# Tag Info

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This solution uses multiple integrals but you don't need to compute them. The final computation is a single integral. The ball has spherical symmetry so its momentum of inertia is the same with respect to the $x$-axis, the $y$-axis and the $z$-axis $I=I_x=I_y=I_z$ with $$I_z=\iiint \rho(r)\left(x^2+y^2\right)\mathrm dx\mathrm dy\mathrm dz$$ now sum up the ...

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As you can see, $\rho$ only depends on a single variable: $r$. Thus, it should be intuitive that one can do this problem by integrating only over the variable $r$. To see what you are supposed to do, consider what happens if you fix $r$: You obtain a spherical shell (as was pointed out in the comments). The moment of inertia of a spherical shell is quite ...

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Whether the upward direction will be taken as positive & the downward direction as negative or vice versa simply depends upon you. It does not bother any physics. Generally, the downward direction is taken as negative. The negative sign indicates the opposite direction only of what you have assigned the positive direction. In Principles of Physics by ...

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It depends on what direction you assign to be positive in your coordinate system. To avoid confusion, just remember which direction acceleration is acting and which direction you assigned to be positive.

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Well, things don't fall up, so gravity would be negative. The longer version: For classical mechanics, the direction of a coordinate system is often arbitrary. For free fall problems I generally choose down to be positive (especially when I'm not dealing with an initial velocity, but it's a matter of preference). What you do need to know, no matter what ...

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If up is positive, and gravity points down, then $a$ (acceleration due to gravity) would be downwards, so it will have a negative magnitude.

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The point at rest in $k$ moves as $(x(t),y(t),z(t)) = (x_0+vt,y,z)$ in the frame in which $k$ itself moves with $v$ in positive $x$-direction. Thus, we have to set $x'(t)=x(t)-vt = x_0 +vt-vt$ to obtain a stationary $x'(t) = x_0$ coordinate for a tupel $(x',y,z)$.

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It's just a drawing convention. Rather than "vertical", time is orthogonal to the x-axis. The reason it is not shown vertical is because the paper surface is 2D, and the author uses the vertical axis for drawing the altitude with respect to the ground. Just recall the way you draw the 3D axis. Here, the author uses X (horizontal),Y (vertical) and time ...

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In the context of general relativity it is often stated that one of the main purposes of tensors is that of making equations frame-independent. Question: why is this true? Actually this isn't quite true. General relativity doesn't have frames of reference (except locally, which is trivially true because GR is the same as SR locally). A better way of ...

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TL;DR version The ICRF +x axis is more or less the direction from the Earth to the Sun at the vernal equinox, or about March 20. Six months later, the direction from the Sun to the Earth is more or less along the +x axis. Keep reading! A bit of a taint of the ancient concept of a geocentric universe remains in modern astronomy. How astronomers specify ...

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Yes. The Hamiltonian is your $H$. $p_r=0$ because $\frac{\partial L}{\partial \dot r}=0$ with $L:=\frac{m}{2}(R^2\dot\theta^2+R^2\sin^2\theta\dot\phi^2)$. The integral is over the 4-dimensional phase space: $(\theta,\phi,p_{\theta},p_{\phi})$, because the particles just move over the 2D surface of the sphere. $J=1$ regardless of the coordinate system you ...

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Instead of working with angles (which are very easy to mess up in 3D), I would just go directly to a transformation matrix. I had to solve a very similar problem to this when tracking charged particles through magnetic fields and all the angles and rotations kept messing me up (probably due to subtle sign errors). So, I went with pure linear algebra. You ...

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I believe you have a math error in your matrix multiply. Substituting the single prime vector into the second rotation and multiplying the matrices doesn't give that third matrix.

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The generalized coordinates are the 2 polar coordinates $r$ and $\phi$. There is 1 constraint. This means we have 2 Lagrange equations of first kind with a constraint force term: Lagrange equation for $r$ is the Newton's 2nd law in the radial direction with a centrifugal force and no constraint force. Lagrange equation for $\phi$ is the angular Newton's ...

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You seem to realize that the rotational analog for Newton's law is important here. This law states that the net torque $\tau$ on an object and its angular momentum $L$ are related by $\tau = \dot{L}$. If I read you question correctly you seem to think that because $\ddot{\phi}=0$, that the angular momentum $L$ must be constant. However, this isn't true. ...

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