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I have this problem relating to a smooth horizontal disk spinning with angular speed $\omega$ with a thin barrier along the diameter of the disk splitting the two masses as shown. The masses are placed onto either side of the barrier at the edge of the disk and have initial speed $v_0$ towards the centre.

Working in the inertial frame centred at the middle of the disk, I have worked out that $\mathbf a=(\ddot{x}-\omega^2x)\hat{\textbf i}+2\omega\dot{x}\hat{\textbf j}$ where $\hat{\textbf i},\hat{\textbf j}$ are vectors that rotate in the rotating frame of reference attached to the centre of the disk.

Here is the bit I am confused about. The barrier exerts a normal force onto the masses and so we can say $\mathbf F=N\hat{\textbf j}$ and since $v_0$ is positive, we have that $2\omega\dot{x}=N>0$ leading to the lower mass in the picture to fly off of the disk. However I cannot get a grasp of how this works physically. I also believed that there would be other forces on the point masses so could this be clarified too?

Image of Rotating Disk

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  • $\begingroup$ From your statement about the force exerted by the barrier acting in a constant direction (i.e. $\hat{\textbf j}$), i'm led to believe that the barrier is not fixed to the disc and doesn't rotate. In this case the disc will keep pushing the upper mass onto the barrier via friction and will push the lower mass away (if there's no friction then the rotating disc is irrelevant). the kinetic frictional force on the masses will be $\mathbf F=-\mu W \hat{\textbf v}$ where $W$ is the weight of the mass and (1/2) $\endgroup$ – alex May 8 '17 at 10:33
  • $\begingroup$ $\hat{\textbf v}$ is the unit vector in the direction of the relative velocity of the mass with respect to the disc.Try using this equation to find out their trajectories rather than the one you have (I do not understand how you derived it but anyway..). For the upper mass you just need to worry about the $\hat{\textbf i}$ component as the barrier will counter any force against it in the $\hat{\textbf j}$ direction.(2/2) $\endgroup$ – alex May 8 '17 at 10:34
  • $\begingroup$ Would upvote if I could...the first comment makes more sense - the barrier being fixed seems blatant now! And well I could solve the equation in the $\hat{\textbf i}$ direction and get that $x=Ae^{\omega t}+Be^{-\omega t}$ and work from there which is a bit messy but works $\endgroup$ – user258521 May 8 '17 at 10:47
  • $\begingroup$ I think the form of the solution may be a little more complicated than that but i can post the actual differential equation we get as the answer if you wish. you can search the web for a solution later. $\endgroup$ – alex May 8 '17 at 11:07
  • $\begingroup$ We know that in an inertial frame $(\frac{d \mathbf r}{dt})^I=(\frac{d \mathbf r}{dt})^R+\omega \times A$ and so using this relation I derived the acceleration formula where Frame I is inertial and Frame R is the rotating one. $\endgroup$ – user258521 May 8 '17 at 11:45
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The force acting on any of the masses is given by $$ \mathbf{F} = -\mu W \hat{\textbf v}$$ where $$\hat{\textbf v}=\frac{ \textbf v_{mass}-\textbf v_{disc}}{|\textbf v_{mass}-\textbf v_{disc}|} $$ $$\hspace{1.8cm}=\frac{v_x\hat{\textbf i}+v_y\hat{\textbf j}-\boldsymbol{\omega}\times{\textbf r}}{\sqrt{v_x^2+(v_y-\boldsymbol{\omega} \times{\textbf r})^2}}$$ where the symbols have their usual meanings. ${\textbf r}$ may be further reduced to $\frac{x \hat{\textbf{i}} +y \hat{\textbf{j}}}{\sqrt{x^2+y^2}}$. In the case of the upper mass, the barrier constraints its motion solely to solely the $x$ direction and we get $v_y=0$ and $\textbf r = x$ (considering the center of the disc as the origin). Therefore ignoring the $\hat{\textbf j}$ component our equation becomes $$\mathbf{F} = -\frac{\mu W v_x}{\sqrt{v_x^2+(\omega x)^2}} $$ which can be written as $$m \frac{d^2x}{dt^2} = -\frac{\mu W v_x}{\sqrt{v_x^2+(\omega x)^2}} $$ writing out the expression for weight, $$m \frac{d^2x}{dt^2} = -\frac{\mu mg v_x}{\sqrt{v_x^2+(\omega x)^2}} $$ $$v_x \frac{dv_x}{dx} = -\frac{\mu g v_x}{\sqrt{v_x^2+(\omega x)^2}} $$ (from now on i'll just write $v_x$ as $v$) cancelling $v$ on both sides: $$\frac{dv}{dx} = -\frac{\mu g }{\sqrt{v^2+(\omega x)^2}} $$ or $$\sqrt{v^2+(\omega x)^2}\frac{dv}{dx}+\mu g =0 $$ This nonlinear equation describes the motion of of the upper mass. Either term may be omitted as an approximation depending on the problem which then makes the equation readily solvable.

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