Solve Pendulum Q: Rot Inertia, COM & Period Oscillation

[MJC8719]

The pendulum consists of a uniform disk with radius r = 10 cm and mass 900 g attached to a uniform rod with length L = 500 mm and mass 100 g.

(a) Calculate the rotational inertia of the pendulum about the pivot.
kgm2

(b) What is the distance between the pivot and the center of mass of the pendulum?
m

(c) Calculate the period of oscillation.
s

First, since this seems like a simple pendulum, I calculated the time of oscillation using the equation T = 2pi SqRt(l/g)
T = 2pi SwRt(0.5/9.81)
T = 1.4185 s which was incorrect.

Once I had the time of oscillation, I was plannning on finding the center of mass using the equation I = T^2MgR/4pi^2

Can someone please help me find the time for oscillation and tell me if my second thoughts are correct.

 

[Doc Al]

physical pendulum, not simple pendulum

First, since this seems like a simple pendulum, I calculated the time of oscillation using the equation T = 2pi SqRt(l/g)

It's not a simple pendulum, but what is called a physical pendulum. There's a reason that parts (a) & (b) come first--those quantities are needed to find the period of a physical pendulum.

 

[m2003]

Hello,

Thank you for your question. I am happy to provide a response to your inquiry.

Firstly, to calculate the rotational inertia of the pendulum about the pivot, we need to consider the individual moments of inertia of the disk and the rod. The moment of inertia of a disk is given by I = 1/2 * MR^2, where M is the mass of the disk and R is the radius. In this case, the moment of inertia of the disk would be 1/2 * 0.9 kg * (0.1 m)^2 = 0.0045 kgm^2. The moment of inertia of a rod rotating about its end is given by I = 1/3 * ML^2, where M is the mass of the rod and L is the length. In this case, the moment of inertia of the rod would be 1/3 * 0.1 kg * (0.5 m)^2 = 0.0083 kgm^2. Therefore, the total rotational inertia of the pendulum would be 0.0045 kgm^2 + 0.0083 kgm^2 = 0.0128 kgm^2.

Moving on to the distance between the pivot and the center of mass of the pendulum, we can use the equation for the center of mass of a system, which is given by xcm = (m1x1 + m2x2 + ... + mnxn) / (m1 + m2 + ... + mn), where m is the mass and x is the distance from the pivot. In this case, we have two masses, the disk and the rod, with their respective distances from the pivot. So, we can calculate the center of mass as follows:

xcm = (0.9 kg * 0.1 m + 0.1 kg * 0.25 m) / (0.9 kg + 0.1 kg) = 0.092 m

Therefore, the distance between the pivot and the center of mass of the pendulum is 0.092 m.

Moving on to the period of oscillation, the equation you have used, T = 2pi * SqRt(l/g), is the correct equation for a simple pendulum. However, in this case, we have a compound pendulum, which is slightly different. The equation for