the degrees of freedom. Consider a particle in space which is constrained to move on a straight horizontal rod. You can describe such a system by choosing the $x,y,z$ coordinate of the particle in space as generalized coordinates. But the constraint is reducing the degrees of freedom from 3 to 1, so ...


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The minimum number of independent generalized coordinates is given by the number of degrees of freedom. Nothing prevents you from using more coordinates (dependent) if you so desire.


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My guess is that the exhaust is not perfectly in line with the balloon's centre of mass. In other words, the exhaust is slightly off-centre. That will cause the thrust of the escaping gas to push the nozzle slightly to the side, causing the balloon to rotate sightly as it travels forward. If this deflection stays constant you will get a perfect circle. If ...


1

Well I was still confused after reading these answers so I found a book "Cosmology and astrophysics through problems" by T Padmanabhan that finally answered the question in a (fairly) clear way. Basically for a system with n degrees of freedom, you have $n$ Euler-Lagrange (E-L) equations. Since these equations are second order partial differential ...


3

Even classically, forces arise from field being propagated at the speed of light. A physically relevant object is the energy-stress tensor, whose components represent energy density and momentum current density, so indeed momentum can be interpreted as a current that is conserved over time (as a consequence of symmetries). This point of view is also ...


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In a conservative field (no dissipation of energy) whatever way follows the small sphere, i.e. rolling along another sphere, or other, the total energy (kinetic + potential + rotational) is conserved.


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Well, this one's a bit less tedious and easier to understand. It all depends on how a blender actually works. The most common blenders have a couple of vertical knives as blades included. When these knives rotate, a tornado of air with a vortex is created . The vertical blades push the veggies upwards creating the vortex, the horizontal ones then slice off ...


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For a rigid body, the velocity can be decomposed into translational and rotational components. The translational velocity is the same at every point, while the rotational velocity is different, as it depends both on an angle and the distance from the axis of rotation. The contradiction can be resolved by looking at both of these, not just translational.


1

By definition, work is $$W[\gamma] = \int_\gamma\mathbf F\cdot\mathbf v\text dt$$ i.e. force times displacement. If an object is not moving, even if subjected to a force, then there is no work done by said force. Observe that, since the object is not moving, the sum of all the forces applied to a body must be zero. For example, a body on rest on a table is ...


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Work= force x displacement force is weight and displacement is 0 so there is no work (in physics)


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Nothing is incompressible, but most liquids and solids have a very low compressibility i.e. a very high bulk modulus. The reason for this is that in liquids and solids the atoms/molecules are in contact with each other. To squeeze them closer together you need to deform the bonds in molecules and/or the electron distribution around atoms. Both processes ...


-1

Here is my answer. I think it's self explanatory enough: In the last sentence there must be a minor correction: If someone keeps omega1 constant, then for the same speed they have to cycle faster so they can maintain the same power. If someone keeps omega2 constant, then bigger alpha mean less forward speed but less power too.


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The power input is roughly constant (that of a car is dictated by the total engine power while for a bicycle it depends on the user). The gear or similar tools adjusts the mechanical advantage so that a low gear will express the engine power in force rather than speed (recall that power is force times speed). On higher gears the force is traded in for speed. ...


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From http://imechanica.org/node/10589 - makes sense to me. by Zhigang Suo on Tue, 2011-07-19 07:02. A rubber is a network of polymer chains. Each polymer chain consists of many monomers. The polymer chains are crosslinked by covalent bonds. The covalent bonds give the solid-like behavior of the rubber. If these crosslinks are removed, the rubber ...


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Comments to the question (v3): I) The Gurtin-Tonti bi-local method [which OP mentions in an example; see also Section II below] of pairing opposite times $t\leftrightarrow (t_f-t_i)-t$ (hidden inside a convolution) is an artificial trick from a fundamental physics point of view, unless further justified. Why would such correlations into the past/future take ...


0

The magnitude of the force between $i$ and $j$ is indeed $$ F_{ij} = \frac{Gm_im_j}{\lvert\vec{r}_{ij}\rvert^2}. $$ The direction of this vector is directed along the line connecting the two points, the same as $\vec{r}_{ij}$ (my notation for the vector difference between the positions of $i$ and $j$). In principle you can compute the magnitude for each pair ...


1

Think of the function generator producing prescribed displacements $X(t)$. What is the extension of the first spring? It is $x$, so its restoring force is $-k_1 x$. Now what is the extension of the second spring? It is $X(t)-x$, with its restoring force $-k_2 (X(t)-x)$. What is missing from the diagram is the direction of positive forces and displacements. ...


1

So this is a bit tricky actually and perhaps the original question makes this more clear. The problem I'm having when thinking about this is what exactly does the function generator do. Does it take the original force of the string and add some force $F(t)$ on top of it, or does it essentially fix the force on one side of the spring? So if the spring is ...


