Calculating Moment of Inertia & Kinetic Energy of Rotating Object

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SUMMARY

The discussion focuses on calculating the moment of inertia and kinetic energy of a system consisting of three point masses (m1=0.2 kg, m2=0.3 kg, m3=0.5 kg) attached to a massless rigid rod. The moment of inertia is calculated using the formula I=Icm+MD², where Icm is the moment of inertia about the center of mass and M is the total mass. The potential energy change when the object swings from horizontal to vertical is also analyzed, leading to the calculation of rotational kinetic energy using E=1/2Iw². The correct approach involves calculating the moment of inertia for each mass separately and summing them.

PREREQUISITES
  • Understanding of moment of inertia calculations
  • Familiarity with potential energy concepts
  • Knowledge of rotational kinetic energy equations
  • Ability to apply the parallel axis theorem
NEXT STEPS
  • Calculate the moment of inertia for each mass using the formula I=mr²
  • Study the parallel axis theorem for additional insights on inertia
  • Learn how to derive potential energy changes in rotating systems
  • Explore the relationship between potential energy and kinetic energy in rotational motion
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Physics students, educators, and anyone interested in mechanics, particularly those studying rotational dynamics and energy transformations in physical systems.

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Homework Statement



An object consists of three point masses (m1=0.2 kg, m2=0.3 kg and m3=0.5 kg) attached to a massless rigid rod. The masses are located at 0 m, 0.4 m, and 0.8 m respectively. (The rod is 0.8 m in length so the 0.5 kg mass is attached to one end and the 0.2 kg mass to the other).
The center of mass is located at 0.52 m.

a) The object is pivoted about an axis so that it rotates from the position of the 0.2 kg mass.

(0.2-------0.3-------0.8 with circular rotation originating at the first mass). What is the object's moment of inertia about that axis?


b) Starting from rest the object is allowed to swing from horizontal to vertical (so the rod is now pointing down). By how much does the potential energy of the object change?

c) from (a) and (b)'s answers, calculate rotational kinetic energy.

Homework Equations


I=Icm+MD2
Icm=?
E=1/2Iw2

The Attempt at a Solution



I attempted 0.522*1 (since one is the total weight) + 0.5*0.522 and got an answer of 0.4 something. I have a study guide that tells me this answer is incorrect. My main problem with this question is calculating the inertia and then figuring out how it relates to the potential energy! Any help would be appreciated.
 
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