What is the moment of inertia of the object

[jdgallagher14]

Homework Statement

An object is formed by attaching a uniform, thin rod with a mass of mr = 6.94 kg and length L = 5.56 m to a uniform sphere with mass ms = 34.7 kg and radius R = 1.39 m. Note ms = 5mr and L = 4R.

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

Homework Equations

I = mr^2
I(rod-end)=1/3 mr^2
I(spherical shell)=2/3 mr^2
I(sphere) = 2/5 mr^2

The Attempt at a Solution

I figured that it would be a spherical shell going around the axis, because the whole sphere is rotating, rather that it rotating at it center of the sphere, and then the rod going around as well. I'm obviously wrong, seeing as I'm reaching out for help, but here's what I had put, I am not sure why it's wrong though, and I don't know what to do.

1/3 (m rod)(L + 2R)^2 + (2/3) (m sphere)(2R)^2
(1/3)(6.94)(5.56+2.78)^2 + (2/3)(34.7)(2.78)^2

 

[Redbelly98]

Welcome to Physics Forums.

Looks like you need to use the parallel axis theorem for this problem. It should be in your textbook or class notes, or you can find it in the PF library https://www.physicsforums.com/library.php?do=view_item&itemid=31".

Are you using the m.o.i. of a hollow sphere? The problem seems to describe a solid sphere.

Also, is the rod attached to the right, left, top, or bottom (or other) edge of the sphere?

 

[jdgallagher14]

oh, i didn't notice that.. you're right, i was using the moment of inertia for a solid sphere. the rod is connected to the left end of the sphere, and the axis is on the right side of the sphere.. like...

===O

 

[Redbelly98]

I meant that I think it is a solid sphere. I saw the "2/3" factor in your calculation, which should be "2/5" if it is a solid sphere.

 

[rwisz]

(1/3)(6.94)(8.34)^2 + (2/3)(34.7)(2.78)^2
I = 12.2 kg*m^2

I would like to provide a response to the calculation of the moment of inertia of the given object. The moment of inertia is a measure of an object's resistance to rotational motion and it depends on the object's mass distribution and the axis of rotation. In this case, the object consists of a rod and a sphere attached together, and the moment of inertia is being calculated about an axis passing through the right edge of the sphere.

The attempt at a solution is incorrect because the rod is not rotating around its end, rather it is rotating around the axis passing through the center of the sphere. In addition, the moment of inertia of a sphere is not equal to 2/3 mr^2, but rather it is 2/5 mr^2. Therefore, the correct calculation of the moment of inertia of the given object should be:

I = (1/3)(6.94)(5.56)^2 + (2/5)(34.7)(1.39)^2
= 20.3 kg*m^2

This is the moment of inertia of the object about the axis passing through the center of the sphere. To calculate the moment of inertia about the right edge of the sphere, we need to apply the parallel axis theorem, which states that the moment of inertia about a parallel axis is equal to the moment of inertia about the center of mass plus the product of the mass and the square of the distance between the two axes.

In this case, the distance between the two axes is equal to the radius of the sphere, which is 1.39 m. Therefore, the moment of inertia about the right edge of the sphere can be calculated as:

I(edge) = I(center) + m(1.39)^2
= 20.3 + (34.7)(1.39)^2
= 65.6 kg*m^2

Therefore, the moment of inertia of the given object about the right edge of the sphere is 65.6 kg*m^2. I hope this explanation helps to clarify any confusion and provides a correct solution to the problem.