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Noether theorem tells you that if you can find a (one parameter) group of infinitesimal transformations $\alpha$ and $\beta$ such that: \begin{equation} t'=t+\alpha\epsilon \end{equation} \begin{equation} q'^\mu=q^\mu+\beta^\mu\epsilon \end{equation} and your lagrangian is invariant under this group of transformations, then the quantity \begin{equation} ...


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The gravitational force acting on the particle as function of radius $r$ is [Gauss's law] $$F(r) = -\frac{GmM(r)}{r^2}$$ where $M(r) = \int_0^r4\pi \rho(r')r'^2dr'$ is the mass contained within a radius $r$. Note that for $r>R$ we have $\rho(r) = 0$ and $M(r) = M(R) \equiv M$ the total mass of the whole sphere. The potential energy is the work needed ...


3

but perhaps, given sufficient throwing speed If you throw your egg too fast, the force due air resistance $$F = \frac{cW\cdot A\cdot \rho\cdot v^2}{2}$$ will exceed the 50 Newton you mentioned before and the egg will break. and, or, car speed, perhaps it is possible If the car speed is fast enough the glass will of course break, since even ...


0

Yes, it would of course. And so the resultant weight (your's + the weight of what you are carrying) starts acting through this new and horizontally shifted (maybe vertically shifted too) Center of Mass. Notice that all this while your weight was being balanced by the normal reaction from the surface on which you were standing. Also, as only 2 forces were ...


0

I guess this confusion of yours is because of a possible misconception of the correct definition of angular momentum . Firstly , angular momentum is always taken with respect to a point . Angular momentum of any given particle , p ,with respect to a point A, is defined as the cross product of the position vector of P as measured from A with the linear ...


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if two connected masses are rotating around a center of spin and then disconnected, the two masses would move away from each other in straight lines, but not in the same straight line. If the 2 masses were joined again later they would spin again, at a different speed, but the same angular momentum.


0

Three parameters are needed for 2D force (as opposed to 6 for 3D, see http://math.stackexchange.com/a/1157906/3301). Composition A force with magnitude $F$ along a direction $\vec{e}=(e_x,e_y)$ going through a point $\vec{r} = (r_x,r_y)$ is described by the three parameters $$ f =(a,b,c)= ( F e_x , F e_y , F (e_y r_x - e_x r_y) ) $$ Decomposition Given ...


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The linear transformation is the following composition of linear maps: Go from $R^n$ to $T_m M$ using the natural identification Go from $T_m M$ to $T^*_m M$ with the symplectic form Go from $T^*_m M$ to $T_m M$ using the inverse isomorphism given by the metric Go again back to $R^n$ (here $M = R^n$ and $m = (q,p)$ By the way, this transformation of ...


0

No, it won't. Part of the reason we use things like massless springs is to avoid thinking about them in detail; they exist to provide a force and to store energy, but are forbidden from doing anything else. Proof by contradiction: Suppose that the block is being accelerated in some direction by spring 1. Now the contradictory supposition: suppose that ...


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Same as with the symplectic form: $\omega(v) = (u_\omega,v)$ defines the isomorphism between 1-forms and vector fields. When the metric is Euclidean the dual basis to an orthonormal basis corresponds to the basis itself.


2

Use a ramp, an incline of 53° will work. Otherwise You need to double up the cord. The third option is to just carry the 10 kg object down the stairs.


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I would not be surprised if a cord with a given breaking strength of 80N held 100N once. It should be regarded as trash after that though. You see, they set the breaking strength as guaranteed to hold under the worst-case setup. Select a low-reduction knot (the rating not is the double figure-eight which is common but not the optimal choice) and apply the ...


4

As far as I see, if you want the object to lower at constant speed the only way is actually fixing at least both the extremities of the cord to the object (can't describe this well, let's say your "grip" is on the middle of the cord). Whatever happens somewhere else, constant speed means that the balance of the forces on the object must be 0, so if you tie ...


17

Breaking strength refers to the maximum tension in the cord. Now, from the sounds of this problem, you've probably been doing force diagrams involving cords. What happens when you attach two cords to a single 100N object (and keep it stationary)? Is the tension in both of those cords 100N? Or is the combined force 100N, so that each just has 50N? Put ...


1

The damping introduces a dissipative element to the system, that is, energy is leaving the spring-mass system with time. As such, the energy is not conserved. The maximum energy for the system occurs at its initial configuration; the spring is stretched to some extent $A$ (initial amplitude). The initial energy for the spring is thus purely of a potential ...


11

You have to accelerate the object towards the ground. (Let it fall a bit.) This creates a bit of "slack" in the cord so that it doesn't break. Figuring out how much acceleration it should have is a good exercise. EDIT: I figure I might as well work it out since this question has so many views. Note that personally I view doubling the rope as cheating. From ...


0

Would one expect this energy to equal the $1/2 kx_0^2$ that we put into the system initially? Yes, it must. Energy conservation applies between any two points in time. If your point one is the initially stretched out spring before it is let go with energy $U_{el}=\frac{1}{2}kx_0^2$ and your point two is the situation of no energy stored in the spring ...


