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Spinning disk help

  1. Apr 22, 2014 #1
    A solid disk with mass M and radius R is moving at t=0 on a horizontal surface in the positive x-direction with speed w_o*R/4. The friction coefficient is b.
    At t=0 the disk is also spinning counter-clockwise with angular velocity w_o. Thus, when viewing the disk at time t=0 you will see it move to the right (I call this the + x-direction) but it is spinning counter-clockwise.
    Question A
    There comes a time, t, that the disk will come to a halt. How far will the disk have moved (since t=0) when it comes to a halt?
    Question B
    What is the angular velocity, w, of the disk at time t when it comes to a halt. If the angular velocity is clockwise indicate that with a + sign, if it is counter-clockwise, indicate that with a - sign. If it is zero, then answer w=0
    question C
    There comes a time (t_1) that the angular velocity will become zero, thus the disk will not rotate anymore. How far will the disk have traveled in the -x direction when t=t_1?
    question D
    What will be the velocity of the center of mass at t_1? Signs in the velocity are important unless the speed is zero.
    question E
    If we compare the w was -w_o and when the velocity of the center of mass was +w_o*R/4, how much kinetic energy has the disk lost at time t_1?
    i have calculated first two parts but couldn't figure out the rest three
    So the first two and the equations I have considered are as follows
    initial conditions
    solid disk has radius R and mass M
    at time t=0
    omega is -w_o (counter clock wise direction)
    v_o=+w_o*R/4 translational motion of the center of mass in the + direction
    Mg*b is the friction between disk and the surface. The disk is rotating
    counter-clockwise AND the center of mass is moving in the + direction.
    Thus the frictional force is in the - direction.
    v_t=v_o+at
    a=-Mg*b/M=-g*b (a is in the - direction)
    ASNSWER TO QUESTION A
    When the object comes to a halt, all translational KE has been converted
    into heat. The frictional force Mg*b is constant throughout the sliding.
    Thus the KE of translation [0.5*M(w_o*R)^2]/16=Mg*b*x
    Thus x (distance traveled) is(w_o*R)^2]/32g*b
    -----------------
    QUESTION B
    The disk will move in the + direction but it will stand still when
    0=w_o*R/4 - g*b*t
    thus the disk will come to a halt at t=w_o*R/4g*b (eq 1)
    If at that time the angular velocity (w) is still negative, the object
    will roll back in the negative direction where it came from. If omega
    is positive the disk will continue to move in the + direction
    (clockwise rotation).
    w_t= -w_o + alpha*t (eq 2)
    alpha is dw/dt
    M*g*b*R=I*alpha (torque equation)
    I_disk=(M*R^2)/2
    alpha=gb*2R)/(R^2)= g*b*2/R (eq 3)
    Substitute t of (eq 1) into (eq 2)
    Substitute alpha (eq 3) in (eq 2)
    ANSWER TO B
    You then find
    w_t= -w_o + w_o/2= -w_o/2
    Conclusion the disk is now standing still and it is rotating at half
    the angular velocity than at time t=0 but it is still rotating counter clockwise as it will roll back
    Please I need some hints to solve for other parts
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 22, 2014 #2

    Simon Bridge

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    Welcome to PF;
    Is this rolling with slipping?
    i.e. what is orientation of the spin axis wrt the translation direction?
     
  4. Apr 23, 2014 #3
    Thanks !
    Yes it is rolling without slipping and spin axis is y (anticlockwise ) and translation direction is + x axis
     
  5. Apr 23, 2014 #4

    haruspex

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    I feel the question is badly worded. Assuming it is a uniform solid disk it will not come to a halt in part A. Its horizontal velocity will become instantaneously zero, but it will still be back-spinning and will then move off to the left. I would not call that coming to a halt.
    Anyway, not only will all of the original linear KE have turned to heat, it will also be rotating more slowly.
     
  6. Apr 23, 2014 #5

    Simon Bridge

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    If it is rolling without slipping - then the rim of the wheel is not in contact with the surface it is rolling against. The spinning makes no difference. The object is slowed by friction until it stops - there it remains. What is making it return?

