Updated 2.8.37 solution

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en/2.8.37.md +21 −22
@@ -52,7 +52,7 @@
<center>
<figure>
<img src="statement.png"
− loading="lazy" width="230" />
+ loading="lazy" width="200" />
<figcaption>
For problem $2.8.37^*$
</figcaption>
@@ -63,40 +63,39 @@
<h3>Solution</h3>
<p>
−
−</p>
<center>
<figure>
− <img src="2.8.37.jpg"
+ <img src="2.8.37.png"
loading="lazy" width="300" />
<figcaption>
− Geometry of the problem
+ Direction of the torque of the rotational force
</figcaption>
</figure>
</center>
−<p>
−From the geometry of the problem
−$$\left\{\begin{matrix}
−\frac{h}{R} = \cos\alpha \\
−\frac{u}{v} = \cos\alpha
−\end{matrix}\right.$$
−Considering the relationship between velocity $u$ and angular velocity $\omega$
−$$\left\{\begin{matrix}
−\frac{h}{R} = \frac{u}{v} \\
−u=\omega R
−\end{matrix}\right.$$
−From where
−$$\boxed{\omega=\frac{hv}{R^2}}$$
+ </p>
+ <p>
+ In order for the wheel to move it is necessary that a moment of force associated with the frictional force from drive belts is transmitted, which is directed along the axis of rotation.
</p>
+ <p>
+ Thus, the momentum will make an angle with velocity $v$ equal to $\pi-\alpha$
+ </p>
+ <p>
+ Projecting onto the linear rotation speed of the wheel:
+ $$u=v\cos(\frac{\pi}{2}-\alpha)=v\sin\alpha\quad(1)$$
+ From the drawing
+ $$\sin\alpha = \frac{h}{R}$$
+ Substituting into $(1)$
+ $$u=v \frac{h}{R}$$
+ From where we find the linear velocity, given that the wheel is circular
+ $$\boxed{\omega=\frac{u}{R}=v \frac{h}{R^2}}$$
+ </p>
+
+
<h4>Answer</h4>
<p>
$$\omega=\frac{hv}{R^2}$$
</p>
− <p style="text-align: right; font-style: italic; font-size: 14;">
− Yuldashev Ulugbek<br>
− </p>
−
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