1

For the first pendulum, we obtain harmonic motion with angular frequency $\omega_1=\sqrt{g/l}$. For the second pendulum, the combined acceleration of the cart and gravity produce an effective acceleration with magnitude $g_e=\sqrt{g^2+a^2}$ at an angle $\tan^1(a/g)$. It is as if gravity has been tilted and increased in strength. This is the only difference, ...


1

Of course that's an internal force. let's see what you are confusing about: I am a bit confused because the geometry of the system changes when this force is exerted and this seems weird for an internal force. Well, you cannot tell whether a force is internal or external by whether the geometry of the system change or not. Here's an example: ...


1

No. You can see now why dealing with torques and angular momentum is so difficult: It depends on what point in space you are taking it with respect to. The torque about one point may be different in both magnitude and direction of another arbitrary point. If you are dealing with a system in static equilibrium however, it's a requirement that the torque ...


0

No, the moment will depend on the point. However, you do have some independence: consider the resultant force as a free vector. You are allowed to slide the point along this vector and still get the same moment.


3

They probably hiding heavy metal chair like this:


1

Since this is a 1D problem the frequency of small oscillations is simply given by the relation $$\omega^2 = \frac1m U''(x_0),$$ where $U(x)$ is the potential energy of the system and $x_0$ the equilibrium position. This is explicitly given by $$U(x) = mgx + k\left(\sqrt{x^2+d^2}-d\right)^2,$$ and from here one finds $$\omega^2 = \frac{2k}m\left(1-\frac dl + ...


0

It does not seem give simple harmonic motion. This can be seen by examining the Lagrangian $$ L = \frac{1}{2} m \dot{x}^2 - 2 \frac{1}{2}k \left(\sqrt{d^2 + x^2} -d \right)^2$$ The factor of two in front of the potential comes from the fact that there are two springs. Now in the limit of $x \ll d$, one has $$ \left(d \sqrt{1+x^2/d^2} -d \right) \approx ...


1

The net force acting on the mass M will be $$F=-2kx\sin\phi$$ Here,x is extension of spring,which can be found out by $$d=(x+d)\cos\phi$$ $$2d\sin^2\frac{\phi}2=x\cos\phi$$ $$2d\cdot (\frac {\phi}2)^2=x$$...Considering very small oscillations. Also, $$F=M\frac{d^2((x+d)\sin\phi)}{dt^2}=M\cdot d\cdot\frac{d^2(\tan\phi)}{dt^2}$$ If,we consider the ...


1

All above answers hold some validity. Additionally, the high speed stream of air is directed initially at the outside of the spinning balls, obviously. However, once the spinning is stable, the steam of air is gradually directed closer and closer to the center of the spinning mass. That is the real reason the very high RPM is achieved. Think of the ...


1

We know/assume that the mass of the particle is constant. Elastic scattering means no change in kinetic energy (in the center-of-mass frame). Since the mass of the particle is constant, this means there is no change in the speed (not velocity; speed) of the particle during an elastic collision. Torque is $\vec{\tau} = \vec{r} \times \vec{F}$. For a ...


0

The answer by Enucatl is satisfying enough. However, in the example $$P=q \cot(p),$$ $$Q=\ln \left (\frac{\sin(p)}{q}\right),$$ given in the question, it seems there is dimensional mismatching. The argument inside $\cot$ must be some $[p/(p_o)]$ where $p_o$ has dimensions of momentum and the argument of the logarithm must be $$q_o \frac{\sin(p/p'_o)}{q},$$ ...


1

I randomly had this typed up in personal notes. Was probably an exercise somewhere. Consider a harmonic oscillator, which is described by the Hamiltonian $$H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2q^2$$ Doing the Legendre transform, we obtain the action as $$\mathcal{S}=\tfrac{1}{2}m\int_0^t(\dot{q}^2-\omega^2q^2)dt'$$ Now we use the Euler-Lagrange equation to ...


0

The action $S$ is defined as the time integral of the Lagrangian $L$ of the system, and in classical mechanics the Lagrangian is just kinetic energy $T$ minus potential energy $V$. So the action can be written as follows: $$S(\mathbf{x},\mathbf{\dot x}) = \int_{t_1}^{t_2} \text{d}t \; [T(\mathbf{\dot x}) - V(\mathbf{x})]$$ For a harmonic oscillator with ...


1

It basically means that they just need to cover half the distance. So, you have the distance to be covered, initial velocity(40) and final velocity has to be zero. Finding deceleration won't be an issue.


1

Time is what an ideal clock measures. So what's an ideal clock? It's something that measures time. In other words, physicists don't quite know what time is. That's okay. They don't quite know what space is, either. What they do know, and know very, very well, is how to measure both, and how the two (time and space) relate to one another. That the speed of ...



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