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How can you find angular speed

  1. Mar 2, 2008 #1
    How can you find angular speed (average), if its a question of rotation about a fixed axis, and all you know really is the rotational inertia of the system? You have no values of the angles (with regards to the reference axis) and no values for time either.
     
  2. jcsd
  3. Mar 2, 2008 #2
    Huh? You know nothing but the rotational inertia and you wanna know the rotational speed? Did I get this right? Because it doesnt really make any sense to me. The inertia is a static attribute which is not related to rotational speed (only through the influence of force).
     
  4. Mar 2, 2008 #3

    Doc Al

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    You'll need more information than just the rotational inertia. (It's like asking: How can I find the speed of an object given its mass?) Do you have a specific problem in mind?
     
  5. Mar 2, 2008 #4
    Yeah, here's the question:
    Four particles, each of mass 0.20 kg, are placed at the vertices of a square with sides of length 0.50 m. The particles are connected by rods of negligible mass. This rigid body can rotate in a vertical plane about a horizontal axis A that passes through one of the particles. The body is released from rest with rod AB horizontal (Figure 1). (the figure is just a square, with particles at each of the four corners and the bottom left particle is labelled A)

    Find the rotational inertia and angular speed.
     
  6. Mar 2, 2008 #5
    Oh the exact phrase is "What is the angular speed of the body about axis A when rod AB swings through the vertical position?"
     
  7. Mar 2, 2008 #6

    Doc Al

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    And I presume that the bottom right particle is labeled B?

    When?
     
  8. Mar 2, 2008 #7

    Doc Al

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    OK. Assuming that it rotates freely about the axis (no friction), what's conserved?
     
  9. Mar 2, 2008 #8
    Lol, no the other particles are not labelled anything. And it doesn't tell you 'when'.

    I'm assuming, kinetic energy will be conserved? (er..I think)
     
  10. Mar 2, 2008 #9
    Certainly. I'd like to help but Im having a hard time imagining the situation. Make up a sketch please.
     
  11. Mar 2, 2008 #10

    Doc Al

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    Sure it does: "when rod AB swings through the vertical position".

    Since the structure is released from rest, kinetic energy had better not be conserved! (Hint: Total mechanical energy is conserved.)
     
  12. Mar 2, 2008 #11
    Well I don't quite understand what "through the vertical position" means, is it when it passes through the line perpendicular to A?

    And yes it is released from rest; mechanical energy is the sum of the kinetic and potential energy right? So this would turn out to be something like:

    Initial Kinetic Energy + Initial Potential energy = Final Kinetic energy + Final potential energy
    which is

    (0.5)(0.2)(0)squared + (0.2)(9.8)(theta1) = (0.5)(0.2)(omega)squared + (0.2)(9.8)(theta2) ?
     

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  13. Mar 2, 2008 #12

    Doc Al

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    When it rotates 90 degrees clockwise, line A-B will be vertical.

    Right.
    Hint: How does the height of each mass change? Use that to find the change in PE.
    Hint2: Write the final KE in terms of rotational KE. What's the rotational inertia of this structure?
     
  14. Mar 2, 2008 #13
    Erm, I found the rotational inertia using the formula I = Sum of m r(squared) where r is the perpendicular distance from the axis, and I got:

    (0.2)[(0)+(0.5)+(0.5)+(0.707)] = 0.3414 And since rotational K.E. is = (0.5)I(omega)squared, it will be (0.5)(0.3414)(omega)squared...

    When you said "how does the height of each mass change", aren't we looking at the entire system as a whole and not really taking into account the mass of each particle? Or since we've been told about the vertical distance travelled by line AB, so theta 2 would be 90 degrees?
     
    Last edited: Mar 2, 2008
  15. Mar 2, 2008 #14

    Doc Al

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    Square those distances. (Looks like it was correct before.)

    Since the system is just a collection of point masses, you'll get the same answer either way: Find the change in PE of each mass or find the change in PE of the entire structure (by tracking the change in height of its center of mass).
    I wouldn't bother using angles to measure change in PE. You'll just have to convert to heights anyway.
     
  16. Mar 2, 2008 #15
    Right...forgot to square the r s, sorry!

    Just wondering, since the rotation in question is from its initial state to when AB is perpendicular to the axis at A (or vertical), wouldn't the distance between the initial centre of mass and the final centre of mass be equal to the length of each side of the square?
     
  17. Mar 2, 2008 #16

    Doc Al

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    Sounds good to me.
     
  18. Mar 2, 2008 #17
    By the way, thank you Doc Al for your help, right now I'm kind of assuming that the c.o.m. is equal to the length and I managed to get an answer for the angular speed: 6.28. I'm not sure if it's correct I hope it is, but I kind of have to go now and I'll check back in the morning. I understand the concept better now atleast!
     
  19. Mar 2, 2008 #18
    Thanks very much!
     
  20. Mar 2, 2008 #19

    Doc Al

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    Sounds good to me. (What are the units?)

    (And you're welcome, of course!)
     
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