Inertia of two masses m2 connected to a rod.

AI Thread Summary
The discussion focuses on calculating the moment of inertia for a system involving a rod with two spherical masses attached at either end. The correct formula for the moment of inertia is identified as I=2 m_2 (d^2 + (2/5) r^2), where d is the distance from the center of the rod to the center of each sphere, and r is the radius of the spheres. Participants clarify the use of the parallel axis theorem to account for the distance between the axes of rotation. Misunderstandings about the placement of the spheres and their contribution to the moment of inertia are resolved, emphasizing that the spheres' centers are actually d away from the central axis. The conversation concludes with a recognition of the importance of understanding the underlying physics principles.
cooev769
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I have never dealt with moment of inertia before, this is a physics lab i need to do some pre planning for which involves topics we have never covered and are expected to learn. I've been busy working two jobs and am struggling to get time to pick up this before my lab tomorrow so some help would be extremely helpful.

We have a rod with length 2d, spinning at its centre. With 2 masses on either side of mass m2, the masses are spherical with radius r. What is the moment of inertia?

The answers is:

I=2 m_2 (d^2 + \frac{2}{5} r2)

By my logic the moment of inertia is the sum of each component, we have two masses m2 and their centre of gravity is d+r away from the centre hence each sphere will contribute, based on the inertia of a sphere:

I_(spheres) =2 \frac{2}{5} m_2 (r + d)^2

And then add the component of the rod, which I believe is ignored as we don't have any value for the mass of the rod. Obviously my assumption is wrong as it isn't the same as the answer. Any tips would be choice. Cheers.
 
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By the way I should emphasise that the first equation is not 4m(etc), it's 2 m2, meaning mass 2. Just thought i'd clarify just in case don't know how to subscript.
 
The spheres have radius r, not r + d. You'll take the moment of inertia of each sphere about its center (which you know) and then compute its moment of inertia about the center of the rod using the parallel axis theorem: Parallel Axis Theorem
 
So how come the rods contribute moment of inertia about themselves and around the centre too? that seems slightly confusing.
 
Ah i see because the axis running through the sphere is parallel to the axis running through the centre of the rod, the parallel axis theorem says that the I through the rod will be equal to the moment of inertia through the axis through the sphere plus the mass multiplied by the distance between the two parallel axis? Is this interpretation correct? So I get.

I_t = I_c + m_2 (d+r)^2

Where I_c is moment of inertia with the axis through the centre of the sphere. Is this correct, i used d+r squared because the distance between the two axis would be d+r?
 
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cooev769 said:
Ah i see because the axis running through the sphere is parallel to the axis running through the centre of the rod, the parallel axis theorem says that the I through the rod will be equal to the moment of inertia through the axis through the sphere plus the mass multiplied by the distance between the two parallel axis? Is this interpretation correct?
Multiplied by the distance squared, I think you mean.

So I get.

I_t = I_c + m_2 (d+r)^2

Where I_c is moment of inertia with the axis through the centre of the sphere. Is this correct, i used d+r squared because the distance between the two axis would be d+r?
Is the distance from the center of the sphere to the axis of rotation r+d or d? If the spheres are just tacked on to the ends of the 2d rod, then you are correct.
 
Okay so to get the answer they arrive at I must therefore assume that somehow the spheres are not just glued on a distance d from the centre but the centre of the spheres are connected to the rod with a hole or something and that the centres of The spheres are actually d away from the central axis?
 
cooev769 said:
Okay so to get the answer they arrive at I must therefore assume that somehow the spheres are not just glued on a distance d from the centre but the centre of the spheres are connected to the rod with a hole or something and that the centres of The spheres are actually d away from the central axis?
Yes, exactly. (Did they provide a diagram?)
 
Nope. Yeah I guess at poorly worded to question. But pretty trivial now I know the theory. Thanks so much.
 
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