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In the situation you gave, it's immediately clear what is meant, and there's no possibility for misinterpretation, so yes, it's perfectly acceptable. (Remember that torque is mathematically defined as a vector for convenience, but the direction of that vector isn't really physical.) The only issue I can see with that is that as you leave the simple ...


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Assume $f$ friction is being applied in direction of F. Direction, now, has no significance as $f$ will come out to be negative if it is opposite direction. $$F+f=ma$$ $$FR-fR=I\alpha$$ symbols have their usual meanings Note that if pure rolling occurs, $f$ is static. Also, $$a=\alpha R$$ You can calculate $f$. If $$|f|> \mu_{static}mg$$ You can ...


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There is a wikipedia article which describes the effect http://en.wikipedia.org/wiki/Equatorial_bulge Basically the bulge is caused by the rotation of the Earth. The centripetal force is given by $F=m\omega^2 r$. Therefore the poles feel a lesser force than the equator which wants to spin out into a disc. This is balanced by gravity which wants the Earth to ...


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I'll answer my own question. Since $\boldsymbol{\omega }=\omega _{z}\hat{\boldsymbol{k}}$ the angular momentum reduces to $ \boldsymbol{L}_{O}=-I_{xz}\omega _{z}\hat{\mathbf{i}}-I_{yz}\omega _{z}\boldsymbol{\hat{j}}+I_{zz}\omega_{z}\boldsymbol{\hat{k}}$. We can split the rod in three pieces, calculate moment(product of inertia for each body and sum up. ...


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As an alternate answer using parallel axis theorem, note that the inertia tensor of a rod pointing in the $y$-direction rotating about its center is $$\mathbf{I}_\text{rod,y}=\left( \begin{array}{ccc} \frac{b^3 \rho }{12} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \frac{b^3 \rho }{12} \\ \end{array} \right)$$ and similar for the $x$ and ...



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