Updated spacing between @ latex expressions

astrosander edited
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revision #10476 parent #9580 GitHub c702ebe ← older newer →
en/2.7.42.md +6 −6
@@ -81,25 +81,25 @@
Angular momentum
$$\left\{\begin{matrix}
L=m_2v\frac{l}{2} \\
−L=I\omega
+L=I\omega
\end{matrix}\right.$$
Parallel axis theorem
$$I=I_0+m_2d^2=\frac{1}{3}m_1l^2+m_2\left(\frac{l}{2}\right)^2$$
$$m_2v\frac{l}{2}=\omega\left(\frac{1}{3}m_1l^2+\frac{1}{4}m_2l^2\right)$$
From where
$$\omega = \frac{6m_2}{4m_1+3m_2}\frac{v}{l}$$
From the figure
−$$h=\frac{l}{2}(1-\cos\alpha)$$
+$$h=\frac{l}{2}(1-\cos\alpha )$$
Law of conservation of energy in rotational motion
−$$\frac{I\omega^2}{2}=(m_1+m_2)\frac{l}{2}(1-\cos\alpha)$$
+$$\frac{I\omega^2}{2}=(m_1+m_2)\frac{l}{2}(1-\cos\alpha )$$
Expressing $\cos\alpha$
−$$1-\cos\alpha=\frac{I\omega^2}{(m_1+m_2)l}$$
−$$\boxed{\cos\alpha=1-\frac{3m_2^2v^2}{gl(4m_1+3m_2)(m_1+m_2)}}$$
+$$1-\cos\alpha =\frac{I\omega^2}{(m_1+m_2)l}$$
+$$\boxed{\cos\alpha =1-\frac{3m_2^2v^2}{gl(4m_1+3m_2)(m_1+m_2)}}$$
</p>
<h4>Answer</h4>
<p>
− $$\cos\alpha=1-\frac{3m_2^2v^2}{gl(4m_1+3m_2)(m_1+m_2)}$$
+ $$\cos\alpha =1-\frac{3m_2^2v^2}{gl(4m_1+3m_2)(m_1+m_2)}$$
</p>
<footer class="row container">
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