| @@ -1,44 +1,71 @@ |
| | | ### Statement |
| | | |
| | − | $2.4.12.$ [Insert problem description here] |
| | + | $2.4.12.$ Two identical uniform cylinders of radius \( R \) each are placed on top of each other next to a wall as shown. After a disturbance, the bottom cylinder slightly moves to the right and the system comes into motion. Find the maximum subsequent speed of the bottom cylinder. Neglect friction between all surfaces. |
| | | |
| | − | __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 |
| | | |
| | + | Consider the moment when the center of the top cylinder has moved down a distance \( x_T \) and the center of the bottom cylinder has moved to the right a distance \( x_B \). At this moment, the cylinders are moving with speeds \( v_T \) and \( v_B \), respectively. From the geometry: |
| | | |
| | − | ### Solution |
| | + | $$ |
| | + | x_T = 2R(1 - \cos \theta). |
| | + | $$ |
| | | |
| | − | [Your solution should be placed here] |
| | + | Energy conservation gives: |
| | | |
| | − | __Example Solution__: |
| | − | The acceleration of the body defined by |
| | + | $$ |
| | + | \frac{1}{2}m(v_B^2 + v_T^2) = mgx_T = 2mgR(1 - \cos \theta). |
| | + | $$ |
| | | |
| | − | $$a(t) = bt$$ |
| | + | From the geometry of the system: |
| | | |
| | − | We know that acceleration is the time derivative of velocity: |
| | + | $$ |
| | + | x_B^2 + (2R - x_T)^2 = (2R)^2. |
| | + | $$ |
| | | |
| | − | $$a(t) = \frac{d v(t)}{d t}$$ |
| | + | Taking the derivative with respect to time yields: |
| | | |
| | − | To find the velocity $v(t)$, we integrate $a(t)$ with respect to time: |
| | + | $$ |
| | + | x_B v_B - (2R - x_T)v_T = 0, |
| | + | $$ |
| | | |
| | − | $$v(t) = \int a(t) \, dt = \int b t \, dt$$ |
| | + | which implies: |
| | | |
| | − | If the initial velocity is $v(0) = 0$, then the velocity becomes: |
| | + | $$ |
| | + | v_T = \frac{v_B x_B}{2R - x_T} = v_B \tan \theta. |
| | + | $$ |
| | | |
| | − | $$v(t) = \frac{b t^2}{2}$$ |
| | + | Substitute \( v_T \) into the energy conservation equation: |
| | | |
| | − | Likewise, integrate $v(t)$ with respect to time: |
| | + | $$ |
| | + | v_B^2 = \frac{4gR(1 - \cos \theta)}{1 + \tan^2 \theta}. |
| | + | $$ |
| | | |
| | − | $$x(t)= \int v(t) \, dt = \frac{b}{2} \int t^2 \, dt$$ |
| | + | Simplify using the trigonometric identity \( 1 + \tan^2 \theta = \sec^2 \theta \): |
| | | |
| | − | From where the coordinate from time, considering the initial conditions: |
| | + | $$ |
| | + | v_B^2 = 4gR(1 - \cos \theta) \cos^2 \theta. |
| | + | $$ |
| | | |
| | − | $$\boxed{x(t)=\frac{bt^3}{6}}$$ |
| | + | Find the maximum speed by setting the derivative with respect to \( \theta \) equal to zero. This occurs at: |
| | | |
| | − | #### Answer |
| | + | $$ |
| | + | \cos \theta_m = \frac{2}{3}. |
| | + | $$ |
| | | |
| | − | [Insert a concise answer or boxed result, like this:] |
| | + | Substitute \( \cos \theta_m \) into the equation for \( v_B \): |
| | | |
| | + | $$ |
| | + | v_B = \sqrt{\frac{16}{27}gR}. |
| | + | $$ |
| | | |
| | − | __Example Answer__: |
| | − | $$ x(t)=\frac{bt^3}{6} $$ |
| | + | The maximum speed of the bottom cylinder is: |
| | + | |
| | + | $$ |
| | + | \boxed{\sqrt{\frac{16}{27}gR}}. |
| | + | $$ |
| | + | |
| | + | #### Answer |
| | + | |
| | + | $$ |
| | + | v_B = \sqrt{\frac{16}{27}gR} |
| | + | $$ |
