A pendulum consists of a uniform disk with radius r = 10 cm and mass 545 g attached to a uniform rod with length L = 500 mm and mass 300 g. (figure: one end of the rod attach to a pivot and the other end attach to the center of the disk)
(a) Calculate the rotational inertia of the pendulum about the pivot.
(b) What is the distance between the pivot and the center of mass of the pendulum?
(c) Calculate the period of oscillation.
Plz help. Thanks.
Let me analyze your thinking. This is straightforward.
I thought I can find (a) by adding the rotational inertia of the rod and the disk.
I = 1/3 ML^2 + 1/2 mr^2
I wander in this case, what's the rotational inertia of the disk? Is that
I=1/2 mr^2?
You're correct in thinking that the rotational inertia of the pendulum can be found by adding the rotational inertia of the rod and the disk.
To calculate the rotational inertia of the disk, you can use the formula:
I_disk = (1/2) * m * r^2
where m is the mass of the disk and r is the radius of the disk. In this case, the mass of the disk is given as 545 g (or 0.545 kg) and the radius is given as 10 cm (or 0.1 m).
Therefore, the rotational inertia of the disk (I_disk) can be calculated as:
I_disk = (1/2) * 0.545 kg * (0.1 m)^2
Now you can calculate the rotational inertia of the rod by using the formula:
I_rod = (1/3) * M * L^2
where M is the mass of the rod and L is the length of the rod. In this case, the mass of the rod is given as 300 g (or 0.3 kg) and the length is given as 500 mm (or 0.5 m).
Thus, the rotational inertia of the rod (I_rod) can be calculated as:
I_rod = (1/3) * 0.3 kg * (0.5 m)^2
After finding both I_disk and I_rod, you can add them together to find the total rotational inertia of the pendulum about the pivot:
I_total = I_disk + I_rod
Now, you can proceed to calculate (b) and (c) based on the information given.