Solve Angular Speed of Rotating Rod: Conservation of Energy

  • Thread starter Thread starter rasensuriken
  • Start date Start date
  • Tags Tags
    Rod Rotating
AI Thread Summary
To find the angular speed of a rotating rod at different angles, the formula \varpi^2=\varpi_0^2+2\alpha\vartheta is not applicable due to the non-constant angular acceleration, α, resulting from varying torque as the rod swings. The torque changes with the angle, making it necessary to use conservation of energy, which considers only initial and final conditions, rather than instantaneous acceleration. The discussion highlights that while initial values can be used correctly, the complexities of angular motion require a different approach for varying conditions. Participants clarify that the torque's dependence on angle complicates the use of constant acceleration formulas. Ultimately, conservation of energy proves to be the more reliable method for this scenario.
rasensuriken
Messages
10
Reaction score
0
The question is as the attachment...
May i know if i want to find the angular speed of the rod when it inclined at certain angle, or at horizontal, can i use the \varpi^2=\varpi_0^2+2\alpha\vartheta?

Attempt: I used it, and used conservation of energy, the latter produces right answer, but i don't know why first one won't work. Any clues?
 

Attachments

  • rotating.jpg
    rotating.jpg
    42.6 KB · Views: 426
Physics news on Phys.org
Hi rasensuriken! :smile:

I can't read your formula … but, looking at the .jpg, I don't think the angular acceleration, α, is constant (because the torque from the weight is different for different angles).

So a formula using constant α won't work! :smile:
 
tiny-tim is right, the author of the JPEG image made a mistake. He forgot that the torque is a cross product of radius and the force, which depends on the angle. And the angle changes as the whole thing swings down, so the alpha should be more like:
\tau = MG\frac{L}{2} \sin{\omega t}
Which, as you can see, depends on the time and so is not constant. The conservation of energy worked because it does not care about the acceleration at any point, but only on the initial and final conditions.
 
Hi kkrizka! :smile:

No, he didn't forget … he only asked for the inital values, for which his equations are correct.

The question in the OP is rasensuriken boldly going further! :smile:
 
tiny-tim said:
Hi kkrizka! :smile:

No, he didn't forget … he only asked for the inital values, for which his equations are correct.

The question in the OP is rasensuriken boldly going further! :smile:

Ah whoops, I've been reading too many phys problems lately and they are starting to get mixed up. :P
 
ya...i also just thought of that...the acceleration is not constant...because the L/2 relative to the center of mass keep on changing...thanks!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top