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| <title>A board of mass m lies on two rollers rotating with high speed towards each other. The distance between the axes of the rollers is L, the coefficient of friction when the board slides on the roller is \mu. Find the frequency of longitudinal oscillations of the board.</title> | | <title>A board of mass m lies on two rollers rotating with high speed towards each other. The distance between the axes of the rollers is L, the coefficient of friction when the board slides on the roller is \mu. Find the frequency of longitudinal oscillations of the board.</title> |
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| <h3 id="back-link"><a href="../../#3.2">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#3.2">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $3.2.14.$ A board of mass $m$ lies on two rollers rotating with high speed towards each other. The distance between the axes of the rollers is $L$, the coefficient of friction when the board slides on the roller is $\mu$. Find the frequency of longitudinal oscillations of the board. | | $3.2.14.$ A board of mass $m$ lies on two rollers rotating with high speed towards each other. The distance between the axes of the rollers is $L$, the coefficient of friction when the board slides on the roller is $\mu$. Find the frequency of longitudinal oscillations of the board. |
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| For problem $3.2.14$ | | For problem $3.2.14$ |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
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| <figcaption> | | <figcaption> |
| Forces acting on the board | | Forces acting on the board |
| </figcaption> | | </figcaption> |
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| <p> | | <p> |
| Newton's second law for the horizontal axis | | Newton's second law for the horizontal axis |
| $$ ma=F_{fr1}-F_{fr2} $$ | | $$ ma=F_{fr1}-F_{fr2} $$ |
| Equilibrium condition for the vertical axis | | Equilibrium condition for the vertical axis |
| $$ mg=N_1+N_2\quad(1) $$ | | $$ mg=N_1+N_2\quad(1) $$ |
| In the equilibrium condition, the sum of the moment of external forces must be equal to zero. | | In the equilibrium condition, the sum of the moment of external forces must be equal to zero. |
| $$ N_1\left(\frac{l}{2}-x\right)-N_2\left(\frac{l}{2}+x\right)=0 $$ | | $$ N_1\left(\frac{l}{2}-x\right)-N_2\left(\frac{l}{2}+x\right)=0 $$ |
| Where does the reaction force on the second support come from? | | Where does the reaction force on the second support come from? |
| $$ N_2=N_1\left(\frac{L/2-x}{L/2+x}\right) $$ | | $$ N_2=N_1\left(\frac{L/2-x}{L/2+x}\right) $$ |
| Substitute into $(1)$ | | Substitute into $(1)$ |
| $$ mg=N_1\left(1+\frac{L/2-x}{L/2+x}\right) $$ | | $$ mg=N_1\left(1+\frac{L/2-x}{L/2+x}\right) $$ |
| | | |
| $$ N_1=mg\left(\frac{L/2+x}{L}\right);\quad N_2=mg\left(\frac{L/2-x}{L}\right) $$ | | $$ N_1=mg\left(\frac{L/2+x}{L}\right);\quad N_2=mg\left(\frac{L/2-x}{L}\right) $$ |
| Newton's second law for the horizontal axis | | Newton's second law for the horizontal axis |
| $$ m\ddot{x}=\mu N_2-\mu N_1= - \mu mg \frac{2x}{L}\quad(2) $$ | | $$ m\ddot{x}=\mu N_2-\mu N_1= - \mu mg \frac{2x}{L}\quad(2) $$ |
| We transform the obtained expression and obtain the equation of harmonic oscillations | | We transform the obtained expression and obtain the equation of harmonic oscillations |
| $$ \ddot{x}(t)+\frac{2\mu g}{l}x(t)=0 $$ | | $$ \ddot{x}(t)+\frac{2\mu g}{l}x(t)=0 $$ |
| Where does the angular frequency of oscillations come from? | | Where does the angular frequency of oscillations come from? |
| $$ \boxed{\omega=\sqrt{\frac{2\mu g}{l}}} $$ | | $$ \boxed{\omega=\sqrt{\frac{2\mu g}{l}}} $$ |
| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| Dzikan Mikita<br> | | Dzikan Mikita<br> |
| Aliaksandr Kanashenka<br> | | Aliaksandr Kanashenka<br> |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$\omega=\sqrt{\frac{2\mu g}{l}}$$ | | $$\omega=\sqrt{\frac{2\mu g}{l}}$$ |
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