Determine the moment of inertia of a bar and disk assembly

In summary, the conversation discusses using the moment of inertia of a compound pendulum to solve a problem, with the formula ## I_P = I_{rod} + M_{rod} \bigg( \frac{S}{2} \bigg)^2 + I_{disc} + M_{disc} (S+R)^2$$ where ##S## is the length of the pendulum and ##R## is the radius of the disc. The conversation also mentions the use of the parallel axis theorem and finding the moment about the mass center.
  • #1
65
2
Homework Statement
A homogeneous 3kg rod is welded to a homogeneous 2kg disc. Determine it's moment of inertia about the axis L which passes through the object's centre of mass. L is perpendicular to the bar and the disk.
Relevant Equations
## I_x = \int y^2\mathrm{d}A ##
## I_y = \int x^2\mathrm{d}A ##
Q.PNG

I have been given an answer for this but I am struggling to get to that point
$$ANS = 0.430\, kg \cdot m^2$$

So I thought using the moment of inertia of a compound pendulum might work where ##I_{rod} = \frac{ml^2}{12}## and ##I_{disc} = \frac{mR^2}{2}## (##l## is the length of the rod and ##R## is the radius of the disc)
$$ I_P = I_{rod} + M_{rod} \bigg( \frac{S}{2} \bigg)^2 + I_{disc} + M_{disc} (S+R)^2$$
where S is the length of the pendulum

$$ I_P = 0.09 + 1.2 + 0.04 + 2 = 2.61\, kg \cdot m^2$$
Much too large for my purposes.
Not really sure where to go after this. Any help is appreciated.
 
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  • #3
It looks like you are defining P as the free end of the rod and applying the parallel axis theorem. You forgot to square the L/2, but maybe that's just a typo in writing the thread. And you wrote that your L is the length of "the pendulum', but you mean the length of the rod. It is rather confusing that you switched from the given "l" to "L", when L was already defined as an axis.

Anyway, you are asked for the moment about the mass centre. To get that you need to find where that is and apply the parallel axis theorem in reverse.
 
  • #4
haruspex said:
It looks like you are defining P as the free end of the rod and applying the parallel axis theorem. You forgot to square the L/2, but maybe that's just a typo in writing the thread. And you wrote that your L is the length of "the pendulum', but you mean the length of the rod. It is rather confusing that you switched from the given "l" to "L", when L was already defined as an axis.

Anyway, you are asked for the moment about the mass centre. To get that you need to find where that is and apply the parallel axis theorem in reverse.
Sorry about that, I changed the post to reflect your corrections. ##l## is for the length of the rod and S is the length of the pendulum now. Thanks for the advice. I'll try giving that a go
 

1. What is the moment of inertia of a bar and disk assembly?

The moment of inertia of a bar and disk assembly is a measure of its resistance to rotational motion. It is a property that depends on the mass distribution of the assembly and the axis of rotation.

2. How is the moment of inertia of a bar and disk assembly calculated?

The moment of inertia of a bar and disk assembly can be calculated by adding the individual moments of inertia of the bar and disk components. The moment of inertia of a bar can be calculated as (1/12) * m * L^2, where m is the mass of the bar and L is its length. The moment of inertia of a disk can be calculated as (1/2) * m * r^2, where m is the mass of the disk and r is its radius.

3. What factors affect the moment of inertia of a bar and disk assembly?

The moment of inertia of a bar and disk assembly is affected by the mass distribution and shape of the components, as well as the axis of rotation. For example, a longer bar or a larger disk will have a greater moment of inertia, while a thinner bar or a smaller disk will have a smaller moment of inertia.

4. Can the moment of inertia of a bar and disk assembly be changed?

Yes, the moment of inertia of a bar and disk assembly can be changed by altering the mass distribution or shape of the components. For example, adding more mass to one end of the bar or increasing the radius of the disk will increase the moment of inertia.

5. Why is the moment of inertia important?

The moment of inertia is an important concept in physics and engineering because it helps us understand and predict the rotational motion of objects. It is used in calculations related to torque, angular acceleration, and angular momentum. It also plays a role in the design and stability of various mechanical systems.

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