# Tag Info

2

$T_\mu^\nu = T_{\mu\sigma}g^{\sigma\nu}$

1

Of course you can specialise the derivation of the three dimensional Euler equations for your special case. However, unless you are expressly asked to do this, I doubt that your teacher means for you to take this path because it is very much an overkill. The yoyo is spinning in a plane about its axis of symmetry. The inertia tensor is now a simple scalar ...

1

Integrating your equations (1) and (2) you get that $(\text i) \ \omega_1 = C_1$ a constant in time, while another constant in time. This equality you can re-write as $(\text {ii}) \ \omega_2^2 + \omega_3^2 = C_2^2$, As to the vector $J_{\Omega}\omega$ you know that its components are $\{J_1\omega_1, J_2\omega_2, J_2\omega_3\}$, therefore $(\text ... 1 I will answer the mass moment of inertia tension question. For an infinitesimal clump of mass${\rm d}m$located at$\vec{r}$its effect on the inertia tensor is $${\rm d}{\bf J} = -[\vec{r}\times][\vec{r}\times]{\rm d}m$$ where$[\vec{r}\times]$is the skew symmetric cross product operator$$\begin{pmatrix}x\\y\\z \end{pmatrix}\times = ... 0 The important point is that both$t^a$and$v^b$are vectoror fields defined (at least) on$C$. The idea is that if$t^a\nabla_av^b=0$and you take$v^b$at a particular point$p$on$C$then$v^b$at all other points on$C$is the parallel transported vector from$p\$. Another way of thinking about it is that the covariant derivative in a given direction is ...

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