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In 3 dimensions you can use the cross product to get an appropriate rotation axis. If $\hat{x}$ and $\hat{n}$ are non-parallel 3-dimensional unit vectors then $\vec{s}=\hat{n}\times \hat{x}$ is non-zero. Since $\vec{s}$ is orthogonal to both $\hat{n}$ and $\hat{x}$ there is some rotation about $\vec{s}$ that takes $\hat{n}$ to $\hat{x}$. You can get the ...


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Let $\theta>0$ denote the angle between the $\hat{\mathbf x}$ and $\hat{\mathbf n}$. Notice that $$ \hat{\mathbf u} = \frac{\hat{\mathbf x}\times\hat{\mathbf n}}{\sin\theta} $$ Is a unit vector perpendicular to both $\hat{\mathbf x}$ and $\hat{\mathbf n}$. The desired rotation is a right-handed rotation around $\hat{\mathbf u}$ by the angle $\theta$. ...


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REVISED ANSWER : The tension in the string will become zero when $v^2/r = gsin\theta$. v is still +ve when this happens. v will only become zero later. So you are correct : first T becomes zero (which cannot happen until $\theta > 0$), then the stone moves as a projectile on a parabolic path, then v beomes zero.


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To answer the generic dimension part, in case it is not self-evident from Valter's answer: In D dimensions, the rotation matrix is the exponential of an angle θ times a matrix K, a normalized generator of the corresponding rotation group SO(D) around some unit axis D-vector k, in the vector representation, so the matrix is D×D. The eigenvectors of these ...


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Rotation of a 3-vector We'll find an expression for the rotation of a vector $\mathbf{r}=(x_1,x_2,x_3)$ around an axis with unit vector $\mathbf{n}=(n_1,n_2,n_3)$ through an angle $\theta$, as shown in Figure . The vector $\mathbf{r}$ is analysed in two components \begin{equation} \mathbf{r}=\mathbf{r}_\|+\mathbf{r}_\bot \tag{01} \end{equation} ...


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If you have found that the initial angular acceleration of the disc is $\alpha = \frac {2g}{3R}$ that must mean that the initial centre of mass acceleration $a = \frac {2g}{3}$ because $a = R \alpha$. If the force acting on the pivot is $X$ upwards then applying Newton's second law for the centre of mass motion gives $mg-X = ma$, so $X = \frac {mg}{3}$. ...


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You have made a model of a viscous fluid coupling which was used in a number of four wheel drive vehicles to transfer torque. The system relies in the fact that adjacent planes of moving liquid experience a viscous force between them.


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It is due to the viscous nature of any liquid. When you stir, the liquid starts spinning and this causes the liquid (the part which is in contact with the pot) to "drag" the pot(due to friction) along with it in the path of its motion.Hope this answers your question.


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A probable explanation for this effect is simply that the bottom of the pot might be a bit bulged out, as to form only one point of contact around which the pot then can rotate relatively freely (with little friction). As you stir the water inside the pot, the moving water molecules exert a frictional force on the walls of the pot, dragging it in the same ...


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Typically there are two kinds of transformation that do not change the outcome of situation. Think of a force vector $\vec{F}$ passing through a point $\vec{r}_A$. Any translation along the line of the force, in the direction $\vec{e} = \frac{\vec{F}}{\| \vec{F} \|}$ will not change the outcome. Any rotation about the line of the force would also not ...


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First thing you need to do is convert the spherical coordinates into cartesian and then derive a 3×3 rotation matrix from the axis angle information. You can manipulate the rotation matrix using elementary rotations and then get the axis angle for the final orientation. Then you convert the axis angle to spherical coordinates for use in your system. Here ...


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In general it changes although the reason is not exactly because its projections changes. For example. You start with a vector (let us say the electric field of a parallel plate capacitor) on the plane $xy$. Then you rotate the coordinate system by an angle. The components of the vector on the new coordinate system is changed. But the vector did not change ...


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Direction of a vector is determined by the components themselves. Now if the components are changed the direction gets changed by the above definition. All this is with respect to one reference frame.


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The frictional force is given by $f = \frac 1 3 mg \sin \theta$ when the no slipping condition ($a=r \alpha$) is satisfied. If the component of weight down the slope, $mg \sin \theta$, was equal and opposite to the frictional force then there would no net force on the cylinder and so the centre of mass of the cylinder would not be accelerating ($a=0$). ...



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