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## Hot answers tagged physics

5

A first good step is multiplying by $2\dot{\theta}$, to get $$\frac{d}{dt} \dot{\theta}^2=2\sin \theta \dot{\theta}$$ which can be integrated to $$\dot{\theta}^2=-2\cos \theta+A,$$ where $A$ is a constant. Thus $$\frac{d \theta}{ \sqrt{A-2\cos \theta}}= \pm dt$$ and a second integration gives $$F_A(\theta)=\pm(t-t_0)$$ where $F_A$ is a primitive ...

3

There are no troibles with units. Look, we find solution of $$\frac{dv}{g-cv} = dt;$$ ok, integrate it: $$\int_{v_0}^v \frac{dv}{g-cv} = \int_{t_0}^{t} dt\Longrightarrow \ln\frac{g-cv}{g-cv_0}=-c(t-t_0).$$ As you can see, both log and $ct$ are unitless. We can express $v$: $$\ln\frac{g-cv}{g-cv_0}=-c(t-t_0) \Longrightarrow v=\frac gc - \left(\frac ... 3 Okay, let the constant yaw-rate be \omega. As the body has a constant linear forward velocity of v, it must follow a perfectly circular path. Let us consider the path of the body from t=0 to t=\Delta t. We see from the figure that:$$\phi_{G} + R = \frac{\pi}{2}\theta + 2R = \pi\Rightarrow \theta = 2\phi_{G}$$From the sine rule in ... 3 A quick google search pulls up this paper by Litvin which includes permutation and matrix representations. For example, it says I is generated as a permutation group by the following elements of S_{12} (and identifies them to more classical geometric generators, which I omit here for brevity):$$g1=(1,2) (3,6) (4,11) (5,7)(8, 10) (9,12)g2=(1,4,3) ...

2

We have $$E=\frac12mv^2=\frac12mv_0^2+mgx$$ from which we can write $$v=v_0\left(1+\frac{2gx}{v_0^2}\right)^{1/2} \tag 1$$ Assuming that $\frac{2gx}{v_0^2}<<1$, we approximate the square root in $(1)$ as $$\left(1+\frac{2gx}{v_0^2}\right)^{1/2}\approx. 1+\frac{gx}{v_0^2} \tag 2$$ where the approximation error is of order ...

2

The force exerted by a small arc of angular width $d\theta$ is, by the Inverse Square Law, equal to $km\frac{M}{2\pi}\frac{1}{y^2+a^2}\,d\theta$, where $k$ is a constant. Integrate from $0$ to $2\pi$. We get $\frac{kmM}{y^2+a^2}$. For the component in the $y$-direction, multiply by $\frac{y}{\sqrt{y^2+a^2}}$.

2

Since $$\frac{dw_1}{dt}w_1+\frac{dw_2}{dt}w_2=\frac12\frac{d}{dt}(w_1^2+w_2^2),$$ your differential equation is equivalent to: $$\tag{1} \frac{du}{dt}=\alpha u,$$ with $$u=w_1^2+w_2^2,\quad \alpha=-2\frac{c}{I}.$$ Solving (1) we get: $$w_1^2(t)+w_2^2(t)=\beta\exp\left(-2\frac{ct}{I}\right),$$ where $$\beta=w_1^2(0)+w_2^2(0).$$

2

The reason that we can decompose the force $mg$ into components $mg\sin(30^\circ)$ and $mg\cos(30^\circ)$ is not because the concept of force has distance embedded in it (through the distance/time$^2$ in $g$). Rather it is simply because force is a vector quantity. Vectors arise often in physics, and it is important to make sense of them. Before we can ...

2

First have a look at: Integral Argument For enough variations: $$\varphi\in\mathcal{C}^\infty:\quad\int_0^T\varphi(s)\delta\varphi(s)\mathrm{d}s=0\quad(\delta\varphi\in\mathcal{C}^\infty_0)\implies\varphi(t)\equiv0$$ That is needed for Euler-Lagrange! Now go carefully through: Detailed Calculation Directional derivative: ...

2

Yes. Renormalization group equations, used both in particle physics and condensed matter physics, often have fixed points with implications for physics. One notable example is asymptotic freedom in QCD, which is based on the observation that the renormalization group equations for the gauge couplings (and the Yukawa couplings as well) have an ultraviolet ...

1

There are some physical limits which are often hidden in the usual presentations. The "physically reasonable" boundary condition is the heat flux boundary condition: there is a thermal conductivity coefficient $k$ for the boundary, and the system is in thermal contact with a temperature $u_{ext}$ at each point of the boundary. Then the boundary condition ...

1

From $$\frac12 t^2 \vec a=\vec p-\vec x-t\vec v$$ we obtain $$\frac14t^4n^2=(\vec p-\vec x)^2-2t\langle \vec p-\vec x,\vec v\rangle+t^2\vec v^2$$ where all but $t$ is given. If we solve this for $t$ (numerically, I suggest), you can obtain your desired $\vec a$ from this as $\vec a=\frac{2(\vec p-\vec x-t\vec v)}{t^2}$.

