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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, ...


0

Torque is a rotational analouge of force.It plays the same role in rotational dynamics as force playes in linear motion.Torque is defined as the moment of force or turning effect of force about the given axis or point.It is measured as the cross product of position and force vector while as angular momentum is the rotational analouge of linear momentum.It is ...


1

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 ...


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 ...


4

You seem to be saying that friction couldn't speed it up, because nothing else is moving that fast. Well, how fast is it moving? We can imagine the gyroscope axis parallel to the z axis, and the casing to be aligned such that the x axis goes through it. If the casing is tipped slightly, the gyroscope resists that turning and one side of the shaft has firm ...


0

If the shape is described with polar coordinates such that the radial distance from some center is $R(t)$ and $t$ is the angle from some datum, then you can define the pressure angle $\alpha$ at each location $t$ as $$ \tan( \alpha) = - \frac{1}{R(t)} \left( \frac{{\rm d}}{{\rm d}t} R(t) \right) $$ The pressure angle will give the direction of the normal ...


6

Backspin! Those shots in which the cue ball "draws" backwards after hitting the target ball involve backspin. Without backspin, the cue ball cannot reverse direction. Consider what happens when the cue ball is not spinning at all when it hits the target ball. The cue ball will come to a dead stop if it hits the target ball straight on. Think of Newton's ...


1

The direction the ball will take depends on the angular momentum. The velocity with which the ball moves or bounces backwards but the chief determinant is the spinning effect of the incoming ball.


5

First of all, if the collision is elastic, the distribution of momentum in between the components is completely determined by momentum and energy conservation! This statement is most obvious in the center-of-mass frame where the total momentum is zero and the two objects are moving in opposite directions. The momentum conservation (the total momentum is ...


1

There is a distinction between points and vectors. Points are positions in space, and vectors are directions. One can easily mix up the two, because in Euklidean space they look rather similar. $\theta$ in this case is a coordinate, i.e. part of the description of a point. The vector associated to that coordinate could be called $\hat{e}_\theta$, and point ...


0

I'm guessing that you understand clearly the effect of precession here. The reason why the wheel starts to fall down is that when we explain the change in angular momentum of the wheel, we say that the angular momentum vector only changes in direction- right? But the angular momentum vector of the wheel doesn't only point outwards; it also points upwards (it ...


1

You are very close. Just to review what is going on, the period is given by \begin{equation} T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{I}{k_{eff}}} \end{equation} where $I$ is the moment of inertia of the system and the torque is proportional to the angle by which the pendulum has been displaced with a coefficient that I'm calling $k_{eff}$ in analogy to ...


0

I understand the question but have not implemented these internals myself; but maybe this observation will help you: when you do the collision response for the inter-penetration contact (with coefficient of restitution > 0) you shouldn't have to apply an impulse so large that the body's CoM actually reflects its velocity - it is enough to apply an impulse ...


-2

The answer is clockwise and you wil generally slice the ball flight... I know the answer mainly from golf experience. As a side comment I fancy golf pros like Wie and Gulbis would know this as well...


0

If we write the equation of rotation about the pivoted point, then moment of inertia of the disk about the pivoted point $P$ dose not equal to the $\frac{1}{2}mR^2$, i.e. from the parallel axis theorem the moment of inertia of the disk about point $P$ will be: $$I_P=I_{cm}+mx^2=\frac{1}{2}mR^2+mx^2=\frac{1}{2}m(R^2+2x^2)$$ On the other hand,the torsional ...


1

By way of analogy, think of what happens when you blow up a balloon and let it go. It spins around, goes this way and that. A balloon rarely goes straight, without spinning. The thrust from a balloon rarely goes through the center of mass. It rotates and translates. Because the thrust vector itself turns with the rotating balloon, the translation is not ...


0

Apparently it's better to ask questions only one-at-a-time on this site, but I understand why you did it like that; these are all related. Now, these are some really interesting questions, hopefully I can help you out (there's some interesting subtleties to consider due to the thrusters). Answer 1 If you fire only B, then it will rotate, and it will ...


1

There is 1 reason; Newton's third law. When you fire a bullet, the bullet has a momentum in one direction (east) and the gun has momentum in the opposite direction (west). Of course, the person stops the gun from moving. When the bullet strikes an object, it imparts its momentum on the object. Neglecting air resistance, it is easy to show that all the forces ...



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