    If it is rolling with slipping then the rim of the wheel may be in contact with the surface - this is why I asked about the orientation of the spin axis - then, off the sense of the description, translational and rotational speeds decrease until the object comes (instantaneously) to translational rest then returns the way it came without slipping.
     
  7. Apr 25, 2014 #6
    I am sorry then it is rolling with slipping
    And since
    "ANSWER TO B
    You then find
    w_t= -w_o + w_o/2= -w_o/2
    Conclusion the disk is now standing still and it is rotating at half
    the angular velocity than at time t=0 but it is still rotating counter
    clockwise and thus it will roll back in the direction where it came
    from."
    It is because of the counter clockwise rotation even when it came to halt is causing the spin to go backwards
     
  8. Apr 25, 2014 #7
    And by halt I mean "When the object comes to a halt, all translational KE has been converted
    into heat. The frictional force Mg*b is constant throughout the sliding."
    Now I can see that it is definitely rolling with slipping / sliding
     
  9. Apr 25, 2014 #8

    haruspex

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    Yes, but as I wrote in post #4, it won't only be the translational KE that's gone into heat. Some of the rotational energy will have gone that way too. So it will have travelled further than the answer you give in the OP.
     
  10. Apr 26, 2014 #9
    Actually this condition is already given in the text of the question and I have assumed it to be correct and calculated my answers based on that
     
  11. Apr 26, 2014 #10

    haruspex

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    Please clarify exactly what the problem says as given to you. I had assumed it said this:
    However, I now realise that while I was right in saying there's more energy gone into heat than just the linear KE, that does not mean it has travelled further than calculated in the OP. The work done by friction is the frictional force multiplied by the distance of relative movement of the two surfaces. Because of the back-spinning, that distance is greater than the distance moved horizontally. So although the answer in the OP is correct the reasoning used is unsafe. Far more reliable is just to use the fact that the linear deceleration is gb and apply SUVAT.

    So, moving on to question C:
    I find the question a little odd. It is certainly true that there will come a time when the disk is no longer slipping, but it is not immediately obvious that there will come a time when it is no longer rotating. That will only happen when the initial rotation has a certain relationship to the initial linear speed. Anyway, to proceed:
    You have shown that when the linear speed reaches zero it is still spinning. What does that tell you about the frictional force at that time?
    What does that imply for (a) the linear acceleration and (b) the angular acceleration?
    Looking at the angular speed at time t, and the angular acceleration, how long before it should stop turning (assuming the angular acceleration stays the same until then)?
    Next, how can you tell whether the angular acceleration will change before that time?
     
  12. Apr 26, 2014 #11
    Yes initial rotation is related to linear speed as
    Initial speed =w_o*R/4
     
  13. Apr 27, 2014 #12

    haruspex

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    Yes, I know that's what's given, but it need not be that special value which can lead to the disk's rotation ceasing.
     
  14. Apr 29, 2014 #13
    I calculated those last questions I got them as
    X= -(wR)^2/8gb
    V of centre of mass as 0 since it come to halt
    And Ke total as 5M(wR)^2/32
    are these correct ?
     
  15. Apr 29, 2014 #14

    haruspex

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    For question C, it's not clear where we're measuring the distance from. Is it from the starting position or from the point where forward motion ceased? Either way, I don't get your answer. Pls post your working.
    (Consider how the force and torque during this phase compare with those during the first phase.)
     
  16. Apr 30, 2014 #15
    Calculated results

    Here it goes
     

    Attached Files:

    Last edited: Apr 30, 2014
  17. Apr 30, 2014 #16

    haruspex

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    No, that's wrong. Reread my post #10 very carefully. The distance you have calculated is the relative movement between the two surfaces. This is much greater than the distance the disk moves horizontally.
    I suggest it will be less confusing to work with torques, accelerations etc. rather than with energy.

    Btw, I believe the question to be wrong in part C. I do not see how it can be true that the disk will stop rotating. It is a mistake to assume that it will continue to slip until it comes to a complete rest. At some point it will start to roll, not slip.
     
    Last edited: Apr 30, 2014
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