Conservation of Energy for a rotating Rod

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SUMMARY

The discussion focuses on the conservation of energy for a rotating rod, specifically addressing the calculation of potential energy as MgL/2. This value arises because the center of mass of a uniform rod is located at L/2 from the pivot point. The torque from gravity is also dependent on the angle of the rod, confirming that the potential energy can be derived from the formula mgh, where h equals L/2. The participants clarify these concepts through iterative questioning and responses.

PREREQUISITES
  • Understanding of rotational kinetic energy and the formula 1/2*I*w^2
  • Knowledge of center of mass for uniform objects
  • Familiarity with torque and its dependence on angle
  • Basic principles of gravitational potential energy
NEXT STEPS
  • Study the derivation of the rotational kinetic energy formula in detail
  • Learn about the calculation of center of mass for various shapes
  • Explore the relationship between torque and angular displacement
  • Investigate the implications of gravitational potential energy in rotational systems
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of rotational dynamics and energy conservation in mechanical systems.

AROD
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Homework Statement


rod%20problem.jpg

Homework Equations



Rotational Kinetic Energy = 1/2*I*w^2

The Attempt at a Solution



I was just wondering if someone explain to me why the potential energy is MgL/2 ?

Is this not the same for the torque from gravity?
 
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As the hint in a) states, you can neglect the mass m. And since the mass M is uniform, you can presume all its weight is in the center, hence MgL/2
 
Because the rod's center of mass is placed at L/2 over the ground.

Edit: oops, already answered :)
 
AROD said:
I was just wondering if someone explain to me why the potential energy is MgL/2 ?
The rod is uniform. Where is its center of mass?
Is this not the same for the torque from gravity?
The torque from gravity about the pivot point depends upon the angle of the rod.

Edit: Already answered twice!
 
ok so then its just mgh with h = L/2 . that makes sense

then as it tips the hypotoneuse would be L/2, and the height would be this times the cos of theta.

thanks
 

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