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A bead is moving along the spoke of a wheel at constant speed u , the wheel rotates with uniform angular velocity w radians per second about an axis fixed in space , at t=0 the bead is in the x axis at the origin, find the velocity in t in POLAR coordinates

In the text book it says : first we have $r(t)= ut$? doesn't that mean $v = u$ is constant?

How did they obtain this result (it changes in time) if $v$ is constant? $v = u (r) + utw(\theta)$ (($r$) and ($\theta$) are the bases of the polar plane)

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You are missing the tangential component.

$$u_t=\omega r = \omega u t $$

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  • $\begingroup$ the bead has the same speed as the wheel? $\endgroup$ – user31731 Jun 18 '14 at 20:51
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    $\begingroup$ The spoke moves with the wheel. $\endgroup$ – Bernhard Jun 18 '14 at 20:52
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In polar coordinates, the speed is $\vec v=\dot r\,\vec u_r+r\omega\vec u_t$ because the rotation rate $\omega$ is imposed by the system. ($\dot r$ is the time derivative of $r$). The acceleration is $(\ddot r-r\omega^2)\vec u_r+2\dot r\,\omega\vec u_t.$ As the only force experienced by the bead is orthogonal to the spoke (it is the reaction of the spoke on the bead that keeps it inside) we must have $\ddot r-r\omega^2=0$. Consequently the solution to this differential equation with $r(0)=0$ is $r(t)=0$ and the speed of the bead is equal to $0$ at all times.

However if the bead is at a distance $r_0$ a $t=0$, we have $r(t)=r_0\cosh(\omega t)$ and then $$ \vec v(t)=r_0\omega\sinh(\omega t)\vec u_r+r_0\omega\cosh(\omega t)\vec u_t.$$ The force acting on the bead to maintain it in the spoke is $\vec F=2mr_0\omega^2\sinh(\omega t)\vec u_t$, where $m$ is the bead's mass.

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