Edits to “Statement”, “Solution”
en/2.4.12.md
+6 −6
| @@ -1,10 +1,10 @@ | |||
| ### Statement | |||
| − | $2.4.12.$ Two identical uniform cylinders of radius | ||
| + | $2.4.12.$ 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. | ||
| ### Solution | |||
| − | Consider the moment when the center of the top cylinder has moved down a distance | ||
| + | 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: | ||
| $$ | |||
| x_T = 2R(1 - \cos \theta). | |||
| $$ | |||
| Energy conservation gives: | |||
| $$ | |||
| \frac{1}{2}m(v_B^2 + v_T^2) = mgx_T = 2mgR(1 - \cos \theta). | |||
| $$ | |||
| From the geometry of the system: | |||
| $$ | |||
| x_B^2 + (2R - x_T)^2 = (2R)^2. | |||
| $$ | |||
| Taking the derivative with respect to time yields: | |||
| $$ | |||
| x_B v_B - (2R - x_T)v_T = 0, | |||
| $$ | |||
| which implies: | |||
| $$ | |||
| @@ -34,25 +34,25 @@Solution | |||
| v_T = \frac{v_B x_B}{2R - x_T} = v_B \tan \theta. | |||
| $$ | |||
| − | Substitute | ||
| + | Substitute $ v_T $ into the energy conservation equation: | ||
| $$ | |||
| v_B^2 = \frac{4gR(1 - \cos \theta)}{1 + \tan^2 \theta}. | |||
| $$ | |||
| − | Simplify using the trigonometric identity | ||
| + | Simplify using the trigonometric identity $ 1 + \tan^2 \theta = \sec^2 \theta $: | ||
| $$ | |||
| v_B^2 = 4gR(1 - \cos \theta) \cos^2 \theta. | |||
| $$ | |||
| − | Find the maximum speed by setting the derivative with respect to | ||
| + | Find the maximum speed by setting the derivative with respect to $ \theta $ equal to zero. This occurs at: | ||
| $$ | |||
| \cos \theta_m = \frac{2}{3}. | |||
| $$ | |||
| − | Substitute | ||
| + | Substitute $ \cos \theta_m $ into the equation for $ v_B $: | ||
| $$ | |||
| v_B = \sqrt{\frac{16}{27}gR}. | |||
| $$ | |||
| The maximum speed of the bottom cylinder is: | |||
| $$ | |||
| \boxed{\sqrt{\frac{16}{27}gR}}. | |||
| $$ | |||
| #### Answer | |||
| $$ | |||
| v_B = \sqrt{\frac{16}{27}gR} | |||
| $$ | |||
| unchanged lines 13 | |||
| @@ -1,10 +1,10 @@ | |||
| ### Statement | ### Statement | ||
| $2.4.12.$ Two identical uniform cylinders of radius |
$2.4.12.$ 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. | ||
| ### Solution | ### Solution | ||
| Consider the moment when the center of the top cylinder has moved down a distance |
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: | ||
| $$ | $$ | ||
| x_T = 2R(1 - \cos \theta). | x_T = 2R(1 - \cos \theta). | ||
| $$ | $$ | ||
| Energy conservation gives: | Energy conservation gives: | ||
| $$ | $$ | ||
| \frac{1}{2}m(v_B^2 + v_T^2) = mgx_T = 2mgR(1 - \cos \theta). | \frac{1}{2}m(v_B^2 + v_T^2) = mgx_T = 2mgR(1 - \cos \theta). | ||
| $$ | $$ | ||
| From the geometry of the system: | From the geometry of the system: | ||
| $$ | $$ | ||
| x_B^2 + (2R - x_T)^2 = (2R)^2. | x_B^2 + (2R - x_T)^2 = (2R)^2. | ||
| $$ | $$ | ||
| Taking the derivative with respect to time yields: | Taking the derivative with respect to time yields: | ||
| $$ | $$ | ||
| x_B v_B - (2R - x_T)v_T = 0, | x_B v_B - (2R - x_T)v_T = 0, | ||
| $$ | $$ | ||
| which implies: | which implies: | ||
| $$ | $$ | ||
| @@ -34,25 +34,25 @@Solution | |||
| v_T = \frac{v_B x_B}{2R - x_T} = v_B \tan \theta. | v_T = \frac{v_B x_B}{2R - x_T} = v_B \tan \theta. | ||
| $$ | $$ | ||
| Substitute |
Substitute $ v_T $ into the energy conservation equation: | ||
| $$ | $$ | ||
| v_B^2 = \frac{4gR(1 - \cos \theta)}{1 + \tan^2 \theta}. | v_B^2 = \frac{4gR(1 - \cos \theta)}{1 + \tan^2 \theta}. | ||
| $$ | $$ | ||
| Simplify using the trigonometric identity |
Simplify using the trigonometric identity $ 1 + \tan^2 \theta = \sec^2 \theta $: | ||
| $$ | $$ | ||
| v_B^2 = 4gR(1 - \cos \theta) \cos^2 \theta. | v_B^2 = 4gR(1 - \cos \theta) \cos^2 \theta. | ||
| $$ | $$ | ||
| Find the maximum speed by setting the derivative with respect to |
Find the maximum speed by setting the derivative with respect to $ \theta $ equal to zero. This occurs at: | ||
| $$ | $$ | ||
| \cos \theta_m = \frac{2}{3}. | \cos \theta_m = \frac{2}{3}. | ||
| $$ | $$ | ||
| Substitute |
Substitute $ \cos \theta_m $ into the equation for $ v_B $: | ||
| $$ | $$ | ||
| v_B = \sqrt{\frac{16}{27}gR}. | v_B = \sqrt{\frac{16}{27}gR}. | ||
| $$ | $$ | ||
| The maximum speed of the bottom cylinder is: | The maximum speed of the bottom cylinder is: | ||
| $$ | $$ | ||
| \boxed{\sqrt{\frac{16}{27}gR}}. | \boxed{\sqrt{\frac{16}{27}gR}}. | ||
| $$ | $$ | ||
| #### Answer | #### Answer | ||
| $$ | $$ | ||
| v_B = \sqrt{\frac{16}{27}gR} | v_B = \sqrt{\frac{16}{27}gR} | ||
| $$ | $$ | ||
| unchanged lines 13 | |||