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$\sqrt{E^2 - p^2}$ is the center-of-mass energy of a system, using units of $c=1$. $E$ would be the total energy ($m + K$) of each particle in the system and $p^2= \vec{p}\cdot\vec{p}$ where $\vec{p}=\Sigma_i \vec{p}_i$ where the sum runs over all particles. This center-of-mass energy is available to the system to form new particles with mass (as in 2 ...


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Set your copy of Halliday, Walker & Resnick on the table so the front cover is parallel to the table and visible to you. Now your right hand flat on the book with your thumb and forefinger forming a ninety degree angle. Orient your hand so your thumb is parallel to the spine of the book and pointing toward the top edge. Your forefinger should be parallel ...


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Angular displacement is define by change in angle i.e. ${\Delta \theta}=({\theta}_1 -{\theta}_2)$, where $\theta$ is taken as positive in anti-clock wise. Hence angular displacement ${\Delta \theta}$ has both magnitude and direction. But let when a particle rotate from a point A in $\theta$ angle in anti-clock wise to point B and then back through $\theta$, ...


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The wikipedia link does an adequate job of explaining this. You can find a very mathematical discussion in this partial duplicate question, but here is my simple take on it. The angular displacement in three dimensions does have a vector nature in the sense of having both a magnitude (the angle through which you turn something) and a direction (the axis ...


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Think of rotation as a manifestation of a change in coordinate direction. Affix a coordinate system at any point on a rigid body and it is going to change direction at the same rate regardless of the point location (even at the rotation axis). This is why angular velocity is shared among the entire rigid body. Mathematically, the rate of change of a 3×3 ...


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I think when a rigid body rotating about any point , the point of rotation remains fixed and all other points of the rigid body always rotate about that fixed point with same angular velocity. Let us suppose a disk rotating about its center with a angular velocity $\omega$ and $r$ is the radius. Now the angular velocity at point A and B must be same. If ...


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Yes, this is certainly true. Mass is defined by $m^2=E^2-p^2$ (in units with $c=1$), where $(E,p)$ is the momentum four-vector built out of the mass-energy and momentum. (This defines what's known as invariant mass, as opposed to "relativistic mass.") Mass as defined in this way is not additive, and depends on the motion of the particles within a system. As ...


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You've almost got it. You would want to express everything in terms of just one variable, otherwise you run into trouble. Right now you have both $\omega$ and $\theta$, but there is a relationship between them... However, since you are asked for an expression in $\theta$, it is simpler to use conservation of energy - you don't actually need to know the ...


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All such things as 'translational energy' or 'vibrational' are more of extrapolating concepts. Yes, there is rough approximation of $C_v$, but that works only for gases with low density. When density in your material increases, the pressure which you 'expect' is defined with not only 'impulse' part, but also 'force' part of pressure. $P = nkT + ...


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Yes translational energy levels are very close together. These can be calculated with particle in a box type quantum mechanics for molecules in the gas phase, which are free to move about. In ice the molecules are held together next to each other and cannot move about - they can vibrate about the positions that they are held in. So their average position ...


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The tricky part in this question is the 'speedometer' part. The speedometer measures the speed of the car if its tyres had the same angular acceleration as in the slipping case, but driven on the road (on which the tyre doesn't slip). It is obvious here, that $a_{com} \neq r\alpha$, where $a_{com}$ is the linear acceleration of the tyre's COM on the ...


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There are two components of acceleration. One tangential and one radial. The friction force must account for both of these components. $$ a_R = \dot{v}= \alpha r $$ $$ a_T = \frac{v^2}{r} = \omega^2 r$$ Your friction condition is $$ \sqrt{ a_R^2 + a_T^2 } \le \mu g $$



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