Rotational Dynamics: Moment of Inertia of 0.44 kg Meter Stick

[PhysicsDud]

Wanted to check this question also:

A meter stick of mass 0.44 kg rotates, in the horizontal plane, about a vertical
axis passing through the 30 cm mark. What is the moment of inertia of the stick?
(Treat it as a long uniform rod)

Answer: I = 0.1628 kg.m^2

Correct??

 

[Doc Al]

Rather than just give your answer, always show how you got your answer. (At least give the formulas that you used.) I assume you meant to calculate the moment of inertia about that axis.

 

[PhysicsDud]

My Work

So I was a little confused but I calculated using the formula I = mr^2 for both sides of the stick and then added them:

I = (0.132 kg)(0.3m)^2 + (0.308 kg)(0.7m)^2
= 0.1628 kg.m^2

 

[Doc Al]

What's the rotational inertia of a rod about one end? It's not $$I = m L^2$$ (where L is the length of the rod).

 

[PhysicsDud]

Inertia through end of uniform rod

That formula is I = 1/3ML^2

So the answer is 0.147 kg.m^2 ??

 

[Doc Al]

You have the correct formula for the rotational inertia, but recheck your calculation.

 

[PhysicsDud]

Is the correct answer 0.0132 kg.m^2

Using 0.3m for the Length?

 

[Doc Al]

In post #3 you showed your work using the incorrect formula for rotational inertia. How would you modify that calculation using the correct formula?

 

[PhysicsDud]

Ok hopefully I've got it this time

I = 1/3 ML1^2 + 1/3 ML2^2
= 1/3 (.132)(.3)^2 + 1/3 (.308)(.7)^2
= 0.064 kg.m^2

 

[Doc Al]

Your method is correct, but I suspect a typo in your final answer. (I get 0.054)

 

[PhysicsDud]

Thank You!

Thanks so much for your help!