# Tag Info

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You'll need to find the individual moment of inertias of the wheels. If they are hollow cylindrical shells with negligible thickness then you can take their moment of inertia about the rotational axis as $mR^2$ where $m$ is the mass of the wheel and $R$ is the radius. Now, as fibonatic mentioned, you will have to assume that the wheel does not slip and ...

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You can calculate it, but some assumptions would have to be made. Namely the wheel does not slip on the surface, air friction can be neglected and the wheel/surface does not deform big inelastic deformations (which would also dissipate energy). The result depends on de (mass) moment of inertia:

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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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As you move from an inertial frame to one non-inertial frame and then to another, you need to invoke various fictitious forces to continue to misapply Newton's Laws to situations where you can't use them. For example, a passenger in a car going around a curve to the left may choose to consider himself as a frame of reference, and explains his constant zero ...

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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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The factor comes from the moment of inertia of the infinitesimal piece. In the disc method, each piece is a filled flat circle (a disc) of radius $r$, and the moment of inertia of a flat circle is $\frac{1}{2}mr^2$. The $\frac{1}{2}$ accounts for the fact that the mass of the circle is distributed between the center and the edge. But in the shell method, ...

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1) 'Gravity is not pulling x down' is a rather confusing way to think about it (as it's always there), but you are right. What's happening is the cross-product, which requires two vectors as an argument. The result is a vector that is perpendicular to both initial vectors. Of course being perpendicular to both still leaves two directions (check it ...

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The translational kinetic energy is simply $\frac{1}{2} m v^2$ where $v$ is the velocity of the center of mass. Rotational kinetic energy is $E_r = \frac{1}{2} I \omega^2$. To solve the problem, we must write the velocity of the rod as function of $\omega$ (or vice versa). Consider the above image. (Note that my convention for $\theta$ is different from ...

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The kinetic enegy at $t=0$ is equal to the friction work done when the wheel has stopped: $E_{kin}=W_f$ that is $1/2J \omega_0^2=F_fs=mg \mu 2 \pi r n$ (with $J=mk^2$) where $r$ is the radius of the bearing bore an n the number of revolutions. Solving for the revolutions gives: $n= \frac{J \omega_0^2}{4 \pi m g \mu r}$ The angular acceleration due to ...

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