1

Let the point at which the child fires the toy cannon from be $P$. Let us consider the situation relative to point $P$. Let $d$ be the distance the shell travels horizontally. Let $T$ be the time for the shell to land after being fired. Vertical motion: $s = ut + \frac{1}{2} at^2$, we get 0 = VT\sin \theta - \frac{g}{2}T^2 \implies T = \dfrac{2V\sin ... 1 Assuming constant acceleration between the time steps, for a time increment of \Delta t, we have for the velocity and displacement at successive times t_n, t_{n+1}: \begin{align} v_{n+1} &= v_n + a \Delta t \tag{1}\\ s_{n+1} &= s_n + \frac{1}{2}(v_{n+1} + v_n) \Delta t \tag{2} \end{align} Then (2) becomes s_{n+1} = s_n + ... 1 Try to formalize what internal stress state should actually mean. What science has come up with is the notion of sectional forces and moments, together with a way to relate them to internal stress state. You need to do a Gedankenexperiment: Imagine a smooth cut passing through the material, including the point p. Remove the material on one side of the cut ... 1 The following should work at least approximately. The granularity of time may cause the planet not to follow a closed orbit. To improve precision you should use a smaller \Delta t and/or learn something about the numerical analysis of handling this type of differential equations more accurately. Initialize variables (run tests to find combos that work for ... 1 There are going to be infinitely many lines that satisfy the condition. For example, as said above, f(x)=\frac{1}{2}+g(x-\frac{1}{2}), where g is an odd function. 1 Is the question is asking for radius of curvature? since that is the reciprocal of curvature. 1 I believe this is a misunderstanding. The text is not dividing a vector \vec A into two components. It is describing how to consistently define a vector potential for the magnetic field of a magnetic monopole despite the fact that no single vector potential can describe this field. It does so by defining two different vector potentials on the two ... 1 I dont know if this is right. My answer is P= -4.0 i + 3.0 j or P= 4.0 i - 3.0 j ? 1 M looks like this: M(u) = (F_{12}(u),\dots,F_{1n}(u),F_{23}(u),\dots F_{2n}(u)),\dots ,F_{nn}(u)) M(u) = (F_{12}(u),F_{21}(u),\dots) = (F_{12}(u),-F_{12}(u),\dots) doesn't make much sense. then the first variation: \delta M(u) = \frac{d}{d\epsilon} M(F_{12}(u+\epsilon v),\dots,F_{nn}(u+\epsilon v))\big|_{\epsilon = 0} using the chain rule: ... 1 This answer is quite late, so I'll make it general for those wondering about how to jump into QM with an undergraduate or higher background in math. A word of caution to mathematicians entering the physics realm: though there is a great overlap in material, the emphasis, pedagogy, and approach of a physicist can be quite different than that of a ... 1 Notice, due to drop through a height x, with initial velocity v_0, the velocity v is increased constantly under earth's gravitational acceleration g. Now, using third equation of the motionv^2=v_o^2+2gx\color{blue}{v=\sqrt{v_0^2+2gx}}$$Again notice,$$v=\sqrt{v_o^2+2gx}=\left(v_0^2+2gx\right)^{1/2}$$... 1 I think it depends on how you organize your problem. Suppose your problem takes place in a time interval a \le t \le b. Some things you know are f'(t) and f(b). Then to find f(t), you may think of integrating backward from b. And maybe some ways to write that would involve thinking of negative dt. 1 You can maybe explain it by a function F : R^3 \rightarrow R. For each point (x,y,z) from the 3-dimenstion space (and which is inside the room), that function would return a number which describes/measures the brightness. 1 This is not really a question about mathematics, it is a problem about physics. The answer is so close to 0 that you can probably say it is 0 in most practical applications. The answer is a function of so many variables that it can only ever be calculated by some pretty advanced modelling. There is no way to calculate it, one of the reasons being that ... 1 You have some confusion about units. You wind up with the generalized eigenvalue problem$$(K-\omega^2 M)X=0$$where \omega^2 is the generalized eigenvalue. This is NOT the problem K=\omega^2M; this would assume that \omega^2 is a generalized eigenvalue of full multiplicity. Also, for this to make sense, K and \omega^2 M must have the same units. ... 1 How would you justify the equation: K = \lambda M? I would make the ansatz: \vec X(t) = \vec A e^{-i\omega t} then you get the algebraic equation: (K - \omega^2M)\vec A = 0. In order for this to have a non trivial solution for \vec A we require that \det (K - \omega^2M) = 0. Thus:$$\begin{vmatrix}K_1 -\omega^2 M & -K_1 & 0\\-K_1 & ...

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