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Consider the reaction impulse $P$ from the rough ground (infinite friction) ruling the impulse $F$ as shown. The equation of motion are: Sum of impulses equal change in momentum (motion of the center of mass) $$P- F = m \Delta v$$ Sum of moments about center of mass equal change in angular momentum $$F (r \sin \theta) + P r = I \Delta \omega$$ Under ...

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Invariably, a good measurement involves comparing something "known" with something "unknown". For example, you put an unknown length next to a ruler. In your case, you have an "unknown" thrust resulting in some acceleration. You could tie a string to your bottle car, run the string over a pulley, and hang a small weight from it. Measure the acceleration for ...

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Can an angle be defined as a vector? It depends on what you mean by "vector". If by "vector" you only mean something that has a magnitude and a direction, then yes, the axis-angle representation qualifies as a "vector". To a mathematician, a vector is something that is a member of a vector space. In this context, the axis-angle representation fails to ...

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Describing a rotation as a vector, with the direction of the vector along the axis of rotation, and the magnitude of the vector as the angle, is known as the axis–angle representation.

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Let $\alpha$ be the angle between the rod and the floor at point $B$ (see my drawing). Then we can write the points $A$ and $B$ with the length $L$ of the rod as $$\vec A = L \sin(\alpha) \begin{pmatrix} 0 \\ 1 \end{pmatrix} \,,\qquad \vec B = L \cos(\alpha) \begin{pmatrix} 1 \\ 0 \end{pmatrix} \,.$$ The center of gravity $M$ of the rod is just the ...

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"Say we have two spinning spheres, each spinning about an axis going through their own centre of mass. I have highlighted a very important statement in your question. If you have a body rotating about its centre of mass then if the centre of mass does not move the angular momentum of the body will be the same about any point. So the angular momentum of ...

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Yes you have to take the frictional force F into account so this exerts a negative impulse moment RFdt on the right hand disk with radius R as well as rFdt on the left hand disk with radius r. So the delta of the rotational moments have the ratio r/R. Assuming the left hand disk spins with rim velocity v0 before impact and after impact they have equal rim ...

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In order for the spinning disc to set the stationary one in motion, friction forces need to act between the contacting surfaces: That requires a Normal force $F_N$ to act between them. Using the kinetic friction coefficient $\mu_k$ we can then state: $$F_F=\mu_k F_N$$ The friction force on the $m,r$ disc causes angular acceleration $\alpha=\frac{d ... 1 When you drop a stationary disc onto a rotating one there must be a time when there is relative motion between the discs as you cannot have an infinite acceleration. If there is no friction then nothing much happens and the spinning disc carries on spinning and the other disc just sits still on top of it. To get an interaction between the discs you need ... 1 yes i think your assumption is correct based on the fact that in equilibrium the discs shouldn't slip on each other. but then you need to change your equation for angular momentum. use parallel axis theorem to write down the moment of inertia for the stationary disc to get the answer in the equation. since you have chosen your axis to be passing through the ... 1 In uniform circular motion,$v = v(r)$so that$a_{c} \propto r$, not$a_{c} \propto r^{-1}$. In other words,$v$is not a constant at all radii while$\omega$is constant for all radii. So the second part of the expression is the one you want to look at in this regard. Though the first half of the expression states that$a_{c}$is only explicitly ... 2 In the first expression$v$is the speed of a single point on the disk, actually any point that is a distance$r$from the center. In the second expression,$\omega$is the angular speed of the entire disk. Since$v$and$\omega$have very different definitions, the interpretation of those two equations is quite different. 0 This may be a result of bad wording and/or misunderstanding. Let me try to clarify - suppose a particle is tied to a string of length$l$and orbits around its center with some angular velocity,$\omega$. That is, in polar coordinates, its position is given by$(r,\theta) = (l, \omega t)\$. From this, we can find the velocity in this coordinate system - ...

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