Angular velocity of suspended rod

Click For Summary
The discussion centers on calculating the angular velocity of a uniform rod when it is released from a horizontal position to a vertical position. One participant suggests using the conservation of energy principle, leading to the formula ω = √(3g/L) for angular velocity. Another participant challenges this approach, arguing that the torque is not constant throughout the motion and that the angular acceleration changes with the angle of the rod. They emphasize the need to integrate the torque to find the correct angular velocity. The debate highlights the importance of considering variable torque and angular acceleration in such problems.
Prathamesh
Messages
20
Reaction score
1

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]
 

Attachments

  • IMG_20160626_173619.jpg
    IMG_20160626_173619.jpg
    21.5 KB · Views: 481
Physics news on Phys.org
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
 
  • Like
Likes 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
 
Beams of electrons and protons move parallel to each other in the same direction. They ______. a. attract each other. b. repel each other. c. neither attract nor repel. d. the force of attraction or repulsion depends upon the speed of the beams. This is a previous-year-question of CBSE Board 2023. The answer key marks (b) as the right option. I want to know why we are ignoring Coulomb's force?

Similar threads