Pivoting Bar and Kinetic Energy

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SUMMARY

The discussion focuses on calculating the kinetic energy of a rotating uniform bar, specifically a 12.0 kg bar that is 2.00 m long and makes 5.00 revolutions every 3.00 seconds. The kinetic energy formula used is K = 1/2 ∫ v² dm, where dm is defined as (M/L)dx. The user expresses confusion regarding the integration process, particularly how to account for the varying velocity of different segments of the bar due to its rotation. The integration approach suggested involves rewriting the integral in terms of dx, but the user is uncertain about its correctness.

PREREQUISITES
  • Understanding of rotational dynamics and kinetic energy
  • Familiarity with calculus, specifically integration techniques
  • Knowledge of uniform motion and angular velocity
  • Basic principles of mass distribution along a length (linear mass density)
NEXT STEPS
  • Study the derivation of kinetic energy for rotating bodies
  • Learn about angular velocity and its relationship to linear velocity
  • Explore integration techniques for variable limits and functions
  • Investigate the concept of moment of inertia for different shapes
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of rotational motion and kinetic energy calculations in rigid bodies.

DanielB
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Homework Statement



A thin, uniform 12.0kg bar that is 2.00m long rotates uniformly about a pivot at one end, making 5.00 complete revolutions every 3.00 seconds.

What is the kinetic energy of this bar? (Hint: Different points in the bar have different speeds. Break the bar up into infinitesimal segments of mass dm and integrate to add up the kinetic energy of all these segments.)

Homework Equations



K = 1/2 integral of v^2 dm
dm = (M/L)*dx

The Attempt at a Solution



I was confused on how to integrate this particular integral as velocity is not a constant and is clearly not in terms of dm. I attempted to integrate in terms of dx rewriting the integral as:

K= 1/2 * (2Pi/t)^2 * M/L integral x^2 dx

I don't think that is the correct direction but I am generally confused with the integration.
[EDIT]: Template was incorrect
 
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I don't see where your velocity takes into account the 5.00 revolutions it does in t=3.00 seconds.
 

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