# Moment of inertia help, please

• *best&sweetest*
In summary: So, in summary, the moment of inertia of the whole system is I = (1/3)ML^2 + m (0.75L)^2 + m1*(L)^2 + (2/5)m1*R^2.
*best&sweetest*
This should be simple but I'm not sure whether I'm doing it right or not...
I have a rod rotating about its edge #1. Rod's length is L and its mass is M. There is a point mass m on the rod, 3/4 L away from the edge #1, and on the other edge (#2) there is a sphere of radius R and mass m1. I need to find the moment of inertia of the whole system. I know that I need to use parallel axis theorem for the point mass, but what should I do with the sphere?
Is it I = (1/3) ML^2 + m (0.75L)^2 + m1*(L)^2 + (2/5)m1*R^2
or I = (1/3)ML^2 + m (0.75L)^2 + (2/5)m1*R^2?
I think it is the first option, but I'm not sure.
Thank you!

*best&sweetest* said:
This should be simple but I'm not sure whether I'm doing it right or not...
I have a rod rotating about its edge #1. Rod's length is L and its mass is M. There is a point mass m on the rod, 3/4 L away from the edge #1, and on the other edge (#2) there is a sphere of radius R and mass m1. I need to find the moment of inertia of the whole system. I know that I need to use parallel axis theorem for the point mass, but what should I do with the sphere?
Is it I = (1/3) ML^2 + m (0.75L)^2 + m1*(L)^2 + (2/5)m1*R^2
or I = (1/3)ML^2 + m (0.75L)^2 + (2/5)m1*R^2?
I think it is the first option, but I'm not sure.
Thank you!
Option 1 loooks correct, but actually, you used the parallel axis theorem for the sphere, not the point mass, which is correct.

I would suggest reviewing the definition of moment of inertia and the parallel axis theorem to ensure you are using the correct equations and understanding the concepts correctly. It is important to accurately represent the physical properties and dimensions of the system in your calculations.

In this case, the moment of inertia for the rod would be (1/3)ML^2, and for the point mass it would be m(0.75L)^2. However, for the sphere, the moment of inertia would be (2/5)m1R^2, not m1L^2. This is because the sphere is rotating about its own center of mass, not about the edge of the rod.

Therefore, the correct equation for the moment of inertia of the entire system would be I = (1/3)ML^2 + m(0.75L)^2 + (2/5)m1R^2. It is important to note that the moment of inertia of the point mass and the sphere cannot be combined into one term, as they are rotating about different axes.

I would also suggest double-checking your calculations to ensure you are using the correct units and that your final answer makes physical sense. If you are still unsure, it may be helpful to consult with a colleague or refer to a trusted resource for further clarification.

Remember, accuracy and attention to detail are crucial in scientific calculations. Good luck!

## 1. What is moment of inertia?

Moment of inertia, also known as rotational inertia, is a measure of an object's resistance to changes in its rotational motion. It is influenced by an object's mass and distribution of that mass around its axis of rotation.

## 2. How do you calculate moment of inertia?

The formula for moment of inertia is I = mr², where I is the moment of inertia, m is the mass of the object, and r is the distance between the axis of rotation and the mass.

## 3. What are the units of moment of inertia?

The SI unit for moment of inertia is kg•m², but it can also be expressed in other units such as g•cm² or lb•ft² depending on the system of measurement being used.

## 4. Why is moment of inertia important?

Moment of inertia is important because it helps us understand and predict how objects will behave when they are rotating. It is also a crucial factor in the design of many mechanical systems, such as engines and turbines.

## 5. Can you provide an example of moment of inertia?

An example of moment of inertia is a figure skater spinning on the ice. When the skater pulls their arms close to their body, their moment of inertia decreases, causing them to spin faster. This is because their mass is now distributed closer to the axis of rotation, reducing their resistance to changes in rotational motion.

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