Edits to “Solution”, “Answer”
ru/1.5.2.md
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| − | КТО СПИСЫВЕТ ТОТ ВРАГ АЛЛАХА | ||
| + | $1.5.2.$ The angular velocity of the coil is $\omega$, the radius of the inner cylinder is $r$, and the radius of the outer cylinders is $R$. What are the velocities of the coil and load axis relative to the ground? | ||
| + | |||
| + |  | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | Consider the instantaneous axis of rotation passing through the point $O_1$, then the speed of the spool axis is | ||
| + | |||
| + | $$ | ||
| + | \fbox{$v_\text{reel} = \omega R$} | ||
| + | $$ | ||
| + | |||
| + | To find the speed of the load, consider the speed point $O$ as a superposition of the velocities of translational motion with the speed of the wheel center and the speed of rotation point $O$ relative to the wheel center. | ||
| + | |||
| + | $$ | ||
| + | \vec{v_O} = \vec{v}_{in} + \vec{v}_{out} | ||
| + | $$ | ||
| + | |||
| + | $$ | ||
| + | v_O = v_{out} – v_{in} = \omega R - \omega r = \omega\cdot (R - r) | ||
| + | $$ | ||
| + | |||
| + | Any point of the thread, due to its inextensibility, has the same speed, therefore, the speed of the load | ||
| + | |||
| + | $$ | ||
| + | \fbox{$v_\text{rope} = v_O = \omega\cdot (R - r)$} | ||
| + | $$ | ||
| + | |||
| + | #### Answer | ||
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| КТО СПИСЫВЕТ ТОТ ВРАГ АЛЛАХА | $1.5.2.$ The angular velocity of the coil is $\omega$, the radius of the inner cylinder is $r$, and the radius of the outer cylinders is $R$. What are the velocities of the coil and load axis relative to the ground? | ||
|  | |||
| ### Solution | |||
| Consider the instantaneous axis of rotation passing through the point $O_1$, then the speed of the spool axis is | |||
| $$ | |||
| \fbox{$v_\text{reel} = \omega R$} | |||
| $$ | |||
| To find the speed of the load, consider the speed point $O$ as a superposition of the velocities of translational motion with the speed of the wheel center and the speed of rotation point $O$ relative to the wheel center. | |||
| $$ | |||
| \vec{v_O} = \vec{v}_{in} + \vec{v}_{out} | |||
| $$ | |||
| $$ | |||
| v_O = v_{out} – v_{in} = \omega R - \omega r = \omega\cdot (R - r) | |||
| $$ | |||
| Any point of the thread, due to its inextensibility, has the same speed, therefore, the speed of the load | |||
| $$ | |||
| \fbox{$v_\text{rope} = v_O = \omega\cdot (R - r)$} | |||
| $$ | |||
| #### Answer | |||