Angular velocity of suspended rod

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SUMMARY

The discussion centers on calculating the angular velocity of a uniform rod of length L and mass m when it is released from a horizontal position to a vertical position. The correct solution utilizes the conservation of energy principle, leading to the formula ω = √(3g/L) for angular velocity. Participants debate the validity of assuming constant torque throughout the motion, with one contributor emphasizing the need to integrate torque and angular acceleration as they vary with the angle θ.

PREREQUISITES
  • Understanding of rotational dynamics and angular motion
  • Familiarity with the conservation of energy principle
  • Knowledge of torque and its relation to angular acceleration
  • Basic calculus for integrating functions related to angular motion
NEXT STEPS
  • Study the conservation of energy in rotational systems
  • Learn about the moment of inertia for different shapes, specifically uniform rods
  • Explore the relationship between torque, angular acceleration, and angular displacement
  • Practice solving problems involving variable torque and angular motion
USEFUL FOR

Students of physics, particularly those studying mechanics, as well as educators and anyone interested in understanding the dynamics of rotating bodies.

Prathamesh
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Homework Statement


A uniform rod of length L & mass m is suspended through an end. It is shifted through an angle of π/2 radian i.e. made horizontal and then released. Find angular velocity when rod is vertical.[/B]

Homework Equations

The Attempt at a Solution


My attempt to a problem is attached
with this post.

But

Solution says
By energy conservation law
1/2 I ω2= mg L/2
(ω = angular velocity)
Hence
ω=√ (3g/L)

Which one is correct? N why other is wrong?[/B]
 

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but in your soln you have assumed that torque remains constant throughout which is wrong
You must correct it and integerate to get the right answer
 
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Likes   Reactions: Prathamesh and Sahil Dhull
Nope...i ve found the torque acting on the rod when it is in hzt position only... and hence angular acceleration (alpha) also changes with theta...so by putting theta =π/2 i ve found w
 

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