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en/3.5.42.md
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| + | ### Statement | ||
| + | |||
| + | $3.5.42.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | From $F_x=F_0\sin(\omega t)$ and $F_y=F_0\cos(\omega t)$ and the initial conditions $v_x=v_y=0$ and $x=y=0$ at , we have | ||
| + | |||
| + | \begin{eqnarray} | ||
| + | a_x&=&\frac{F_0}{m}\sin(\omega t),\\ | ||
| + | a_y&=&\frac{F_0}{m}\cos(\omega t),\\ | ||
| + | v_x&=&\frac{F_0}{m\omega}-\frac{F_0}{m\omega}\cos(\omega t)\\ | ||
| + | v_y&=&\frac{F_0}{m\omega}\sin(\omega t)\\ | ||
| + | x&=&\frac{F_0}{m\omega}t-\frac{F_0}{m\omega^2}\sin(\omega t)\\ | ||
| + | y&=&\frac{F_0}{m\omega^2}-\frac{F_0}{m\omega^2}\cos(\omega t) | ||
| + | \end{eqnarray} | ||
| + | |||
| + | The last two equations are parametric equations of a cycloid, which is the trajectory of a point on the rim of a disk of radius $R=F_0/(m\omega^2)$ that rolls without slipping on the $x$-axis (shown in blue). Its average velocity is $F_0/(m\omega)$ in the positive $x$ direction. If the initial velocity is $v_0=F_0/(m\omega)$ in the negative $x$ direction, which is perpendicular to the initial force, then the time-dependent force will keep it in a circular orbit (shown in red) as the equation $F_0=mv_0^2/R$ for centripetal force is satisfied. | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $3.5.42.$ [Insert the problem statement] | |||
| ### Solution | |||
| From $F_x=F_0\sin(\omega t)$ and $F_y=F_0\cos(\omega t)$ and the initial conditions $v_x=v_y=0$ and $x=y=0$ at , we have | |||
| \begin{eqnarray} | |||
| a_x&=&\frac{F_0}{m}\sin(\omega t),\\ | |||
| a_y&=&\frac{F_0}{m}\cos(\omega t),\\ | |||
| v_x&=&\frac{F_0}{m\omega}-\frac{F_0}{m\omega}\cos(\omega t)\\ | |||
| v_y&=&\frac{F_0}{m\omega}\sin(\omega t)\\ | |||
| x&=&\frac{F_0}{m\omega}t-\frac{F_0}{m\omega^2}\sin(\omega t)\\ | |||
| y&=&\frac{F_0}{m\omega^2}-\frac{F_0}{m\omega^2}\cos(\omega t) | |||
| \end{eqnarray} | |||
| The last two equations are parametric equations of a cycloid, which is the trajectory of a point on the rim of a disk of radius $R=F_0/(m\omega^2)$ that rolls without slipping on the $x$-axis (shown in blue). Its average velocity is $F_0/(m\omega)$ in the positive $x$ direction. If the initial velocity is $v_0=F_0/(m\omega)$ in the negative $x$ direction, which is perpendicular to the initial force, then the time-dependent force will keep it in a circular orbit (shown in red) as the equation $F_0=mv_0^2/R$ for centripetal force is satisfied. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||