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Let's do this using angular momentum as a vector. This should clear up the question on using both axes separately or one new one. The spinning around the y-axis will give an angular momentum in the y-direction: $\vec{L_{y}} = \hat{y} L_{y}$, while the spinning around z-axis gives angular momentum in the z-direction: $\vec{L_{z}} = \hat{z} L_{z}$. We can get ...

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That looks lie a homework style question. So here the necessary hints. The tensors for the base shapes are listed here. It should be clear how to get the tensors if you put two objects together. The same applies if subtracting the inner cone or cylinder for a hollow object. For additional details check the parallel axis theorem

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Calculate the MI (moment of inertia) of the bowling ball around the axis of rotation , i.e - shoulder. Assuming the bowling ball to be a solid sphere (it is actually a shell?), since the shell thickness isn't given, this would be- Icm = (2/5)MR^2 (MI about it's center of mass) Hence MI around shoulder, using parallel axis theorem - Ishoulder = Icm + Md^2 ( ...

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After further research, I found that the moment of inertia of a system consisting of multiple objects, like the arm-bowling ball system in the problem, can be found simply adding the moments of inertia of each object 1, though use of the horizontal axis theorem may be necessary if the object is rotating around an axis parallel to its typical axis of rotation ...

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I'll make an example, to make things clear. Take a two body system, in which the particles are seperated by a constant distance $d$ and have mass $m_1 = m_2 = m$. This is a holonomic constraint, since $$| \vec{r}_1 - \vec{r}_2 | = d$$ with the particle-positions $\vec{r}_1$ and $\vec{r}_2$. This system is therefore reduced to 5 degrees of freedom (6 minus ...

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I think this should help clear things up. Suppose you take a rod at rest and apply a force $F$ perpendicular to the rod at a distance $r$ away from its center of mass for a short time $\delta t$ - short enough that the orientation of the rod does not change much during the time the force is applied. The rod's linear momentum will become $$F\delta t$$ (from ...

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Acceleration of the center of mass is always $F/m$, so if force and mass are the same, the center of mass will accelerate the same way, regardless of the point where the force acts. After the same time of experiencing the same force, the body in the rotating case has greater kinetic energy than in the non-rotating case. This is due to greater work done by ...

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The only way to purely rotate a rigid body about its center of mass is to apply a pure torque (no net force). If the net force applied is zero then the center of mass is not accelerating. However and combination of translation and rotation of the center of mass can be viewed as a pure rotation about the instant center of rotation. So to effectively answer ...

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Your approach is wrong because the ball is slipping - therefore the relationship between $a_t$ and $r\alpha$ is not constant (your expression $a_t = r\alpha$ does not hold). Instead, you need to look at the torque on the ball: you know the normal force and the coefficient of friction, so you know the force being applied off center $$F = \mu m g$$ which ...

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When you fix a reference point (take it to be the origin of your reference frame) you can write the position as $$\vec{r} = r \hat{r}$$ where $\hat{r}$ is the unit vector pointing toward the particle. Deriving you obtain $$\vec{v} = \frac{dr}{dt} \hat{r}+ r \frac{d \hat{r}}{dt}$$ The first term is the radial component of the velocity, the second one is ...

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I do not believe the pin has any influence in the motion of the gears. At least not if the pin is light enough. The reason is that in such a case internal forces will be zero. Let us look at the two extreme cases: 1) Minimum $x$: $x=0$, then the pin (and the arm) will simply rotate parallel to the bearing at $\omega_c$ 2) Maximum $x$: $x=r_0$, then the ...

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