Rotational Inertia of two particles

[PhysicsPhun]

Two particles, each with mass m = 5.4 g, are fastened to each other and to a rotation axis at P, by two thin rods, each with length L = 0.68 m and each with a mass of 7.9 g, as shown. The combination rotates around the rotation axis with an angular velocity of 10.4 rad/s. Find the rotational inertia (i.e. the "moment of inertia") of the combination about P.


The diagram is like this...


p------o------o

p being the pivot and the o's being the masses.

Rotational Inertia = The sum of all... mi+ri^2

There's also

The sum of all .. 1/2mi(wri)^2 = 1/2(The sum of all .. miri^2)w^2.

Sorry i don't really know how to write out the formulas hehe.

I don't understand how to add the sums of the masses*radius^2. like where do you start and stop? Do you just add up the lenth after the pivot and add up the mass of that and plus those into the first equation?

Any help would be awesome.

 

[Janitor]

If all you are trying to do is calculate the moment of inertia, you can ignore the rotational speed.

The moment of inertia, ignoring the contribution from the "thin" rods, is just the contribution from the two masses. Just add m r^2 to m r^2, or since the m and the r are the same for each rod's mass and length, multiply m r^2 by 2, and you've got your answer, in gram-meters squared. If they want the answer in some other units that they have specified, then of course you will need to apply a units conversion factor to the answer.

 

[Chen]

The moment of inertia, ignoring the contribution from the "thin" rods, is just the contribution from the two masses. Just add m r^2 to m r^2, or since the m and the r are the same for each rod's mass and length, multiply m r^2 by 2, and you've got your answer, in gram-meters squared. If they want the answer in some other units that they have specified, then of course you will need to apply a units conversion factor to the answer.

I don't think it would be wise or correct to ignore the rods in the calculation of the moment of inertia. They may be thin, but certainly not massless, and in fact they both weigh more than the balls. Also, r is the distance of each mass from the axis P, so it is not the same for both masses. One is distanced L from the axis, and the other is distanced 2L from it. All in all the moment of inertia should be:

I = \Sigma m_ir_i^2 = m_1L^2 + m_1(2L)^2 + m_2(\frac{L}{2})^2 + m_2(\frac{3L}{2})^2 = (5m_1 + 2.5m_2)L^2

Where m1 is the mass of each ball, m2 is the mass of each rod and L is the length of each rod. The terms in the equation are, respectively, for the first (left) ball, the second (right) ball, the first (left) rod and the second (right) rod. Since the rods are uniform, we take their center of mass to be at the middle of them.

 

[Doc Al]

I = \Sigma m_ir_i^2 = m_1L^2 + m_1(2L)^2 + m_2(\frac{L}{2})^2 + m_2(\frac{3L}{2})^2 = (5m_1 + 2.5m_2)L^2

Where m1 is the mass of each ball, m2 is the mass of each rod and L is the length of each rod. The terms in the equation are, respectively, for the first (left) ball, the second (right) ball, the first (left) rod and the second (right) rod. Since the rods are uniform, we take their center of mass to be at the middle of them.

Chen, you are absolutely correct that the rotational inertia of the rods cannot be neglected. But you must model them as rods, not as point masses.
So your 3rd and 4th terms need to be corrected by adding the rotational inertia of a rod about its center of mass, which is:
I_{rod, cm} = \frac{1}{12}ML^2

 

[Chen]

Fair enough... to my defense we never covered inertia at school though. Would you mind correcting my expression please?

 

[Doc Al]

I = \Sigma m_ir_i^2 = m_1L^2 + m_1(2L)^2 + m_2(\frac{L}{2})^2 + m_2(\frac{3L}{2})^2 = (5m_1 + 2.5m_2)L^2

The correct expression is:
I = m_1L^2 + m_1(2L)^2 + [m_2(\frac{L}{2})^2 + \frac{1}{12}m_2L^2] + [m_2(\frac{3L}{2})^2 + \frac{1}{12}m_2L^2]
The bracketed [] terms represent the use of the parallel axis theorem to calculate the rotational inertia of the rods about the axis of rotation.

This reduces to:
I = (5m_1 + \frac{8}{3}m_2)L^2

 

[PhysicsPhun]

Thanks so much guys, i think i follow what you're saying.

 

[Janitor]

I concur.

I was being exceptionally sloppy in suggesting ignoring the contribution of the rods to the moment of inertia.

 

[Doc Al]

I was being exceptionally sloppy in suggesting ignoring the contribution of the rods to the moment of inertia.

Yes. Now go stand in the corner, next to Chen. (Just kidding guys! )