http://www.smartphysics.com/images/content/mechanics/ch15/momentofinertia3.png

An object is formed by attaching a uniform, thin rod with a mass of m = 7.07 kg and length L = 5.16 m to a uniform sphere with mass M = 35.35 kg and radius R = 1.29 m. What is the moment of inertia of the object about an axis at the right edge of the sphere?

I think the expression would be 0.4MR^2+MR^2+mL^/3, is it?


Compare the three moments of inertia calculated above:
I,CM < I,left < I,right
I,CM < I,right < I,left
I,right < I,CM < I,left
I,CM < I,left = I,right
I,right = I,CM < I,left

how can I calculate the moment of inertia of the object about an axis at the center of mass of the object?

http://www.smartphysics.com/images/c...ofinertia3.png

An object is formed by attaching a uniform, thin rod with a mass of m = 7.07 kg and length L = 5.16 m to a uniform sphere with mass M = 35.35 kg and radius R = 1.29 m. What is the moment of inertia of the object about an axis at the right edge of the sphere?

I think the expression would be 0.4MR^2+MR^2+mL^/3, is it?


Compare the three moments of inertia calculated above:
I,CM < I,left < I,right
I,CM < I,right < I,left
I,right < I,CM < I,left
I,CM < I,left = I,right
I,right = I,CM < I,left

how can I calculate the moment of inertia of the object about an axis at the center of mass of the object?

Western, were you given a formula for the moment of inertia of a sphere by itself around an axis at its edge (like in the picture)? If not, are they expecting you to do the volume integral to derive it yourself? Otherwise, just out of curiosity, how did you come up with your answer?

Thanks.

I'm not sure on how to calculate the moment of inertia at the center of mass either. How would one go about doing this?

Quote:

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Originally Posted by western50

What is the moment of inertia of the object about an axis at the right edge of the sphere?

i think the expression would be 0.4MR^2+MR^2+mL^/3, is it?

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Not quite. The parallel axis theorem states that if you know the moment of inertia of an object about its center of gravity, then to find the moment of inertia about a point that is d distance from the CG you add md^2. So consider each piece of the problem:

For the rod, its moment of inertia about its CG is I=mL^2/12 (presumably you looked that up in a book or on line). But the point you are interested in is a distance d = (L/2 + 2R) away from the rod's CG. So the total moment of inertia for the rod is mL^2/12 + m(L/2 + 2R)^2

Now for the sphere: the moment of inertia for a sphere about its center is MR^2. Using the parallel axis theorem, you add MR^2 to that to find its moment of inertia about a point on its surface.

Now simply add the two results together. What do you get?

Quote:

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Originally Posted by western50

Compare the three moments of inertia calculated above:
I,CM < I,left < I,right
I,CM < I,right < I,left
I,right < I,CM < I,left
I,CM < I,left = I,right
I,right = I,CM < I,left

---

I don't understand what this is trying to say - can you be a bit more specific please?

Quote:

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Originally Posted by western50

how can i calculate the moment of inertia of the object about an axis at the center of mass of the object?

---

I is calculated as follows:



Is that what you're asking? Or are you asking about the parallel axis theorem? You can use the parallel axis theorem if you know I for an axis at one end of an object. For example: the moment of inertia for a uniform rod measured at one end of the rod is mL^2/3. So to find the moment of inertia about its center of gravity you subtract m(L/2)^2:



Hope this helps.