Need confirmation with a question answer

1. Dec 22, 2004

Sanosuke Sagara

I have my question and solution in the attachment that followed.However I still can't find the answer that match that textbook.Is my answer wrong or the textbook answer wrong ?Do comment if I have done any calculation error.

Thanks for anybody that spend sometime on this question.

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2. Dec 22, 2004

Nylex

The question says find the moment of inertia about the horizontal axis through O. You need to take the angle into account. I got an answer of 0.24975 kg m^2, but I don't know where I've gone wrong :/.

3. Dec 22, 2004

Sanosuke Sagara

I have seen a question like this before in a book with the solution is by adding the total moment of inertia on the string.

That is , I = m_1 L_1 + m_2 L_2 without the claculation being interrupt by the angle 30 degree .But using this solution cannot find the correct answer.

4. Dec 22, 2004

Staff: Mentor

Your method and answer seem correct to me. The moment of inertia does not depend on the angle.

5. Dec 22, 2004

Sanosuke Sagara

Thanks for mentor confirmation .Thanks for mentor mentor help.

6. Dec 22, 2004

Staff: Mentor

Just to be clear: That equation should have the lengths squared:
$$I = m_1 L_1^2 + m_2 L_2^2$$

7. Dec 22, 2004

learningphysics

Are you sure about this? The moment of inertia is different depending on which axis of rotation you choose isn't it... mL^2 ... here L is the perpendicular distance from the mass to the axis of rotation... if you change the axis of rotation the distance is different...

8. Dec 22, 2004

Staff: Mentor

The moment of inertia depends on the axis of rotation, but not on the angle.

9. Dec 22, 2004

learningphysics

The distance of the first object from the horizontal axis through O is 0.3cos30, and the second is 0.9cos30.

Here's what I get for the moment of inertia:

$$I=m_1r_1^2+m_2r_2^2$$
$$I=(0.2)(0.3cos30)^2+(0.3)(0.9cos30)^2$$
$$I=0.19575kgm^2$$

10. Dec 22, 2004

Staff: Mentor

No. The distances from the axis are given as 0.3m and 0.9m. For some reason you are calculating the vertical component of those distances, but that has nothing to do with finding the moment of inertia. The moment of inertia of a point mass depends only on its distance from the axis of rotation, not on the angle it makes.

Just as the mass of an object does not depend on its position, the moment of inertia of an object about a given axis does not depend on its orientation.

This is incorrect.

11. Dec 22, 2004

learningphysics

I might have misinterpreted the diagram. But it seems like 0.3 is the distance between O and P. And 0.9 is the distance between O and Q. So if we take the axis as one through O that is perpendicular to OPQ then we can use these distances, but that's not what the question asks, it asks for the moment of inertia about the "horizontal axis" through O.

I don't think the distance of each mass from the horizontal axis through O is given.

Am I wrong when I say 0.3cos30 is the distance between P and the horizontal axis through O?

I apologize for any errors I've made. I might have misinterpreted the original question.

12. Dec 22, 2004

Staff: Mentor

Ah... Now I see what you are talking about. You are assuming that the horizontal axis goes left to right, while I assume it goes perpendicular to the paper. Since it says "freely hinged" at O, and an angle is given, I just assumed that the angle was with respect to the axis of rotation! I picture it as a freely swinging rod (swinging left to right in the picture).