Moment of intertia for rod with weight on one end

[smegmaster]

Hello,

I'm trying to derive the moment of inertia for a long rod with a point-weight on one end (the rod rotates through its midpoint). Could anyone offer any help? I know the formula for the rod is normally (1/2)ML^2, but I'm not sure how the weight on one end alters that.

Thanks!

 

[Doc Al]

Add the moment of inertia of the point mass to that of the rod; that will give you the moment of inertia of the entire object.

FYI: I assume you meant that the moment of inertia of a rod about its center is (1/12)ML^2.

 

[smegmaster]

Add the moment of inertia of the point mass to that of the rod; that will give you the moment of inertia of the entire object.

FYI: I assume you meant that the moment of inertia of a rod about its center is (1/12)ML^2.

1/12, right. So I can just add the mass of the extra weight to the rod's moment of intertia?

Thanks for the help.

 

[smegmaster]

The length of the rod I have is 4.9m, with its mass 10.5kg. Without the weight, the moment of inertial is 21.00875. If the mass of the weight on the end is 5.18kg, would I be correct in saying the new moment of inertia is just 21.00875 + 5.18 = 26.18875kg?

 

[Doc Al]

So I can just add the mass of the extra weight to the rod's moment of intertia?

No. (For one thing, mass and moment of inertia are different kinds of quantities--with different units--so you can't add them.) Add the moment of inertia of the point mass to the rod's moment of inertia. (What's the moment of inertia of a mass about some point?)

 

[smegmaster]

Ahh, sorry. Moment of inertia for a point is its mass times the square of the radius. Thus the moment of inertia for the weight is 5.18*(4.9/2)^2 = 31.09295, bringing the total moment of inertia to 52.1017. Thanks!