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If the billiard ball does not move when struck by the ping pong ball, it means that the impulse from the ping pong ball was not sufficient to overcome the static friction between the billiard ball and the table. If the billiard ball is spinning, the friction is kinetic and smaller.


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The inertia tensor is a bit more descriptive in the spherical tensor basis (so instead of having nine basis dyads made from combinations like $\hat x \hat y$, you have 9 basis tensors that transform like the nine $Y_l^m$ for $l \in \{0,1,2\}$). Since $I_{ij}$ is antisymmetric, all $l=1$ spherical tensors are zero. The $l=0$ portion is: $$ I^{(0,0)} = \frac ...


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Using the inertia tensor $I_{jk}$ of an object, you can construct an ellipsoid satisfying the equation $$1=x_{j}I_{jk}x_{k} .$$ As you said, the eigenvectors of $I_{jk}$ are the principal axis $\vec{r}_{j}$ of your object with the respective moment of inertia $\theta_{j}$ as eigenvalue. Thus, changing coordinates to the principal axis of the object ...


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Try to make your question more clear, I'm answering you, but I'm not sure, because I didn't understand your question well. Let us Express $\mathscr M$ first in function of $\mathbf r$ the vector ray, note that : $$\mathbf r^2=x_k x_k$$ Where $k=1,2,3\dots $ the composants of $\mathbf r$ and let considered an axis $D$ characterized by its vector $\mathbf ...


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Nice question you got here! Let's go through this step by step: 1) Embedding of the body To understand the coordinates "rigidly fixed" on the rotating body, think of the object as a manifold $B$ embedded in space $M=\mathbb{R}^3$, i.e. an injective immersion $$\imath:B\rightarrow M.$$ Since we want to discribe a motion of this body, we let this mapping be ...


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I will use Einsteins-summation convention to make it more readable, this means I will sum over indices that occur twice within a product, i.e. e.g. $I_{ij} \Omega_j \equiv \sum_{j=1}^3 I_{ij} \Omega_j$. So, the way to get your equation is as follows: $\Omega_i \; I_{ji} \; \Omega_j \enspace = \enspace M_j \; \Omega_j \qquad \text{(1)}$ Now use: $\Omega \...


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They have solved the problem with respect to the center of mass(hence I=mr^2/2)as, though we know that there is friction acting on the system which is responsible for pure rolling it does not do any work because by definition of pure rolling the point remains instantaneously at rest. So while applying the work energy theorem we consider only the work done by ...


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In the above question, they are analysing the situation from Centre of Mass. That's why they have taken the moment of inertia as $\frac{1}{2}mr^2$. However you can do the analysis from the instantaneous centre of rotation. The results obtained will be the same.


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The way I would explain it is non-rigorous, but intuitively clear: The moment of inertia $MI$ of a mass $M$ with respect to an axis is closely approximated by $ MI = M R^2$, where R is the perpendicular distance from the axis to the center of mass of $M$. (This is very nearly true when the mass M occupies a small roughly spherical or cubical volume and R is ...


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I believe that the Steiner principle can be used to show that axis through the centre of mass has the lowest moment of inertia (across the parallel axes), but it doesn't explain why. First, the principle is not intuitive. Second, to prove it you would use a construction similar to what is needed to show the statement in the first place. Here is my atempt to ...


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There seems to be some information in the question which you have not provided. We are not told whether the disk is spinning when the square frame is set swinging. If the disk is not spinning about its centre C initially, and if there are no torques which start it spinning when the square frame swings (because the axle is frictionless) then the disk doesn't ...


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The rings that fly off retain the same angular momentum they had right before they flew off so that the total angular momentum remains the same without the rod speeding up. Once the rod is by itself it would require a torque that does not exist to change the rate of rotation of the rod.


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The three inequalities follow from $$I_{xx}=\sum_k m_k(y_k^2+z_k^2),$$ $$I_{yy}=\sum_k m_k(x_k^2+z_k^2),$$ and $$I_{zz}=\sum_k m_k(x_k^2+y_k^2).$$ They do not follow from the symmetry and positive-definiteness conditions that you mention. For example, the diagonal matrix with diagonal elements 1, 3, and 5 is symmetric and has positive-definite ...


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The perpendicular axis theorem actually has two statements or necessities if you must say: $$ $$ 1) The rigid body to be considered should be two dimensional $$ $$ 2) If there exist three mutually perpendicular axis I1, I2, I3, out of which any two(lets say I1 and I2) are in the plane of the 2D body then: $$ I1 + I2 = I3 $$ Now out of I1,I2,I3 which ...


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As Wikipedia states: The perpendicular axis theorem states that the moment of inertia of a plane lamina about an axis perpendicular to the plane of the lamina is equal to the sum of the moments of inertia of the lamina about the two axes at right angles to each other, in its own plane intersecting each other at the point where the perpendicular axis ...


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