Finding KE of a rod with two particles and angular velocity?

[rockchalk1312]

In the figure, two particles, each with mass m = 0.79 kg, are fastened to each other, and to a rotation axis at O, by two thin rods, each with length d = 5.9 cm and mass M = 1.2 kg. The combination rotates around the rotation axis with angular speed ω = 0.32 rad/s. Measured about O, what is the combination's kinetic energy?I=Ʃm_ir_i²
KE=1/2Iω²(.79kg)(.059m²)+(.79kg)(.118m²) = .0137 = I
1/2(.0137)(.32rad/s²) = 7.01E-4 = KE

Do I need to convert cm to m? Also how do you factor in the mass of the rod? I assume we need that, otherwise they wouldn't have given its mass, etc. I'm extremely new at all of this. Thank you!

 

[collinsmark]

Hello KU fan,

Do I need to convert cm to m?

Well, yes, if you want to work in SI units (kilogram, meter, second units), you'll need to change the centimeters to meters. Doing so means the units of moment of inertia are the standard [kg·m²], and the kinetic energy is in units of Joules [kg·m²/s²].

You can work in some other system of units, but your final answer won't end up in units of Joules so easily.

Also how do you factor in the mass of the rod? I assume we need that, otherwise they wouldn't have given its mass, etc.

Yes, you do need that. You need to find the rods' contribution to the total moment of inertia.

Here is a useful link that might help (I often find it very helpful):
http://en.wikipedia.org/wiki/List_of_moments_of_inertia

In the figure, two particles, each with mass m = 0.79 kg, are fastened to each other, and to a rotation axis at O, by two thin rods, each with length d = 5.9 cm and mass M = 1.2 kg. The combination rotates around the rotation axis with angular speed ω = 0.32 rad/s. Measured about O, what is the combination's kinetic energy?I=Ʃm_ir_i²
KE=1/2Iω²(.79kg)(.059m²)+(.79kg)(.118m²) = .0137 = I

I think you missed the order of the squaring and the closing parenthesis when you typed in the numbers. I think you meant:

(.79 [kg])(.059 [m])² + (.79 [kg])(.118 [m])² = .0137 [kg·m²] = I

Anyway, besides that, you're missing something. Your approach would be correct if the rods were of negligible mass. But they're not. (The mass of each rod is heaver than the mass of each particle! The masses of the rods are definitely not negligible.)

You'll need to incorporate the contribution of the rods' moments of inertia into the total.

 

[rockchalk1312]

Anyway, besides that, you're missing something. Your approach would be correct if the rods were of negligible mass. But they're not. (The mass of each rod is heaver than the mass of each particle! The masses of the rods are definitely not negligible.)

You'll need to incorporate the contribution of the rods' moments of inertia into the total.

Thank you for the reply!

So I went to your link and attempted to use the moment of inertia formula for the rod, I=mL²/3 which gave me (2.4kg)(.118m)²/3=.0111.

Then I added this to my original inertia for the two particles, .0111+.0137=.0248 and submitted this into the KE formula which gave me .00126 J and this was still the wrong answer.

Was it correct to use this formula and just double the mass and length for the whole rod?

Thank you again!

 

[collinsmark]

Thank you for the reply!

So I went to your link and attempted to use the moment of inertia formula for the rod, I=mL²/3 which gave me (2.4kg)(.118m)²/3=.0111.

Then I added this to my original inertia for the two particles, .0111+.0137=.0248

That's pretty much what I got (ignoring any minor rounding differences). (Don't forget your units though! )

and submitted this into the KE formula which gave me .00126 J

That's roughly what I got too (ignoring any minor rounding differences).

and this was still the wrong answer.

I'm not quite sure what to tell ya. Maybe it's the rounding errors that are the issue. You lost a little precision in your calculations. You might want to try again, but keep more significant figures in your intermediate calculations.

Was it correct to use this formula and just double the mass and length for the whole rod?

Yes, treating the two rods as one big rod is one valid way to solve this particular problem. (If the rods were not in a straight line, or were not connected at a point, it wouldn't work. But since both rods are behaving/configured as one, big, straight rod for this problem, it's okay to treat them as such.)

There is another way to do this problem too (and this is important for future, more complicated problems). You could treat both rods separately. The rod connected to the rotation axis has a moment of inertia of I=mL²/3. The other rod is a little more complicated though, but you can find its moment of inertia by using the parallel axis theorem.