Edits to “Solution”, “Answer”
en/2.7.25.md
+7 −7
| ### Statement | |||
| $2.7.25.$ Three identical cylinders were spun up to an angular velocity $\omega$, and connected in such way, that left and right are pressd to the central cylinder with the same force. Axes of cylinders are porallel and fixed. Which angular velocities will cylinders have in the end? | |||
| ### Solution | |||
| @@ -8,21 +8,21 @@Solution | |||
|  | |||
| Firstly we can say that moduls of friction forces is the same(force of connecting and matereals are same): | |||
| − | $$F1 = F2 = F$$ | ||
| + | $$F_1 = F_2 = F$$ | ||
| Whats more in the end: | |||
| − | $$ \omega1= \omega3= -\omega2$$ | ||
| + | $$ \omega_1= \omega_3= -\omega_2$$ | ||
| From second newton's law for rotation(for the first and second cylinders): | |||
| − | $$I\frac{\omega1 - \omega}{t} = -F\cdot R $$ | ||
| − | $$I\frac{\omega2 - \omega}{t} = -2F\cdot R $$ | ||
| + | $$I\frac{\omega_1 - \omega}{t} = -F\cdot R $$ | ||
| + | $$I\frac{\omega_2 - \omega}{t} = -2F\cdot R $$ | ||
| Solving this system: | |||
| − | $$ \omega2 = 2\omega1 - \omega $$ | ||
| + | $$ \omega_2 = 2\omega_1 - \omega $$ | ||
| So we have: | |||
| − | $$ \omega1=\omega3=\frac{\omega}{3}; \omega2=-\frac{\omega}{3}$$ | ||
| + | $$ \omega_1=\omega_3=\frac{\omega}{3}; \quad \omega_2=-\frac{\omega}{3}$$ | ||
| #### Answer | |||
| − | $$ \omega1=\omega3=\frac{\omega}{3}; \omega2=-\frac{\omega}{3}$$ | ||
| + | $$ \omega_1=\omega_3=\frac{\omega}{3}; \quad \omega_2=-\frac{\omega}{3}$$ | ||
| ### Statement | ### Statement | ||
| $2.7.25.$ Three identical cylinders were spun up to an angular velocity $\omega$, and connected in such way, that left and right are pressd to the central cylinder with the same force. Axes of cylinders are porallel and fixed. Which angular velocities will cylinders have in the end? | $2.7.25.$ Three identical cylinders were spun up to an angular velocity $\omega$, and connected in such way, that left and right are pressd to the central cylinder with the same force. Axes of cylinders are porallel and fixed. Which angular velocities will cylinders have in the end? | ||
| ### Solution | ### Solution | ||
| @@ -8,21 +8,21 @@Solution | |||
|  |  | ||
| Firstly we can say that moduls of friction forces is the same(force of connecting and matereals are same): | Firstly we can say that moduls of friction forces is the same(force of connecting and matereals are same): | ||
| $$F1 = F2 = F$$ | $$F_1 = F_2 = F$$ | ||
| Whats more in the end: | Whats more in the end: | ||
| $$ \omega1= \omega3= -\omega2$$ | $$ \omega_1= \omega_3= -\omega_2$$ | ||
| From second newton's law for rotation(for the first and second cylinders): | From second newton's law for rotation(for the first and second cylinders): | ||
| $$I\frac{\omega1 - \omega}{t} = -F\cdot R $$ | $$I\frac{\omega_1 - \omega}{t} = -F\cdot R $$ | ||
| $$I\frac{\omega2 - \omega}{t} = -2F\cdot R $$ | $$I\frac{\omega_2 - \omega}{t} = -2F\cdot R $$ | ||
| Solving this system: | Solving this system: | ||
| $$ \omega2 = 2\omega1 - \omega $$ | $$ \omega_2 = 2\omega_1 - \omega $$ | ||
| So we have: | So we have: | ||
| $$ \omega1=\omega3=\frac{\omega}{3}; \omega2=-\frac{\omega}{3}$$ | $$ \omega_1=\omega_3=\frac{\omega}{3}; \quad \omega_2=-\frac{\omega}{3}$$ | ||
| #### Answer | #### Answer | ||
| $$ \omega1=\omega3=\frac{\omega}{3}; \omega2=-\frac{\omega}{3}$$ | $$ \omega_1=\omega_3=\frac{\omega}{3}; \quad \omega_2=-\frac{\omega}{3}$$ | ||