Moment of inertia of a rod with balls attached

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SUMMARY

The discussion centers on calculating the moment of inertia for a rod with attached point masses. The correct approach involves using the Parallel Axis Theorem (PAT) to adjust the moment of inertia when the axis of rotation is shifted. The final calculation for the total moment of inertia is I_total = 12.62 kg*m², derived from I_total(rod) = I_cm + md², where I_cm is calculated as (1/12) * 3.5 * (2.6)² + 3.5 * (1.3)². This highlights the importance of correctly applying the theorem to account for the distribution of mass relative to the new axis.

PREREQUISITES
  • Understanding of moment of inertia and its formulas, specifically I=mr² and I=(1/12)ml² for rods.
  • Familiarity with the Parallel Axis Theorem (PAT) for shifting axes of rotation.
  • Basic knowledge of physics concepts related to rotational motion.
  • Ability to perform calculations involving mass and distance in the context of rotational dynamics.
NEXT STEPS
  • Study the Parallel Axis Theorem in detail to understand its applications in various scenarios.
  • Practice calculating moment of inertia for different shapes and configurations of mass.
  • Explore advanced rotational dynamics concepts, including torque and angular momentum.
  • Review examples of real-world applications of moment of inertia in engineering and physics.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators and anyone involved in teaching or learning about rotational dynamics and moment of inertia calculations.

AfronPie

Homework Statement


Attached.

Homework Equations


I=mr^2, I=(1/12)m*l^2 for a rod.

The Attempt at a Solution


Part A I got by doing I=(1/12)3.5*(2.6)^2+2*.7*1.3^2 (I added the moment of inertia of the rod and the balls). Part B since the axis is on one of the balls, I thought we don't include that ball in the calculations. So the rod stays, (1/12)m*l^2 and we add the other .7 kg ball which is 2.6m away. So I_total=(1/12)3.5*2.6^2+.7(2.6)^2=6.7 kg*m^2. However that isn't the right answer. Can someone please explain to me where I went wrong. Any help is greatly appreciated.
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AfronPie said:
So the rod stays, (1/12)m*l^2
Think about that.
 
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TSny said:
Think about that.
Can you explain your hint. Why wouldn't you include the I of the rod?
 
You do need to include I for the rod. But when you move the axis of rotation to one end of the rod, does I for the rod remain the same?
 
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TSny said:
You do need to include I for the rod. But when you move the axis of rotation to one end of the rod, does I for the rod remain the same?
Im confused. The only reason why I knew I=1/12ml^2 is because I looked up moment of inertia for a rod. The mass is the same and the length of the rod is the same? Can you please explain what it would change to and why?
 
Moment of inertia depends on how the mass of the object is distributed around the axis of rotation. When you switch the axis, you change how the mass is distributed relative to the new axis. Consult PAT (Parallel Axis Theorem).

http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html
 
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TSny said:
Moment of inertia depends on how the mass of the object is distributed around the axis of rotation. When you switch the axis, you change how the mass is distributed relative to the new axis. Consult PAT (Parallel Axis Theorem).

http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html
Alright so I'm using the theorem. So I_total(rod)=I_cm+md^2=(1/12)3.5*2.6^2+4.9(1.3)^2=10.25
10.25+.7(2.6)^2=14.98 kg*m^2. Is this right?
 
AfronPie said:
Alright so I'm using the theorem. So I_total(rod)=I_cm+md^2=(1/12)3.5*2.6^2+4.9(1.3)^2=10.25
10.25+.7(2.6)^2=14.98 kg*m^2. Is this right?
It looks OK except for the factor of 4.9. How did you get that number?
 
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TSny said:
It looks OK except for the factor of 4.9. How did you get that number?
It's the M in the parallel axis equation which I thought meant the sum of all the masses.
 
  • #10
You should apply the theorem to just the rod. You are taking care of the point masses separately.
 
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  • #11
TSny said:
You should apply the theorem to just the rod. You are taking care of the point masses separately.
I_total(rod)=I_cm+md^2=(1/12)3.5*2.6^2+3.5(1.3)^2=7.89
7.89+.7(2.6)^2=12.62 kg*m^2. I just put it in and its right, thanks a lot for your help Tsny.
 
  • #12
Great! Good work.
 

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