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Let us consider the following setup:

We have a rotating disk of radius $r$ and with angular velocity $\omega$. At the periphery of the disk, we have a projectile with some mass that's sent with a velocity $v$ in some direction $\theta$, my question then becomes, is it possible to find this value of $\theta$ such that the projectile passes the center of the disk?

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So, my strategy for solving this problem is to somehow insert an inertial system XYZ, and a coordinate system that rotates with angular velocity given above. Now, since we can easily express the velocity of the projectile on the periphery by $\vec{v} = \vec{v}_O + \vec{\omega} \times \vec{r} + \vec{v}_{rel}$ where $O$ is the center of the disk. From this, we can relate the XYZ coordinates with the xyz coordinates using a rotation transformation, hence we can express our velocity $\vec{v}$ in XYZ, which then is easier to integrate since we don't have to bother with integrating basis vectors that change with time. After that, we can find the expression $X(t)$ by just considering the X - component of the velocity vector, and then solving for $X(t) = 0$ for some time $t$ given that $X(0) = R$, but this seems rather complicated. I wonder whether there's some more intuitive and easier way to solve the problem that won't involve taking integrals.

Thanks.

Tanamas
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  • Is the disk horizontal, vertical, or are we neglecting gravity – Abhiram Cherukupalli May 28 '22 at 18:16
  • Yes, the disk is horizontal and gravity is neglected. – Tanamas May 28 '22 at 18:18
  • So the ball is thrown in the plane of the disk? – Abhiram Cherukupalli May 28 '22 at 18:18
  • Does the ball have mass? if so the problem would become much morecomplicated – Abhiram Cherukupalli May 28 '22 at 18:28
  • We can assume it has no mass then to make it as simple as possible, and yes its thrown in the plane of the disk – Tanamas May 28 '22 at 18:33
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    Ok I posted an answer check – Abhiram Cherukupalli May 28 '22 at 18:33
  • I have a comment that is not a comment to this question specifically, but to questions about uniform rotation generally. It appears that for the human brain ongoing rotation is an Achilles heel of understanding. Comparison: navigation using maps printed on paper. When using a physical map like that: while it's more convenient to align the map with the geographic north/south line, it's not a necessity. If the angle is constant we can reorient in our head. But I see over and over that with ongoing rotation the human brain is very prone to losing track. I find that puzzling, and very interesting. – Cleonis May 28 '22 at 19:31

2 Answers2

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So, let us see it in the frame of the disk. What do we want? We want there to be no net movement in the tangential direction, we want it to be radial. We need the net relative motion in the tangential direction to me do. Let's say v makes an angle $\theta$ with the radius, in the frame of the rotating disk there is a tangential $\omega r$, the tangential component of v must cancel it.

So, $$\omega r=v \sin \theta \implies \theta=\arcsin \left(\frac{\omega r}{v}\right)$$

Abhiram Cherukupalli
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  • Oh I see, very clever to think of that. That matches exactly what I sought. Thank you – Tanamas May 28 '22 at 18:36
  • This method works perfectly well even if the object has mass, since the object will still move in a straight line in an inertial frame. If it has no tangential velocity in the inertial frame then it will pass through the origin, and if it passes through the origin in the inertial frame it will also go through the origin in the rotating frame. – Michael Seifert May 28 '22 at 18:40
  • @MichaelSeifert Ok yes, I was only thinking about the rotating frame, yep the mass doesnt matter. Though the problem is a lot more interesting in the rotating frame – Abhiram Cherukupalli May 28 '22 at 18:42
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Any initial velocity vector with a tangential velocity equal to $-r_0 \omega$ and an inward radial component will eventually pass through the center. (This is because in an inertial frame, it will be traveling with a purely radial velocity.) If the initial magnitude of the velocity in the rotating frame is $v_0$, then the magnitude of the radial velocity will be $$ v_r = \sqrt{v^2 -r_0^2} $$ and so in the inertial frame the particle's position will simply be $$ X(t) = r_0 -v_r t $$ with $Y(t) = Z(t) = 0$. This can then be straightforwardly transformed into the rotating frame if desired.

Michael Seifert
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