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1

The fastest way is to compare kinetic energies in the two cases: \begin{align*} KE &= \tfrac{1}{2}I_{\text{cm}}\omega^2_{\text{cm}} + \tfrac{1}{2}M(R\omega)^2_{\text{cm}} \\ KE &=\tfrac{1}{2}I_{\text{inst}}\omega_{\text{inst}}^2 = \tfrac{1}{2} (I_{\text{cm}} + MR^2)\omega^2_{\text{inst}} \end{align*} So $\omega_{\text{inst}}=\omega_{\text{cm}}$. The ...


0

You haven't given enough information: What you are asking for is that missing bit. I think I can guess just what that is: If $\vec{A}$ is a position vector (e.g. points from the origin to a point that is represented in both the inertial and the non-inertial frame of reference) and your non-inertial frame of reference rotates at a constant angular velocity ...


0

Can we say that the force of this rotating rigid body is given by $F=MV^2/R$, where R is the radius and M is the mass of the rotating rigid body? No, you can't. I believe you have a fundamental misunderstanding of what $F=MV^2/R$ represents. It is the magnitude of force required to maintain a particle of mass $M$ on a curved path with instantaneous ...


2

No, every point on a non-slipping ball moves at its own speed. For every point, the motion is made up of rotation and translation. If the center of the ball with radius $R$ moves at velocity $v$, then the motion of a point at a distance $r$ can be thought of as the superposition of rotation with angular velocity $\omega = \frac{v}{R}$ translation with ...


1

Just an addition to John Rennie's answer. The equipartition theorem can only be derived in classical statistical physics. In quantum statistics it is not correct. For each degree of freedom there is a characteristic temperature below which the quantum effects are significant. This temperature is very high for rotation around the axis of the molecule; I guess ...


9

The energy levels of a diatomic molecule are $E = 2B, 6B, 12B$ and so on, where $B$ is: $$ B = \frac{\hbar^2}{2I} $$ Most of the mass of the molecule is in the nuclei, so when calculating the moment of inertia $I$ we can ignore the electrons and just use the nuclei. But the size of the nuclei is around $10^{-5}$ times smaller than the bond length. This ...


0

However as seen by A , B remains at fixed distance and also doesn't rotate (relative angular velocity is zero). But it does rotate, if your reference frame does not rotate, but just gets centered on your point of interest. If you want to consider a rotating reference frame, then all points (that are fixed to the disc, or to the frame) are obviously ...


1

A rotating reference frame is an accelerated reference frame so $A$ and $B$ are at rest in an accelerated reference frame. Assume an inertial reference frame $S_0$ and another reference frame $S$, with a common origin and rotating with respect to $S_0$. Let the (constant) angular velocity vector of $S$ be $\mathbf \Omega$. Then, the time rate of change of ...


5

If I understand your question correctly you are saying that: $$ v = r\omega $$ and therefore: $$\begin{align} v_A &= r\omega \\ v_B &= \tfrac{1}{2}r\omega \\ v_A &= 2v_B \end{align} $$ but how can $A$ and $B$ have different velocities when they are both attached to the disk so the separation between is fixed? The answer is that $A$ and ...


5

Yes, $B$ does rotate when seen from a static frame of coordinates outside the disk: As to velocities and accelerations, see the article in Wikipedia. It says, $$\vec {v_s} = \vec {v_r} + \vec {\Omega} \times \vec r,$$ where $v_s$ is the velocity in the static frame and $v_r$ in the rotating. If you apply this formula for both points $A$ and $B$, their ...


4

There are a number of ways to represent a 720 degree spin of a particle, but the particle has to be a little more complex then a spinning ball. Imagine this wiki image (from http://en.wikipedia.org/wiki/Spin-1/2) as the inside of a larger ball and you can see an example of 720 degree spin. My favorite representation is the idea of breaking the electron into ...


6

You have fallen prey to a popular simplification of spinors. The statement "you have to turn electron by 720 degrees in order to get the same spin state" does not refer to an actual rotation of an actual electron. In quantum mechanics, we describe the states of objects as elements of a Hilbert space $\mathcal{H}$. The crucial thing is that not all elements ...



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