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0

If you are just interested in the rate of precession, then $r$ just has to be the distance from the support (tip of the top) to the center of mass - this is the distance that gives rise to the torque through $mgr\sin\phi$ However, if you want to do this as a vector equation, then you have the instantaneous angular momentum (vector) of the top $I\vec\omega$ ...


2

The "missing" kinetic energy is still there ... it's now in the rotation of the yo-yo.


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For me the second case was more intuitive. The friction and gravity both induce torques of the same sign on the body. Gravity acts through the center of mass, which in this case aligns with the center of rotation. So gravity is not inducing any torque. All the torque is coming from friction. If you imagine the same scenario where the ramp is ...


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Friction always opposes relative motion of the point of contact. To know the direction of friction assume that no friction is present and see the direction in which point of contact is moving: friction is opposite to that direction .


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If the sphere is not spinning, and the axis about which you are taking angular momentum stays through the same position in the sphere, then the relative velocity of the sphere to the axis will always be zero regardless of translational motion. Therefore, the angular momentum will be zero.


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If you are looking for quantitative values, look to further than the equations of motion $$ \begin{align} \sum_{i} \left( \vec{F_i} \right) & = m \vec{a}_{cm} \\ \sum_i \left( \vec{\tau}_i + (\vec{r}_i-\vec{r}_{cm}) \times \vec{F}_i \right) & = I_{cm} \vec{\alpha} + \vec{\omega} \times I_{cm} \vec{\omega}\end{align} $$ where the left hand side is ...


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The angular equivalent of the impulse-momentum theorem states The change in angular momentum of a system is equal to the product of the (average) external torque time the time it is applies In math that is (finite version): $$ \langle \vec{\tau}_{ext} \rangle \,\Delta t = \Delta \vec{L} \,,$$ or (infinitesimal version): $$ \vec{\tau} \, \mathrm{d}t = ...


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I would say three degrees of freedom, and you can use, e.g., the Euler angles (http://en.wikipedia.org/wiki/Euler_angles ).


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Answer to first question: At the instant being considered, the space and body axes are identical, so at that moment the matrix $a$ that relates the two sets of axes is simply the identity matrix. $dG'$ is a vector, so $a_{ji}dG_j' = dG_i'$ is simply equivalent to the statement that with $I$ the identity matrix and $V$ an arbitrary vector, $IV=V$. Answer ...


5

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.


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A simple picture here would be that dissipation leads loss of mechanical energy. But angular momentum has to be conserved. The lowest possible mechanical energy with a given value of the angular momentum of fluid inside a container is that of rigid body rotating around the axis of largest moment of inertia.


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You apparently mean "simulate" when you used the word "model". You'll need two things to accomplish this: A better rotational integrator than the one presented in this answer to your other question, and A physical model of a system that loses energy while conserving angular momentum. Regarding the first item, that integrator is not bad. It has the ...


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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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 ...


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I guess, in layman's terms, this can be described using only moment of inertia and center of mass. In example (a), the book would be held with the axis of rotation horizontally, yielding a low moment of inertia about other axes because it is very narrow in this direction. This would minimize rotation about a different axis and the book would only rotate ...


0

Both cases are the same. Let's assign some axes. Say $z$ points up through the helicopter, $y$ forward, and $x$ to the right. And let's agree that the blades are spinning counterclockwise when seen from above. That is, the angular frequency $\vec{\omega}$ and angular momentum $\vec{L}$ of the blades are initially both in the $+z$-direction, and the ...


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As @NeuroFuzzy pointed out, since the blade will rotate with that pitch for 180 degrees before you change the pitch again, the average position in which the upward forces act during those 180 degrees will be behind axis of rotation, not to the left of it, which only is where the force acts in the beginning of those 180 degrees.


1

We are not told whether the static coefficient of friction is infinite, of between an infinity and zero. It is a poorly written question if this information is not provided. However, it reminds me of a paper that I read last year (which addresses this occurrence very thoroughly.) The link to the paper can be found here. Instead of regurgitating this paper ...


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.


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 ...


0

Ball A has linear momentum pointing in the direction of its flight path. But it is not conserved, because you have a force acting on it (centripetal force), that as no counter part (unless you specify where the axle is mounted and allow for the mount to move as well). Observe, that you can ascribe an angular momentum to B as well even when it does not move ...


0

I think that the easiest way to see this is by considering a reaction wheel. This device consists in a motor with a flywheel attached. When the motor starts spinning, the wheel gains some angular momentum, equal and opposite momentum is gained the cage of the motor and its holder (a ship, a rocket, the astronaut...) that counter rotate. When the desired ...



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