# Finding the minimum radius of the pivoted disc

Here is a question based on Simple Harmonic Motion that I tackled just now. However I think I am having an approach to tackle this but I am not sure about it.

Ouestion: A uniform disc of radius $R$ is pivoted about point $P$ such that it is free to oscillate in the vertical plane. Distance between the pivot and centre of disc $x$, such that the time period of oscillation is minimum.

So till yet I know that for a circular disc like this one, we have time period, $$T=2\pi\sqrt{\frac{I}{\kappa}}$$ where $I$ is moment of inertia of the disc and $\kappa$ is the torsional constant.

So after substituting the values knowing that $\kappa=\frac{\tau}{\theta}$ and $I=mr^2$ I have,

$$T=2\pi\sqrt{\frac{mr^2\theta}{\tau}}$$

Now next to this I am thinking of finding the minima of this function so that I can get the least value of r. However I am still not sure that this will help me in anyway.

Also I want to know that if we can differentiate this function to get minimum value of $r$ then do we really need to substitute $\tau$ as $\tau=r\cdot F$.

• The formula for the time period is actually a solution to the pendulum equation, see The Simple Pendulum Thus, it might not be trivial to adjust it to other cases. I suggest you set up the equations of motion for your specific case, and then solve these equations, and come up with your own formula for the time period. Hint: take care of the moment of inertia, it is different as the rotation point is not the CoG. – ROIMaison Jul 2 '14 at 16:56
• Do you mean equations of rotational motion? Do you really think it would help me? I am asking this because this question is given in the exercise of Simple Harmonic Motion. – Saharsh Jul 2 '14 at 16:59
• Harmonic motion simply implies that the solution is periodic. Actually, now that I see it, the formula for the Time period is not coming from the simple pendulum, but the principle is the same. Compare a normal spring mass system but then with theta instead of x, and tau instead of k. I think for educational purposes it's always better to start from elementary equations (e.g. EoM, mass balance), and then work your way to the answer, rather than simply substituting values. This way you can find out if the assumptions made are still valid. – ROIMaison Jul 2 '14 at 17:26

## 1 Answer

If we write the equation of rotation about the pivoted point, then moment of inertia of the disk about the pivoted point $P$ dose not equal to the $\frac{1}{2}mR^2$, i.e. from the parallel axis theorem the moment of inertia of the disk about point $P$ will be: $$I_P=I_{cm}+mx^2=\frac{1}{2}mR^2+mx^2=\frac{1}{2}m(R^2+2x^2)$$ On the other hand,the torsional constant is: $$\kappa=\frac{\tau}{\theta}=\frac{mgx\sin{\theta}}{\theta}\approx mgx$$ for small $\theta$. So we have the time period as follow: $$T=2\pi\sqrt{\frac{I_P}{\kappa}}=2\pi\sqrt{\frac{m(R^2+2x^2)}{2mgx}}=2\pi\sqrt{\frac{(R^2+2x^2)}{2gx}} \tag{1}$$ If your design parameter is $x$, for minimizing the period you must solve the necessary condition for extremum, i.e.: $$\frac{\partial T}{\partial x}=0$$ and then checking the suﬃcient conditions for minima, i.e.: $$\frac{\partial^2T}{\partial x^2}\geq0$$ in the extremum point.

But if your design parameter is $R$ and you want to find a radius that minimize the period, it's very clear from equation $(1)$ that the radius must be zero i.e. the all disk mass must be placed at the center of the disk (in this case you don't have a rigid body, you have a point mass).