Правка разделов «Statement», «Solution», «Answer»
en/2.7.25.md
+14 −30
| @@ -1,44 +1,28 @@ | |||
| ### Statement | |||
| − | $2.7.25.$ [Insert problem description here] | ||
| + | $2.7.25.$ 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? | ||
| − | __Example Statement__: | ||
| − | $1.1.1.$ Determine the coordinate $x(t)$ of a body as a function of time $t$, given that its acceleration is defined as $a(t) = bt$, where $b$ is a constant. | ||
| − | |||
| ### Solution | |||
| − | [Your solution should be placed here] | ||
| + |  | ||
| − | __Example Solution__: | ||
| − | |||
| + | Firstly we can say that moduls of friction forces is the same(force of connecting and matereals are same): | ||
| + | $$F1 = F2 = F$$ | ||
| − | $$ | ||
| + | Whats more in the end: | ||
| + | $$ \omega1= \omega3= -\omega2$$ | ||
| + | From second newton's law for rotation(for the first and second cylinders): | ||
| − | We know that acceleration is the time derivative of velocity: | ||
| + | $$I\frac{\omega1 - \omega}{t} = -F\cdot R $$ | ||
| + | $$I\frac{\omega2 - \omega}{t} = -2F\cdot R $$ | ||
| + | Solving this system: | ||
| − | $$ | ||
| + | $$ \omega2 = 2\omega1 - \omega $$ | ||
| − | To find the velocity $v(t)$, we integrate $a(t)$ with respect to time: | ||
| + | So we have: | ||
| + | $$ \omega1=\omega3=\frac{\omega}{3}; \omega2=-\frac{\omega}{3}$$ | ||
| − | $$v(t) = \int a(t) \, dt = \int b t \, dt$$ | ||
| − | |||
| − | If the initial velocity is $v(0) = 0$, then the velocity becomes: | ||
| − | |||
| − | $$v(t) = \frac{b t^2}{2}$$ | ||
| − | |||
| − | Likewise, integrate $v(t)$ with respect to time: | ||
| − | |||
| − | $$x(t)= \int v(t) \, dt = \frac{b}{2} \int t^2 \, dt$$ | ||
| − | |||
| − | From where the coordinate from time, considering the initial conditions: | ||
| − | |||
| − | $$\boxed{x(t)=\frac{bt^3}{6}}$$ | ||
| − | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result, like this:] | ||
| − | |||
| − | |||
| − | __Example Answer__: | ||
| − | $$ | ||
| + | $$ \omega1=\omega3=\frac{\omega}{3}; \omega2=-\frac{\omega}{3}$$ | ||
| @@ -1,44 +1,28 @@ | |||
| ### Statement | ### Statement | ||
| $2.7.25.$ [Insert problem description here] | $2.7.25.$ Three identical cylinders were spun up to an angular velocity $\omega$, and connected in such way, that left and right are pressd to the central cylinder with the same force. Axes of cylinders are porallel and fixed. Which angular velocities will cylinders have in the end? | ||
| __Example Statement__: | |||
| $1.1.1.$ Determine the coordinate $x(t)$ of a body as a function of time $t$, given that its acceleration is defined as $a(t) = bt$, where $b$ is a constant. | |||
| ### Solution | ### Solution | ||
| [Your solution should be placed here] |  | ||
| __Example Solution__: | Firstly we can say that moduls of friction forces is the same(force of connecting and matereals are same): | ||
| $$F1 = F2 = F$$ | |||
| $$ |
Whats more in the end: | ||
| $$ \omega1= \omega3= -\omega2$$ | |||
| From second newton's law for rotation(for the first and second cylinders): | |||
| We know that acceleration is the time derivative of velocity: | $$I\frac{\omega1 - \omega}{t} = -F\cdot R $$ | ||
| $$I\frac{\omega2 - \omega}{t} = -2F\cdot R $$ | |||
| Solving this system: | |||
| $$ |
$$ \omega2 = 2\omega1 - \omega $$ | ||
| To find the velocity $v(t)$, we integrate $a(t)$ with respect to time: | So we have: | ||
| $$ \omega1=\omega3=\frac{\omega}{3}; \omega2=-\frac{\omega}{3}$$ | |||
| $$v(t) = \int a(t) \, dt = \int b t \, dt$$ | |||
| If the initial velocity is $v(0) = 0$, then the velocity becomes: | |||
| $$v(t) = \frac{b t^2}{2}$$ | |||
| Likewise, integrate $v(t)$ with respect to time: | |||
| $$x(t)= \int v(t) \, dt = \frac{b}{2} \int t^2 \, dt$$ | |||
| From where the coordinate from time, considering the initial conditions: | |||
| $$\boxed{x(t)=\frac{bt^3}{6}}$$ | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result, like this:] | $$ \omega1=\omega3=\frac{\omega}{3}; \omega2=-\frac{\omega}{3}$$ | ||
| __Example Answer__: | |||
| $$ |
|||