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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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Your question implicitly assumes that a any object will "boomerang" (you mean return to thrower I guess) only depending on the rotation speed. This is not the case. In very simplistic terms, the boomerang motion depends on its shape and material, besides on its speed. Actually even the speed is not as simple, because it needs a proper combination of ...


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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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Your claim is false (in $\mathbb R^3$). If $\vec{a}$ and $\vec{b}$ are fixed vectors (I assume they are not co-linear and satisfy $|\vec{a}|=|\vec{b}| \neq 0$) there are infinitely many rotations $R \in O(3)$ such that $R\vec{a}=\vec{b}$. One is the rotation $R$ of the angle between $\vec{a}$ and $\vec{b}$ performed around an axis orthogonal to the plane ...


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There's another component to the solution: nonuniform friction. The blade, even if untaped, has different coatings than does the shaft. The butt end may or may not have a tape roll applied. All this means that, even if you throw the stick with zero spin applied, when it hits the ice it's almost guaranteed that there'll be a nonzero net torque due to the ...


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It is called conservation laws, conservation of momentum and conservation of angular momentum. Because friction on the ice is very small, the geometry of the stick is a line with non uniform mass, there will be angular momentum and linear momentum that will be transferred to the ice at the points of contact. To only rotated there should be no linear ...


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Expanding on the comment, here is an answer. Getting all the microcosmic force correct is a bit outside of the scope of this answer, so let me answer the question with a toy model instead. Assume $N$ particles on a ring, attached to each other with perfect springs. The force of the springs is then $F=kx$ where $x$ is the displacement from equilibrium. We ...


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When there is curvature, tensile forces (in a ring, in a string, ...) will give rise to a net force as shown in the following sketch:



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