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2

Here, in the above picture $M \ge m$. Note that this is the most general case. We can have $M = m$ and the angle $\theta$ can vary anywhere between $[0;\cfrac{\pi}{2}]$ (Actually, the most general case would have been to take 4 different masses but we will be going out of the bandwidth of your problem, and it would be a pointless discussion.) Now, ...

2

usually linear velocity is the velocity of a point rotating around the axis of rotation given by $$\vec v = \vec{\omega} \times\vec{r}$$ when object has no translational motion but if the object has both translational and rotational motion then $\vec v$ will be measured from Center-of-momentum frame. in the frame where the C.M is translating at velocity ...

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Suppose we have a particle moving with a velocity $\vec{v}_1$ and a object which has no linear velocity, but is rotating with angular velocity $\vec{\omega}$. Now suppose we look at a point at a displacement $\vec{r}$ from the axis of rotation of this rotating velocity. The velocity of this point is $\vec{v}_2 = \vec{\omega} \times \vec{r}$. Now the ...

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If you have a center of mass C located at $\vec{r}_C$ and a force $\vec{F}$ passing through an arbitrary point A (at $\vec{r}_A$) then the net moment of the force about C is $$\vec{M}_C = (\vec{r}_A-\vec{r}_C)\times \vec{F}$$ where × is the vector cross product. Your equations of motion sum up all the forces and moments at the center of mass  \sum_i ...

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You more or less know the answer (but perhaps you need to rephrase your question). If the question is "for a given force vectors $\vec{F}$, when should I consider its effects on linear velocity $\vec{v}$ /linear acceleration $\vec{a}$? and when should I consider angular velocity $\vec{\omega}$ / angular acceleration $\vec{\alpha}$?" (1) If a force vector ...

3

The answer to that is because the moment of inertia is not the same for the solid cylinder than for the hollow one. As you write the formula for the moment of inertia, it depends on the distribution of the mass. The further away the mass is from the rotation axis, the more contributes to the moment of inertia (as in distance squared $r^2$). So, since the ...

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You've failed to take into consideration that $r$ is the radius of a piece of mass $\delta m$ rotating about an axis. So that the product of $\delta m$ and $r$ must be done first before the sum, or more probably the integral.

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The angular momentum of a massive sun may cause the freely falling spaceship to start spinning in the direction of the sun's angular momentum for an effect of frame dragging. You can take a look at the Kerr metric which describes the behaviour of the spacetime near a massive spinning object. If you're not familiar with general relativity it could be ...

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The only way for a falling object to be made to rotate and translate is if there was a separate force causing this rotation. In an atmosphere this is a net force on one side of the craft whose surface area (and therefore drag) is the highest, causing this part of the craft to rotate away from the direction the entire craft is translating. Essentially ...

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Start with \begin{align} \alpha(t) = \frac{d\omega}{dt}(t). \end{align} Integrate both sides from $t_i$ to $t_f$; \begin{align} \int_{t_i}^{t_f}dt\,\alpha(t) = \int_{t_i}^{t_f}dt\, \frac{d\omega}{dt}(t) = \omega(t_f) - \omega (t_i) \end{align} The second equality on the right follows from the fundamental theorem of calculus which basically says that if ...

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