# Thread: find moments of inertia and its period

1. ## find moments of inertia and its period

2. For part (i), are you familiar with the parallel axis theorem? As to part (ii), a circular hole in a circular disk can be treated as the disk without the hole plus a disk corresponding to the hole with negative mass equal to the subtracted mass of the hole. You should also be familiar with the physical pendulum.

--Kevin C.

3. ## then?

Originally Posted by TwistedOne151
For part (i), are you familiar with the parallel axis theorem? As to part (ii), a circular hole in a circular disk can be treated as the disk without the hole plus a disk corresponding to the hole with negative mass equal to the subtracted mass of the hole. You should also be familiar with the physical pendulum.

--Kevin C.

i got :
I= (m_a * a^2 + 2 m_c * c^2 - m_b * b^2) /2

then how can we get the period?

4. Originally Posted by szpengchao
i got :
I= (m_a * a^2 + 2 m_c * c^2 - m_b * b^2) /2

then how can we get the period?
Where are you getting an $m_c$ from? The moment of inertial of the main disc is
$I_a = \frac{1}{2}Ma^2 + Mc^2 = \frac{1}{2}\rho (\pi a^2 - \pi b^2)(a^2 + c^2)$
and the "moment of inertia" of the hole is
$I_b = \frac{1}{2}mb^2 = \frac{1}{2} \rho \pi b^2 \cdot b^2$
where $\rho$ is the density of the material.

So
$I_{tot} = \frac{1}{2}\rho (\pi a^2 - \pi b^2)(a^2 + c^2) - \frac{1}{2} \rho \pi b^2 \cdot b^2$

To find the period for small oscillations use
$T = 2 \pi \sqrt{\frac{I_{tot}}{MgL_{cm}}} = 2 \pi \sqrt{\frac{I_{tot}}{(\pi a^2 - \pi b^2)gL_{cm}}}$
where $L_{cm}$ is the distance from the center of mass to the axis of rotation. I leave it to you to find $L_{cm}$.

-Dan