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I think you are confussed between mass moment of inertia and area moment of inertia. The first is an equivalent of mass in angular direction and is defined as $\int_V{r^2\rho dV}$. An angular equivalent of $F=ma$ is: $$\tau=I\alpha$$ where $\tau$ is torque (angular equivalent of force, with units $[Nm]$), $I$ is mass moment of inertia (angular equivalent of ...

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In the frame of reference of the body, is the centripetal force felt or is only the centrifugal force felt? It depends on what you mean exactly. Consider, for example, the amusement park ride Dumbo at Disneyland: . On this ride, passengers sit in mini Dumbo replicas and are swung around in a circle. What forces do they feel? Well, firstly, they ...

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The centrifugal force on the ring is the pseudo force when in the ring's reference frame, which causes it to move outwards, given by $$\vec{F} = m\frac{v^2}{r} = mr\omega^2$$ Where m is the mass of the object, v is the tangential velocity of the object, and omega is the angular velocity To find the time required for the ring to fall off, you need ...

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Here's a motivation for where the inertia tensor $I=(I_{ij})$ (and by extension moments of inertia) comes from. It's a quantity that's analogous to mass for rotational motion in the sense that the kinetic energy of a rotating object is essentially proportional to the inertia tensor times the square of the body's angular velocity. More precisely ...

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The angular momentum of a single particle with mass $m$ in motion about an axis, with angular speed $\omega$, a distance $r$ from the axis, is $L = r (m v) = m r^2 \omega$. When we consider an extended body, the sum up the contribution ($m r^2$) from each particle in motion inside the body, and this is the moment of inertia. More generally, \begin{align} ... 1 If \vec{p} the vector connecting the center of mass of b1 to the center of mass of b2 then you must have \vec{v}_2 = \vec{v}_1 + \vec{\omega}_1 \times \vec{p} \\ \vec{\omega}_2 = \vec{\omega}_1  \vec{a}_2 = \vec{a}_1 + \vec{\alpha}_1 \times \vec{p} + \vec{\omega}_1 \times \vec{\omega}_1 \times \vec{p} \\ \vec{\alpha}_2 = \vec{\alpha}_1 

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