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The centrifugal force acting on a revolving particle with negligible size is $\frac{mv^2}{r}$. What if the size is not negligible? Say we are talking about a large homogeneous circular disc, so its radius is not negligible when compared to the radius of its revolution.

For the sake of simplicity, assume the disc is perpendicular to the plane of revolution. For each point on the disc, the centrifugal force acting on it is the same, namely $\frac{mv^2}{r}$. So, if I am correct, we have to integrate these small contributions of each small point on the disc over the total disc. I have ignored the tangential velocity, since by symmetry the centrifugal force acting on a point is equal to that acting diametrically opposite to it, and they cancel each other out. (Do they? that's what I think).

I will calculate the centrifugal acceleration to get rid of the $m$ term in the integration. After the result is obtained, I shall multiply it by the total mass later and get the centrifugal force. To do this, it suffices to simply calculate the following double integral:

$$ \int_{0}^{r} \int_{0}^{2\pi} \frac{v^2}{r} d \theta dR $$

My question is: how do we solve the scenario when the plane of the disc is not perpendicular to the plane of revolution? Say it makes an angle $\theta$ with the plane of revolution (to those who haven't read the above text, we are not talking about a particle, we are integrating over the whole circular disc). Then, the contributions of the tangential velocities will not cancel out. That is freaking me out.

I have also drawn a picture, which basically takes a small shell of radius $m$ and takes an angle $X$, then tries to find the angle of the tangent to the point $(m, X)$ (in polar coordinates), but I cannot go further because it gets really complicated. Any help will be appreciated.

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  • $\begingroup$ Hint, look up the parallel axis theorem and mass moment of inertia $\endgroup$ Jul 30, 2013 at 20:27
  • $\begingroup$ What is your end goal? Total force, Internal forces due to differential acceleration, Kinematics, Full Dynamics? $\endgroup$ Jul 30, 2013 at 20:29
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    $\begingroup$ Generally helpful may be the Wikipedia articles on "Rigid body dynamics" together with the moment of inertia tensor of a cylinder (which is just a disk of finite thickness, of course) given there. If a disk is rotating around an axis (through its center) which is at an (less than 90°) angle to the symmetry axis, then there is a torque acting to reduce this angle, trying to turn the symmetry axis closer to the rotation axis. $\endgroup$
    – user12262
    Jul 30, 2013 at 20:30
  • $\begingroup$ What do you mean by "plane of revolution"? Do you mean the plane perpendicular to the axis of revolution? $\endgroup$
    – Dan
    Jul 31, 2013 at 7:25

1 Answer 1

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Bear with me for this image-less description of the situation. I'm calling the center of the disk $D$ and the center of revolution $O$.

Consider a diameter on this disk, whose projection on the plane of revolution lies along the radius vector of the center of mass of the disk.Now consider a point $R$ on this diameter at a distance of $r$ from $D$.

Take two points lying equidistant from $R$, and on opposite sides of $R$, on a chord passing through $R$, perpendicular to the diameter. The sum of the acceleration vectors of these two points will be along $\vec {OR}$(by symmetry, very similar to the diametrically opposite case). This will be true for all pairs of points on that chord. So we can say that the net acceleration of each chord or rod element will lie along the radius vector of the rod's center.

You can easily obtain this acceleration as a function of $r$ using integration. I'm calling it $a(r)$(I'm leaving most of the math for you, sorry!).

The integral you get is $$\int a(r) \frac{\vec{OR}}{|\vec{OR}|}$$ $$= \int a(r) \frac{\vec{OD}+\vec{DR}}{\sqrt{OD^2+r^2+2.OD.r\cos\theta}}$$

Again I leave the math on you, Hope this much helps!

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