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If you take for rigid body the Rodrigues rotational Matrix which is: $$S=I_3 +\tilde{d}\,\sin(\varphi)+\tilde{d}\tilde{d}\,(1-\cos(\varphi))\tag 1$$ where $\vec{d}$ is a $3\times 1$ the instantaneous rotation axes with $\vec{d}\cdot\vec{d}=1$ , $~\varphi$ is the rotation angle about this axis $~,I_3$ is $3\times 3$ identity matrix and tilde operator is: ...

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According to your diagram the particle looks like is traveling in uniform circular motion. We can find the radius by using the relation $v_{tangential}=\omega r$

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In this answer I am referring to top position only.Similar arguments can be made for any position. Well think it like this.You know that radius of curvature of circular motion is given as: R= F/(m ω^2) where F is net force. Notice if we keep angular speed decreases, radius decreases. However you know that such a thing is not happening here.The reason of this ...

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Basically, if we know the concrete is traveling in a circle with frequency $\omega$, we know its trajectory - that is, its position vector as a function of time. So, by taking the second derivative of that function, we know its acceleration. For circular motion, that works out to be a vector of magnitude $R\omega^2$, always pointing toward the center of ...

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Torque, angular velocity and angular momentum are not inherently vectorial quantities. One way to demonstrate that is to consider motion in some space with more spatial dimensions than our familian three spatial dimensions. In a space with 4 spatial dimensions the following applies: to specify orientation of a state of rotation you need to specify the plane ...

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The vectors describing rotation are chosen along the axis of rotation because that is the only fixed direction in a rotating system. Given that choice, they do a good job of representing the magnitude and direction of rotational quantities, and as long as the axle remains fixed, they relate to each other and the various linear quantities in a ...

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Based on your free body diagram, yes, the center of mass would not experience any acceleration, but the object would rotate about the center due to the torque from force $f$. However, based on what you have drawn, perhaps you have misunderstood how the forces are at work here. It looks like to me that you have a ball, (or disk, etc.) that is on an incline. ...

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