Find the period of a pendulum consisting of a disk of mass M

[LCSphysicist]

Homework Statement

Find the period of a pendulum consisting of a disk of mass M and
radius R fixed to the end of a rod of length I and mass m. How does
the period change if the disk is mounted to the rod by a frictionless bear·
ing so that it is perfectly free to spin? See figure above right

Relevant Equations

t = rf
Jcm = mr²/2

I did the first step pretty well, and the answers match, but i don't understand why in the second case the period changes.

 

[etotheipi]

I think this exact same question was asked a few weeks ago. Perhaps if you have a look you can find some replies there.

As for the second question, my recommendation would be to first write the angular momentum of the compound pendulum about the axis of rotation as the sum of that of the rod and the disk, knowing that the angular momentum of the disk is the angular momentum of its centre of mass plus that relative to its centre of mass.

Does the disk spin about its centre of mass in the frictionless case?

Edit: Whoops, here it is... https://www.physicsforums.com/threads/period-of-pendulum.988481/, name is familiar

 

[etotheipi]

I think it's worth me explaining a bit more of it considering that it's been a little while. In the previous thread you made the correct observation that if the disk was attached via a frictionless bearing, it would not spin about its centre of mass (since there is no torque to accelerate it about its centre of mass).

Now, the angular momentum of the whole configuration about the axle is $\vec{L} = \vec{L}_{\text{rod}} + \vec{L}_{\text{CM, disk}} + \vec{L}_{\text{disk w.r.t. CM}}$. Now we know that $\vec{L}_{\text{disk w.r.t. CM}} = \vec{0}$, and $\vec{L}_{\text{CM, disk}}$ is just the angular momentum if the entire mass of the disk were at a point at the position of the centre of mass, i.e. $(ml^2) \omega \hat{z}$.

So this new frictionless scenario is identical to a rigid body consisting of the rod with a point mass attached at the end. This means the new effective moment of inertia is lower.

 

[LCSphysicist]

I thought I understood, my senses deceive me hahaha
I am trying to distinguish the things here.
A consequence of putting the disk on a frictionless bearing is that now the system see the disk as a simple particle, so?
That's mean, don't matter the radius of the disk, if it is on a frictionless bearing, the system "see" just a particle?

 

[etotheipi]

You can think of it like that, yes. The actual moment of inertia of the system doesn't change, since the critical sum $\sum m_i r_i^2$ to the axis of rotation has not changed. But $\omega$ and $\alpha$ are not defined for a non-rigid body, so we can't use this approach anyway.

Instead, we need to consider angular momenta. With this approach, we have then shown that the dynamics of the system are equivalent to that of having a point mass attached to the end of the rod. And you can go further to define an effective moment of inertia, which has the property that if you plug it through the equations for rigid body rotation you will get out the right results, even though this body is not rigid.

 

[LCSphysicist]

Seeing this again, a doubt raised to me.
In real life, to study a pendulum consisting of a light bar (negligible mass) with a massive sphere on the end, it's enough to consider the moment of inertia as
"ML²" being L the length of the bar
or we need to consider "ML² + 2MR²/5", being R the radius of the sphere.?]

The second seems right to me.

 

[etotheipi]

Just so I'm thinking of the same thing, we have a massive rod with a sphere attached to the end, and we are calculating the MoI of the system about an axis through the other end of the rod.

The MoI of the rod about this axis is $\frac{1}{3}m_{rod}l^2$. The MoI of the sphere about its centre of mass is $\frac{2}{5}m_{sphere}r^2$; by the parallel axis theorem, the MoI of the sphere about the main axis is $\frac{2}{5}m_{sphere}r^2 + m_{sphere}l^2$.

That would make the total MoI $\frac{1}{3}m_{rod}l^2 + \frac{2}{5}m_{sphere}r^2 + m_{sphere}l^2$.

 

[haruspex]

consisting of a light bar

we have a massive rod

 

[LCSphysicist]

Just so I'm thinking of the same thing, we have a massive rod with a sphere attached to the end, and we are calculating the MoI of the system about an axis through the other end of the rod.

The MoI of the rod about this axis is $\frac{1}{3}m_{rod}l^2$. The MoI of the sphere about its centre of mass is $\frac{2}{5}m_{sphere}r^2$; by the parallel axis theorem, the MoI of the sphere about the main axis is $\frac{2}{5}m_{sphere}r^2 + m_{sphere}l^2$.

That would make the total MoI $\frac{1}{3}m_{rod}l^2 + \frac{2}{5}m_{sphere}r^2 + m_{sphere}l^2$.

So as i suppose that the mass of the rod if negligible, i think i concluded right this time... Funny, i never saw a book approaching the pendulum by this way, probably it always consider the sphere to small to be considered the new term...

 

[kuruman]

Seeing this again, a doubt raised to me.
In real life, to study a pendulum consisting of a light bar (negligible mass) with a massive sphere on the end, it's enough to consider the moment of inertia as
"ML²" being L the length of the bar
or we need to consider "ML² + 2MR²/5", being R the radius of the sphere.?]

The second seems right to me.

The length $L$ of the light rod is from the point of support to the surface of the sphere. To apply the parallel axis theorem correctly you need to use the distance from the point of support to the CM of the sphere: $I=\frac{2}{5}MR^2+M(L+R)^2$.

 

[etotheipi]

a light bar

Ja massive rod

I guess this is why they say don't do Physics at 00:30 in the morning

 

[Frodo]

The effect is quite significant.

Take a nominal one second pendulum with a length of about 40 inches and a spherical bob of radius 3 inches. It has a period some 100 seconds/day longer than a 40 inch simple pendulum which is equivalent to lowering the bob by 0.1 inch.