Updated spacing between @ latex expressions

astrosander edited
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revision #10414 parent #9583 GitHub c702ebe ← older newer →
en/2.7.41.md +2 −2
@@ -84,14 +84,14 @@
Determening the moment of inertia, using parallel axis theorem
$$I=\frac{1}{2}mL^2+m\left(\frac{L}{4}\right)^2=\frac{7}{48}mL^2\quad(2)$$
From the $(1)$ and $(2)$
−$$v=\frac{5}{6}\omega L\Rightarrow \boxed{\omega=\frac{6v}{5L}}$$
+$$v=\frac{5}{6}\omega L\Rightarrow \boxed{\omega =\frac{6v}{5L}}$$
Also, in addition to rotational motion, translational motion will be observed. From the law of conservation of momentum
$$mv=2mu\Rightarrow u=v/2$$
$u$ — centre of mass velocity</p><p>
THe initial energy was as a kinetic energy
$$E_0=\frac{mv^2}{2}$$
The total energy after the collision:
−$$E_f=\frac{2mu^2}{2}+\frac{I\omega ^2}{2}+\frac{m(\omega\cdot L/u)^2}{2}$$
+$$E_f=\frac{2mu^2}{2}+\frac{I\omega^2}{2}+\frac{m(\omega\cdot L/u)^2}{2}$$
Conservation of energy
$$E_0=E_f+Q$$
Where, $Q$ — the released energy
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