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Angular momentum and inertia problem

  1. Dec 14, 2011 #1
    Hi this is some coursework i have for an assignment is there anyway you could shed some light on what steps to take?

    So far i have worked out the inertia with the children to be 552 by working out the mass of the roundabout to be 200KGS plus the weight of the children and using 0.5mv^2, not sure where yo go from here but i will post more info later but am posting this from my phone so will be more helpful later. thanks


    Figure 3 shows a plan view of an object, that is, the view looking down on the object from directly overhead.
    The object is a children’s roundabout which is free to rotate about a vertical axle through its centre. The axle is shown by the small circle in the centre of Figure 3.
    Figure 3 For use with Question 4.
    The roundabout has a radius of 2.0 m and a moment of inertia about the central axle of 400 kg m2. Four children, each of mass 19 kg, then stand on the roundabout, each at a distance 1.9 m from its centre. The children are equally spaced, that is, the angular separation of one child from the next is 90°. For the purposes of this question you should treat each child as a point mass. Ignore friction between the roundabout and the axle and any effects due to air resistance.

    (a) What is the relationship between the total external torque acting on a body and the rate of change of the body’s angular momentum? Under what conditions does the angular momentum of a body remain constant in time?

    (b)
    The roundabout is initially at rest with all the children on it, and a constant force F of magnitude 94 N is appliedtangentially at the rim of the roundabout, as shown in Figure 3, for 5.0 s. Find the following quantities for
    the roundabout at the end of this time:
    (i) the magnitude of the angular momentum,

    (ii) the direction of the angular momentum,

    (iii) the angular speed,

    (iv) the rotational energy.

    (c)
    At the end of this interval, i.e. at t = 5.0 s, the force F ceases to act, and at the same time each child moves radially inwards to a new position 20 cm from the axle. Find the following quantities for the roundabout:
    (i)
    the new magnitude of angular momentum, (1 mark)
    (ii)
    the new angular speed, (3 marks)
    (iii) the new rotational energy. (1 mark)
    (d)
    A constant frictional retarding force of 94 N is now applied to the outer edge of the roundabout. How long will it take for this force to bring the roundabout to rest?
    (e)
    Comment on the relationship between the time that you obtained in part (d) and the time that the force acted in (b). Interpret this relationship in terms of the torque acting and the rate of change of a rotational quantity (angular speed, angular momentum or rotational energy)
    ]




    3. The attempt at a solution
     
    Last edited: Dec 14, 2011
  2. jcsd
  3. Dec 14, 2011 #2

    Delphi51

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    Welcome to PF, Tom!
    It is a little difficult to follow the question without the diagram, but I think you have an error in your calc for the total moment of inertia.
    So the I due to the children is ½mr² = ½(4*19)*(1.9)². That just adds to the given 400 for the merry-go-round itself.
     
  4. Dec 14, 2011 #3
    Yeah i see the logic now i used 2 instead of 1.9 so i did wrong. i can't post the diagram at present but its basically a solid disc with radius of 2m laying flat with an axis of rotation perpendicular to the centre, with I of 400kgm, the disc is stationary at start but a force 95N pushing tangently on the side of the disk for 5 seconds, the children all 4 of them stand each 90 degrees apart so 1 in each quadrant at a radius of 1.9m from centre
     
  5. Dec 14, 2011 #4
    Here you go Diagram
     

    Attached Files:

  6. Dec 14, 2011 #5

    Delphi51

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    Go ahead and calculate the torque, angular acceleration, angular velocity, etc.
    I can help you avoid dead ends and check for errors, but it is extremely important that you do all the work yourself.
     
  7. Dec 14, 2011 #6
    Yeah I understand was just looking for some guidance was not at home earlier so can only start working on it again now, was just after pointers, will see how I get on working the above out.

