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Rotational inertia & tangential force

  1. Mar 20, 2007 #1
    1. The problem statement, all variables and given/known data

    A horizontal disc of diameter 10 cm is spinning freely about a vertical axis
    through its centre at an angular speed of 80 revolutions per minute. A piece
    of putty of mass 7.0 g drops on to and sticks to the disc a distance of 3.0 cm
    from the centre. The angular speed reduces to 65 revolutions per minute.

    Calculate the moment of inertia of the disc.


    2. Relevant equations

    I= MR2

    3. The attempt at a solution

    R= 0.03
    M= 0.007

    but if i use the formula I= MR2,,,, what about the informaton they have given of rev/ min this is the angular speed....

    not sure the method to do this???
     
  2. jcsd
  3. Mar 20, 2007 #2

    Doc Al

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    Staff: Mentor

    conservation

    When the putty is dropped onto the rotating disc, some things change and some things remain the same. What stays the same?
     
  4. Mar 21, 2007 #3
    mass remains same, diamiter remains the same
    angular velocity decreases
     
  5. Mar 21, 2007 #4

    Doc Al

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    what's conserved?

    All true, but what important dynamical quantity is conserved?
     
  6. Mar 21, 2007 #5

    SGT

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    Mass of the disk remains the same, but the mass and moment of inertia of disk + putty is not the same.
     
  7. Apr 11, 2007 #6
    so calculating this....

    is my formula corect?
     
  8. Apr 11, 2007 #7

    hage567

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    Homework Helper

    What formula are you referring to?? I=MR^2 is correct for the putty if that's what you mean. But you haven't yet taken into account the moment of inertia of the disk.
     
  9. Apr 25, 2007 #8
    mass remain the same-

    how do i calcualte the moment of inertia...with the informaiton given>>
     
  10. Apr 25, 2007 #9

    Doc Al

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    Staff: Mentor

    More importantly, angular momentum remains the same. Compare the angular momentum of the system before and after the putty is dropped.
    Since the moment of inertia of the disc is what you are trying to find, just call it I_d. But realize that the moment of inertia of "disc + putty" is the sum of the individual moments of inertia. You can calculate the moment of inertia of the putty using I_p = MR^2.
     
  11. May 1, 2007 #10
    tangential force

    1. The problem statement, all variables and given/known data

    A horizontal disc of diameter 10 cm is spinning freely about a vertical axis
    through its centre at an angular speed of 80 revolutions per minute. A piece
    of putty of mass 7.0 g drops on to and sticks to the disc a distance of 3.0 cm
    from the centre. The angular speed reduces to 65 revolutions per minute.

    2. Relevant equations

    A constant tangential force is now applied to the rim of the disc
    which brings the disc to rest in 8.0 s. Calculate the magnitude of
    this force.

    3. The attempt at a solution

    need help..dont know how to start
     
  12. May 1, 2007 #11

    cristo

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    Science Advisor

    What do you know about circular motion? You are given the angular speed of the disc, and the time in which it is brought to rest. Can you see how this will help you find the force?

    Please show some work, even if it's just relevant equations/ knowledge from your text or course notes. You must know how this forum works by now!
     
  13. May 1, 2007 #12
    didnt have a clue wear to start,,thats for the tip...

    f=m*r*w^2

    = 7000*0.05*w^2

    w= 80-65 / 2= 7.5

    f= 3.5*7.5^2
    = 196.875N
     
  14. May 2, 2007 #13
    can anyone tell me if this correct?
     
  15. May 2, 2007 #14

    hage567

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    No, that doesn't look right.

    How did you get 7000 for the mass? You shouldn't even be using mass here, you should be using moment of inertia, I. Why are you solving for w? You know that already. You're dealing with angular momentum and torque, right? How do you think you can relate things like force, torque, and angular acceleration?

    Also, be careful of your units.
     
  16. May 4, 2007 #15
    torque= force *lever arm

    angular acceleration= a/r

    a=acceleration
    r=radius

    a= v-u/ t

    change in velocity= 65-80= -15 rev/ min

    -15 rev/min= -15 rev every 60 sec = -0.25 rev/ sec

    -0.25/8= -0.03125 rev/sec^2

    angular acceleration= -0.03125 rev/sec^2

    tangential force= mass*angular acceleration
    = 0.07 * -0.03125
    0.0021875 N

    this answer dosnt look right.
    Need help please.
     
  17. May 4, 2007 #16

    hage567

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    OK first of all, the disc is coming to a STOP when the force is applied, so that means your final angular velocity is ZERO. Take the time to look at what the question is asking you!

    It is best to convert angular velocities into rad/sec.

    You STILL didn't convert 7g to kg properly, but you need to consider more than just the mass, you are dealing with rotational quantities.

    Remember that torque can be expressed as

    [tex]\tau = I\alpha[/tex]

    where I is the moment of inertia. That is what you need to consider here (I pointed this out in my last post as well). Use this along with the first equation you listed. I know there was a first part to this question that you posted before, so you should already know the moment of inertia you need to use to solve this.
     
  18. May 5, 2007 #17
    7g= 7kg/1000
    = 0.007kg

    Torque= I*alpha..

    what is alpha...

    the quesiton also ask to calcualte tan...force...why do we need torque?
     
  19. May 5, 2007 #18

    Hootenanny

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    alpha is the angular acceleration.
    What is torque... :wink:
     
  20. May 5, 2007 #19

    hage567

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    alpha is the angular acceleration. The force is creating a torque on the disc that is bringing it to a stop.
     
  21. May 6, 2007 #20
    is torque a force.......

    if so,

    then

    torque= I* alpha

    I= MR^2

    angular accelereation= change in angular velocity/ time
    = (65- 80) / 8= -1.875 rev/sec^2
     
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