Concept
The period of a physical pendulum of length $L$ and pivoted about its end is:
$$T = 2 \pi \sqrt{\frac{I}{mgd}}$$
where $I =$ moment of inertia about the pivot and $d =$ distance from the pivot to the center of mass. For the hoop, $I = \frac{1}{2}mD^2$ and $d = \frac{D}{2}$. Therefore,
$$T = 2 \pi \sqrt{\frac{D}{g}} $$
Thus, a simple pendulum of length $D$ will have the same period.
Procedure
- Verify that the pendulum’s length is 2/3rds the length of the bar. The bob should align with the red tape on the hoop.
- Slowly displace the pendulum bob and hoop to one side and release them at exactly the same time.
- Notice that the pendulum bob and the hoop oscillate at the same frequency.
Equipment
- Hoop (0.5m)
- Physical Pendulum Rod
- Large Rod Clamp
- Pendulum Bob
- Large Rod Stand