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4

Consider the following diagram: This shows a mass $m$ moving past a point $P$ in a straight line. Note that the mass isn't connected to $P$ in any way - it's just moving past in a straight line. The angular momentum of $m$ about $P$ is given by: $$ \vec{L} = \vec{r} \times m\vec{v} $$ So the direction of $\vec{L}$ is normal to the screen and the ...


3

I think it might be this $$mgh=\frac{1}{2}mv_{cm}^2+\frac{1}{2}I_{cm}\omega^2=\frac{1}{2}I\omega^2.$$ I mean, it seems that you added the kinetic energy due to center of mass velocity to the rotational kinetic energy respect to the rotational point. It should have been respect to the center of mass.


3

$\vec\omega = I^{-1} \vec L$, and $\vec L$ is constant in the absence of external forces. The bit that I think you're missing is that $I$ rotates with the rigid body, so it is not constant in general and neither is $\vec\omega$. I played with your online example, and the angular velocity does seem to always remain constant when I'm not poking the block, ...


3

The length of the arc PQ is $r\Delta\theta$, as Feynman says, but the difference from $\tan\Delta\theta$ or $2\tan (\Delta \theta/2)$ or something else is negligible because $\Delta \theta$ is assumed to be infinitesimal (infinitely small) in the argument, anyway. For that reason, both angles OPQ and OQP should be taken to be 90 degrees.


2

A fluid is modelled as a vector field and therefore we use vorticity to describe its spinning motion. Angular momentum is more often used for a single object or particle, but not so often for a vector field (even though it is still applicable in principle). For a fluid in general, vorticity is twice the mean angular velocity and this fact to me makes it less ...


2

We don't need to talk about angular momentum because the conservation law is summed up by vorticity. Consider the vorticity equation (in the context of a rotating frame as well): $$ \frac{D\boldsymbol\omega}{Dt}=\boldsymbol\omega\cdot\nabla\mathbf u $$ (ignoring all other terms that are normally contained in this term). If we take the coordinate system where ...


1

I think I understand what you want. Let us call $z$ the axis along the cylinder and $x$, $y$ the other two directions. If the center of the rod is at point C and the end at A you want $I_{xx}^A = \int \rho ({y^2+z^2}) {\rm d} V$ With $m=\int \rho \,{\rm d}V =\rho \pi R^2 H $ and ${\rm d}V = r\,{\rm d}\theta{\rm d}r{\rm d}z$ and $(x,y,z)=(r \cos\theta, r ...


1

Simplifying the problem slightly, I think it can be solved. Assumptions: Pushing force is applied at the bottom of the box - so there is no net torque about the horizontal axis Weight distribution in the box is even Force distribution (normal force) is even - imagine 1000's of tiny springs touching the ground Coefficient of friction is constant Now we ...


1

I think you can calculate the electric field it generates using the Maxwell equations and it should be a constant field. Then with the field you can calculate the torque exerted on the disc and thus you can get the $\omega(t)$.



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