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Rotational Kinetic Energy Problems

  1. Mar 9, 2007 #1
    Hello. I'm an intro physics student, who needs help with physics problem, because my graduate student refuses to help anybody in my class anyway... here is my first question on rotational kinetic energy

    1) "A uniform disk of mass M and radius R can rotate freely about a horizontal axis through its center and perpendicular to the plane of the disk. A small particle of mass m is attached to the rim of the disk at the top, directly above the pivot. The system is given a gentle nudge, and the disk begins to rotate."

    a) "What is the angular velocity of the disk when the particle is at its lowest point?"
    b)"At this point, what force must be exerted by the disk on the particle to keep it stuck to the disk"

    ..... First of all where should I start with part a)? And how does rotational kinetic energy apply to this problem. Second, how do I find the force for part b)

    ....

    My second question is....

    2) "A uniform ring 1.5 m in diameter is pivoted at one point on its perimeter so that it is free to rotate about a horizontal axis. Intially, the line joining the support point and the center is horizontal."

    a) "If the ring is released from rest, what is its maximum angular velocity?"
    b) "What minimum initial angular velocity must it be given if it is to rotate a full 360 degrees?"

    ... Once again any tips would be helpful... I appreciate it. Thanks.
     
  2. jcsd
  3. Mar 9, 2007 #2

    Doc Al

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    My hint for both problems: What's conserved?
     
  4. Mar 9, 2007 #3
    Rotational Kinetic Energy is conserved... but how does the small particle apply in the first problem and how can I formulate into an equation to solve for the angular velocity?

    Also how does one for problem #2, when the maximum angular velocity occurs?
     
  5. Mar 9, 2007 #4

    Doc Al

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    No, total mechanic energy is conserved. What other form of energy is relevant here?

    Both involve conservation of energy. To express the rotational KE, you'll need to figure out the rotational inertia of both systems about the required axis.
     
  6. Mar 9, 2007 #5
    ..... So it
    U(final) + K(final) = U (initial) + K(initial)

    ... So, in the first problem, do you treat U(initial) as zero for both the disk and the particle, and the U(final) for the particle as mg(-2*pi*R)? Is this correct? And for the first problem, does the equation for solving for angular velocity like this.....

    mg(-2*pi*R)+((MR^2*angular velocity^2)/2)+((m*v^2)/2)=0

    ... Is this correct? for the second part, of this problem, is it referring to torque... because I'm a little confused to what force must be applied to the particle?

    Thank you. I appreciate your help and knowledge greatly.
     
    Last edited: Mar 9, 2007
  7. Mar 9, 2007 #6

    Doc Al

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    Good.

    That would work fine.

    Almost. What's the rotational inertia of a disk? Also, express the KE of the particle in terms of angular velocity--since disk and particle rotate together.
     
  8. Mar 9, 2007 #7
    Sorry.. I really don't understand what you mean by this. I though that the
    rotational inertia of a disk is "1/2*M*R^2" and wouldn't this correlate to final KE of the disk as "1/4*M*R^2*(angular velocity)^2" as the KE? Would the disk have an intitial KE?
    And for the particle, would the KE= be 1/2*m*R^2*(angular velocity)^2? Is this right....
    so would the equation be

    Ui=0
    Ki=0
    Uf+Kf=0... So.......

    mg(-2*pi*R)+(1/4*M*R^2*(angular velocity)^2)+(1/2*m*R^2*(angular velocity)^2)=0?

    Is this right?.. Once again thank you.
     
    Last edited: Mar 9, 2007
  9. Mar 9, 2007 #8

    Doc Al

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    Right! (Check your equation and you'll see that you had it wrong before.)
    Ignore the "gentle nudge".

    Good.

    Now you've got it.
     
  10. Mar 9, 2007 #9
    Is is meant by what force must be exerted... do they mean torque? And how do I go about it... hints?
     
  11. Mar 9, 2007 #10

    Doc Al

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    They mean force. Hint: Is the particle accelerating?
     
  12. Mar 9, 2007 #11
    ... No, it is not accelerating, because it is held in place, but if you plug into Newton's second law.. F=ma, a=0, so what force are they referring to, meaning it just be the normal force, Fn=mg???..

    Additionally, for the second problem, I had stated previously......I really don't how you can relate the rotational kinetic energy for the ring, if I don't have the stated mass..... then how do I find the maximum angular velocity??
     
  13. Mar 9, 2007 #12

    Doc Al

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    Hint: The particle is moving in a circle. (It's attached to the rotating disk.)

    Try it and see. Maybe you won't need the mass. :wink:
     
  14. Mar 9, 2007 #13
    ... Do you mean that I take my answer from part a) in the first problem and relate that the centripetal acceleration, a=r*(angular velocity^2), so I mutiply the angular velocity found in part a) and then take m*a=F, so ....

    F= R*(angular velocity ^2)*(M + m), is this right...?
     
  15. Mar 10, 2007 #14

    Doc Al

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    Absolutely.
    Good. Realize that here F is the net force. There may be more than one force acting on the particle. You want the force exerted by the disk on the particle.

    Two problems: (1) The one I mentioned above and (2) what's M doing in there?
     
  16. Mar 10, 2007 #15
    So.. Doc Al.. you're saying that it is only the centipetal acceleration that is keeping the particle on the disk??
     
  17. Mar 10, 2007 #16

    Doc Al

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    I'm saying that the particle is being centripetally accelerated, which requires a certain net force on the particle. (Given by F = ma.) Use that information to figure out how strongly the disk must pull on the particle. (Maybe its glued to the disk! But if it wasn't attached somehow, the particle would have slipped off long ago.) Don't neglect any other forces acting on the particle. (I only see two forces acting.)
     
  18. Mar 10, 2007 #17
    Is the other force, the force of gravity meaning...

    The disk must exert a force=F

    F= m*R*(angular velocity^2)+(m*g)???

    .... Additionally I still don't quite understand the second part to the question I previously posted. The line connecting the pivot point and center, has me real confused...???
     
  19. Mar 10, 2007 #18

    Doc Al

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    Exactly!
    You mean the hoop problem? (If you discuss multiple problems in a thread, be sure to be clear about which one you are referring to at any time.)

    A picture would help, but I assume you have a hoop free to rotate about a pivot point somewhere on its perimeter. The center of the hoop starts out being at the same level (on a horizontal line) with the pivot point, then the thing is allowed to swing freely.
     
  20. Mar 10, 2007 #19
    Just wanted to say thank you, Doc Al, for your help. You really were a good teacher and helper. Thanks.
     
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