Problem with Rotational Inertia

In summary, the conversation discusses a problem involving the rotational inertia of a disc with a hole cut out and its center of mass being off-center. The parallel axis theorem is mentioned and a calculation is provided for the full disc minus the missing disc. The question of finding the period for small oscillations about the vertical position is also brought up, with the solution involving finding the center of mass and using it to calculate the potential energy.
  • #1
Lancen
17
0
I am having problem with a relatively simple problem that I probably have done before but because my Physics have become so rusty recently I just can't get the right answer. You have a disc that sitting about some point P on the ground - so its upright - and there is a hole cut out on the top of the disc so that the top of the hole touches the top of the disc. The hole has a radius of a/2 and the disc and radius of a. The center of mass of the disc c is a/6 from the center of the disc. And the disc has a uniform mass density of v. What is the rotational inertia of the disc about the point P that it touches on the ground?

So parallel axis theorem right? Ip = Icm + ML^2. My problem is what is Icm? 1/2M*L^2 by any means. But what is L? Usually if the disc' center of mass is about its center then its just the radius but here the center of mass is a/6 from the center of the disc.
 
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  • #2
Hint: If the disk were solid, you could find its moment of inertia. By definition the moment of inertia is a sum of the products (dm)r^2 where dm is the mass of a tiny piece of the disk. Since it is just a sum, you could divide the disk up any way you found convenient and find the moments of inertia of each piece and add them together. Think of a solid disk as being made up of two parts. One part is the thing you have, and the second part is the disk that was cut away. The total angular momentum of a solid disk is the sum of those two parts.
 
  • #3
I got that far. I know the cut out hole is presented as "negative" mass, but the problem is Ip = Icm + M*h^2 = M*L^2+M*h^2=1/2((pi*a^2*v)-pi*(a/2)^2*v))*L^2+((pi*a^2*v)-pi*(a/2)^2*v)*(5a/6)^2.

Here Icm is the rotational inertia about the center of mass C which is not at the center of the circle, and h is the distance from the Center of mass to the point at which the circle touches the ground P. My problem is what's L?
 
  • #4
Lancen said:
I got that far. I know the cut out hole is presented as "negative" mass, but the problem is Ip = Icm + M*h^2 = M*L^2+M*h^2=1/2((pi*a^2*v)-pi*(a/2)^2*v))*L^2+((pi*a^2*v)-pi*(a/2)^2*v)*(5a/6)^2.

Here Icm is the rotational inertia about the center of mass C which is not at the center of the circle, and h is the distance from the Center of mass to the point at which the circle touches the ground P. My problem is what's L?
There are a couple of ways to approach this. Here is a calculation based on the "full" disk about point P minus the "missing" disk about point P. It does not require you to find the CM of the object or an L related to that CM. It can be done that way, but it involves more steps. This way uses separate Ls for the full disk and the missing disk.

[tex] I_{FCM} = \frac{1}{2}Ma^2 [/tex]

[tex] I_{FP} = \frac{1}{2}Ma^2 + Ma^2 = \frac{3}{2}Ma^2 [/tex]

[tex] I_{MCM} = \frac{1}{2}\left( {\frac{M}{4}} \right)\left( {\frac{a}{2}} \right)^2 = \frac{1}{{32}}Ma^2 [/tex]

[tex] I_{MP} = \frac{1}{{32}}Ma^2 + \frac{M}{4}\left( {\frac{{3a}}{2}} \right)^2 = \frac{1}{{32}}Ma^2 + \frac{9}{{16}}Ma^2 = \frac{{19}}{{32}}Ma^2 [/tex]

[tex] I_{NP} = I_{FP} - I_{MP} = \frac{3}{2}Ma^2 - \frac{{19}}{{32}}Ma^2 = \left( {\frac{{48}}{{32}} - \frac{{19}}{{32}}} \right)Ma^2 = \left( {\frac{{29}}{{32}}} \right)Ma^2 [/tex]
 
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  • #5
Ah, I see. I have one more problem with the question. If the disc is upright and is free to roll without slipping along the surface on which it rests under gravitational forces, what's the period T for small oscillations about the vertical position? I am having a bit of trouble understanding the problem. Is it saying that under some condition instead of rolling forward it will just rock back and forth about an imaginary line down the middle? And is this a simple or physical pendulum? Physical I think because it is represented as a rigid body. Do I just go straight ahead and try to solve for T=2*pi/w, where w=sqrt(mgh/I)? Where w is the angular velocity and I is the moment of inertia, and solve this using the moment of inertia that was just solved for?
 
  • #6
Lancen said:
Ah, I see. I have one more problem with the question. If the disc is upright and is free to roll without slipping along the surface on which it rests under gravitational forces, what's the period T for small oscillations about the vertical position? I am having a bit of trouble understanding the problem. Is it saying that under some condition instead of rolling forward it will just rock back and forth about an imaginary line down the middle? And is this a simple or physical pendulum? Physical I think because it is represented as a rigid body. Do I just go straight ahead and try to solve for T=2*pi/w, where w=sqrt(mgh/I)? Where w is the angular velocity and I is the moment of inertia, and solve this using the moment of inertia that was just solved for?
Now you tell me :rolleyes: I avoided calculating the CM in the I calculation, but now we need it. You had quoted it earlier as a/6 below the center of the big disk, and that is correct. Any displacement of the disk from point P being in contact with the floor will raise the CM, so there will be a restorative force that will bring it back to equilibrium. It's not quite a physical pendulum because there is no fixed point of rotation. Still, if we can show that for small oscillations the potential energy is a quadratic function of diplacement, then we will have shown the motion to be harmonic and we can find the period. See if you can write an expression for the potential energy in terms of the angle of the roll from equilibrium. The potential energy can be taken as zero at equilibrium and Mgh(θ) elsewhere; h(θ) is the height of the CM relative to the lowest point.
 
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Related to Problem with Rotational Inertia

1. What is rotational inertia?

Rotational inertia, also known as moment of inertia, is a measure of an object's resistance to change in its rotational motion. It depends on the object's mass and distribution of mass around its axis of rotation.

2. What causes a problem with rotational inertia?

A problem with rotational inertia can occur when there is a mismatch between the applied torque and the object's rotational inertia. This can lead to difficulties in changing the object's rotational motion or maintaining its stability.

3. How does rotational inertia affect objects in motion?

Rotational inertia affects objects in motion by determining how easily they can rotate or change their rotational motion. Objects with a higher rotational inertia will require more torque to change their motion compared to objects with a lower rotational inertia.

4. Can rotational inertia be changed?

Yes, rotational inertia can be changed by altering the distribution of mass within an object or by changing the object's shape. For example, a figure skater can decrease their rotational inertia by pulling their arms in closer to their body, allowing them to spin faster.

5. What are some real-life examples of rotational inertia?

Some common examples of rotational inertia include a spinning top, a spinning figure skater, a spinning gyroscope, and a spinning bicycle wheel. These objects exhibit rotational inertia due to their mass and distribution of mass around their axis of rotation.

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