# Tag Info

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The air higher up is moving faster to the ground. Yes, this is mostly correct - at least in principle. However, the effect is tiny compared to the natural speeds of the wind, particularly in the higher atmosphere. This means that in practice everything you say is true but it can be ignored. More precisely, a point on the surface of the Earth on the ...

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"If you go straight up into the sky without accelerating in any direction except up, so you stay directly above the point you left off and just go straight up, you are travelling at the same angular velocity at your new altitude, but you are changing speed." Granting the first part, the second does follow. But that's not what happens. Ignoring atmospheric ...

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Because your arm is effectively rotating, along with the ball, the latter will continue to rotate at the same speed after it leaves the hand. In reality, the fingers usually impart a spin though friction as the ball is released and slides out of the hand.

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You are on the right track with your line of thinking, but as per our comments above, it may be better to think about applying forces to each ball as opposed to velocities. If you apply a force to each ball you can find the center of mass motion by summing all of the forces on all of the balls and using $\vec{F}_{net}=m_{tot}\vec{a}_{CM}$, where $F_{net}$ ...

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v = omega * r omega = v / r In these equations, v refers to tangential velocity; that is, velocity that is perpendicular to the radius of the circle the object is travelling in. To find angular velocity, create a vector from any ball to the center of mass (this will be r) and then get the velocity component of that ball that is perpendicular to that ...

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The frame is only instantaneously aligned with the body frame. The measuring frame is not moving, but the body frame is. So the motion and momentum measure non zero because is it only the alignment that is used and not the motion for measuring. The equations of motion are still on an inertial frame, just not aligned with the world coordinate system. The way ...

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Thank you both for the help. I were able to find a, better I think, solution myself, but only partially why it works. Maybe you could help me with that? If I break the rotation down into three components that each will satisfy the rotation for each vector, I can add the components together to find the total rotation axis.I have tested this in geogebra and it ...

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The rotation velocity vector $\vec \omega$ must be in the plane perpendicular to the vector difference $\vec a - \vec a_0$ and also be in the plane perpendicular to the vector difference $\vec b - \vec b_0$. Consequently, unless these two vector differences are parallel to each other (and provided neither is zero), the rotation velocity vector must be ...

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When a single vector is rotated from one angle to another, and the angle between them is not 180°, then there are infinitely many axes of rotation that could have brought that about - and these axes all lie on a plane that bisects the two vectors. When you have two pairs of vectors, then for each pair there would be such a plane, and the actual axis of ...

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By uniform motion in a straight line, the law refers to both its magnitude and direction. You can't go back and forth without a force.

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I'm guessing you're misinterpreting this diagram: as meaning $\vec \omega$ is "along the plane of motion". That's not what the curved arrow is meant to denote, it is meant to denote the rotation around $\vec \omega$. The angular velocity $\vec \omega$ itself, since $\vec \omega\propto\vec r \times \vec v$, is always orthogonal to $\vec r$ and $\vec p$ and ...

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You don't explain what you are doing in the experiment, but I would guess it is supposed to demonstrate conservation of angular momentum. If we calculate the angular momentum $L$ using: $$L = I\omega$$ then unless some external force is applied $L$ will be a constant. When you start the experiment the disk has moment of inertia $I_\text{Disk}$ and is ...

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Center or rotation isn't located where those two gears make contact, it's in a point closer to the smaller gear's center

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The issue here is that there is a force acting on the axle to counter act gravity on that part of the disk, so not all parts of the disk will be accelerating uniformly at $9.81 \text{ m/s}^2$. Note also that this force acts at the axis of rotation, and therefore doesn't contribute to the torque. That is why you don't need to consider it when calculating ...

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