Time period of oscillation of a physical pendulum and spinning disk

[Avi Nandi]

Homework Statement

Find the period of a pendulum consisting of a disk of mass M and radius R fixed to the end of the rod of length l and mass m. How does the period change if the disk is mounted to the rod by a friction less bearing so that it is perfectly free to spin? The centre of the disk is attached to the rod.

The Attempt at a Solution

I can find the position of the centre of mass of the system, the torque due to gravity and the moment of inertia of the system about the pivot. From this quantities i shall form the equation of motion and thus i can find the time period.

Now i see no reason why the period will change if the disk is free to spin. Firstly i think the disk will not spin since there is no torque acting on it. Both gravity and the force exerted by the rod pass through the bearing. If it also spins the position of centre of mass doesn't change. The moment of inertia of the system too remains unchanged.

Am I correct since I am feeling that I missed something?

 

[TSny]

You are correct that the disk will not rotate if the disk is free to spin.

But, how would you express the disk's angular momentum about the pivot of the pendulum
(a) if the disk rotates with the pendulum?
(b) if the disk does not rotate?

 

[Avi Nandi]

If the spin velocity is constant then the time period will be the same. But if the spin has some acceleration then the oscillation may be damped or lose it's oscillatory character. Am i correct?

 

[TSny]

You have two situations. The first is where the disk and rod form one rigid body so that the disk rotates with the pendulum as the pendulum swings. The second is where the disk does not rotate about its center as the pendulum swings. See picture. The red dot is just a reference mark painted on the disk.

 

[Nickie]

I would like to provide a thorough response to this content.

Firstly, to find the period of oscillation of a physical pendulum, we can use the equation T = 2π√(I/mgd), where T is the period, I is the moment of inertia about the pivot, m is the mass of the pendulum, g is the acceleration due to gravity, and d is the distance from the pivot to the center of mass.

In the given scenario, we have a disk of mass M and radius R fixed to the end of a rod of length l and mass m. The moment of inertia of this system about the pivot can be calculated as I = ml^2 + 1/2MR^2. Substituting this in the equation for period, we get T = 2π√((ml^2 + 1/2MR^2)/mgd).

Now, if the disk is mounted to the rod by a frictionless bearing, it means that there is no external torque acting on the disk. This would result in the disk remaining in its initial orientation and not spinning. Therefore, the position of the center of mass and the moment of inertia would remain unchanged, leading to the same period of oscillation as before.

In conclusion, the period of oscillation of a physical pendulum with a disk attached to the end of the rod will not change if the disk is mounted to the rod by a frictionless bearing, as there is no external torque acting on the disk. However, if there is friction present in the bearing, it could affect the period of oscillation by causing the disk to spin and altering the moment of inertia of the system.