    The one thing that is holding me back though it think is how do I relate the 95N of force applied for 5 secs to the velocity or angular motion of the disk, obviuos it will involve acceleration (maybe torque???) but am still stuck, any help here would be great as I'm struggling to work out angular speed etc (omega)

    Thanks
     
  8. Dec 14, 2011 #7
    I seem to have found an equation for Torque which is T=Fr sin (feta) as the force of 94N is applied tangentially to the rim of the disc it must be at 90 degrees to the centre of rotation
    So T=94N x 2 sin (90) = 188 Rad/s as the force is applies for 5 seconds can i use this to work out the acceleration and velocity to be 188 x 5 = 940 rad/s ????????
     
  9. Dec 14, 2011 #8
    finding angular frequency or Omega using Omega(w)=2pi/T so w=2pi/188 i get 0.033 rads/s

    Am I going along the right lines?

    Thanks
     
  10. Dec 14, 2011 #9
    at this point I don't really now what to do with the 5 secs the force of 94N was applied so i times 0.033 rad/s by 5 to get an angular speed of 0.033 x 5 = 0.1671

    then using this to find angular momentum i get L=IW(omega) so L=537.18 x 0.1671 = 89.76

    B) ii) force is in the anti clockwise direction with torque facing up

    B) iii is just the result i have already form omega

    B) iv rotational energy is 1/2 I W(omega) = 0.5 x 89.76 x 0.1671 = 1.2533 J

    onto

    C) i and i have found the new Moment of inertia using I=1/2 mr^2 to be 1/2(4*1.9)0.2^2 (for children) = 1.52 so 1.52+400 = I = 401.52 now it states at this point that the 94N of force stops so i take it that the Torque stops??? but then i cannot work out W(omega) as i need W=2pi/T i know W will stay the same with torque stopping but i know that the children moving from radius of 1.9m to 0.2m will increase speed so im stuck again.
     
  11. Dec 14, 2011 #10
    OK firstly this B) iv rotational energy is 1/2 I W(omega) = 0.5 x 89.76 x 0.1671 = 1.2533 J

    Is meant to be B) iv rotational energy is 1/2 I W(omega) = 0.5 x 537.18 x 0.1671 = 7.500128 J

    C)ii is used L= mvr instead and got L = 184.4784 using L=276(m) x 0.33(v) x 2(r)

    C)iii 1/2 IW^2 i got 0.5 x 401.52 x 0.459450^2 = 42.3792 J

    D i got 28.252317 seconds

    I need to sig fig it all yet and im pretty sure 90% of it is wrong, im going to write up a bit now and then time for bed

    Cheers
     
  12. Dec 14, 2011 #11

    Delphi51

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    Sorry, I've been away for a few hours.
    You've got the torque right on with your 188!
    You will have to use a formula to find the rotational acceleration from the known torque.
    Check this list of formulas and pick one that has torque and angular acceleration:
    http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#rlin

    Once you find the rotational or angular acceleration, you can use it in another formula to get the rotational velocity, ω. It is not .033 or .167.
     
  13. Dec 15, 2011 #12
    Ok thanks so is use (alpha)A=a(normal a)T/r

    or alpha = circular acceleration (linear) times Torque divided by Radius

    I take it i need to work out Linear acceleration which is a=v^2 / r

    I don't know how to get velocity, is it just 94N times 5 secs to be 470 Rad/s


    Also I have found an equations saying Alpha(a)=T/I or Alpha = Torque divided by Inertia in which case I would have Alpha = 188 / 537.18 = 0.3499757 Is this correct??

    Sorry to be so clueless
     
  14. Dec 15, 2011 #13
    all done to C) feeling confident but dont have a clue about c)
     
  15. Dec 15, 2011 #14

    Delphi51

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    Good on alpha = .34996!
    It corresponds to having the acceleration and time in a linear problem.
    You would then use v = a*t in the linear problem to get velocity.
    Use the corresponding rotational formula to get ω.

    Next use the rotational analog of Ek = ½mv² to get the energy.
    After that, the linear formula to get momentum would be p = mv.
    Use the rotational analog to get angular momentum in this problem.
     
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