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Angular motion question

  1. Jun 26, 2014 #1
    1. A mine cage of mass 1.5 tonnes is accelerated uniformly upwards from a speed of 1.1 m/s to 4.6 m/s in a time of 20 seconds. The winding drum which is 1.5 metres in diameter has a mass of 1.2 tonne and has the dynamic characteristics of a disc.
    There is a bearing friction torque of 48 Nm.

    For a disc the moment of inertia is given by I = mr^2/2

    Determine:
    a.) the angle turned by the winding drum[/b]



    2.
    I = mr^2/2[/b]



    3. Struggling to begin any attempt on this question.
    alpha = w2^2 - w1^2/2 x sigma?
     
    Last edited by a moderator: Jun 27, 2014
  2. jcsd
  3. Jun 26, 2014 #2

    CAF123

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    There is multiple steps involved here, requiring you to analyse the components of the system in turn and then collectively. Can you visualise what is going on? From the given data, what pieces of information can you deduce? You will need equations from uniformly accelerated motion and rotational dynamics.

    BTW, was there an answer provided for this question?
     
  4. Jun 26, 2014 #3
    the information required I think is the speeds, time and diameter. I have examples of the other questions that accompany this question I'm sure but this part I am lacking examples/information. No there was no answer provided.

    the formula to work out the disttance the cage has moved:
    s = 0.5 x (u + v) x t
    s = 0.5 x (1.1 + 4.6) x 20 = 57m.

    Formulas
    Equations of linear motion:
    v = u + at
    s = ut + 0.5at^2
    s = 0.5(u+v)t
    v^2 = u^2+2as

    Equations of angular motion:
    w2 = w1 + alpha t
    omega = w1 t+0.5 alpha t^2
    omega = 0.5 (w1+w2)t
    w2^2 = w1^2+2 alpha omega
     
    Last edited: Jun 26, 2014
  5. Jun 26, 2014 #4

    CAF123

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    I am imagining a pulley like set up with the cage attached to one extreme of the string wrapped around a disc. The tension force in the string causes the cage to accelerate upwards. This induces a net torque about the centre of the disc with the effect of turning the disc. After computing the linear acceleration of the cage, what is the magnitude of this tension force?
     
  6. Jun 26, 2014 #5
    found an equation: to convert arc length (s) to angle turned (omega) = omega = s/r

    using my previous answer... s = 0.5 x (1.1+4.6) x 20 = 57
    is it as simple as dividing 57 by the radius? (0.75m) to find the answer to my question?
    if so, ive done that calculation 57m/0.75m = 76. is this answer in radians or degrees?
     
    Last edited: Jun 26, 2014
  7. Jun 26, 2014 #6

    CAF123

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    I guess it would be if we were to ignore the physical description of the drum. But you know some geometry of it, so I suppose you are to use it.

    First try what I suggested at the end of #4.
     
  8. Jun 26, 2014 #7
    Sorry I don't understand that post completely, could you elaborate please?
    Thank you very much for helping.

    found this..? (top 1)

    http://media.wiley.com/Lux/25/329725.image3.png
     
  9. Jun 27, 2014 #8

    CAF123

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    For the cage to be uniformly accelerated upwards, there exists a net force on it in the upwards direction. Write Newton's second law for the cage ##\sum F = ma##. You can find the right hand side of this equation using the information provided in the first sentence of the question.

    Therefore, you can find the tension force in the string that causes the cage to accelerate up. There is no information given about the string, so let us assume a typical set up, that is to say a light, massless string. The tension is then the same at all points in the string.

    Have you covered rotational dynamics in your class yet (i.e torques, etc..)?
     
  10. Jun 27, 2014 #9
    The question states the mass of the lift cable can be neglected. Is that formula you've posted mass x alpha?
    Yeah I have. There are other parts to the question than (a.)
     
  11. Jun 27, 2014 #10

    CAF123

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    It is Newton's second law in the familiar form. a here is the linear acceleration of the mass.
    Ah, this made me realize perhaps I am reading too much into the question. Your analysis in your previous post is correct (sorry). The frictional torque acts in the bearing, not opposing the string. So yes, the height gained by the mass is directly related to the change in displacement of a point on the rim of the disc in the obvious way assuming no slip of the string (no relative friction between string and groove of disc).
     
  12. Jun 27, 2014 #11
    the answer i got of 76 is that in degrees or radians? if its in degrees it can be left like that and will be the final answer correct?
    but for the second part to find omega it will need to be changed to rads if its in degrees correct?

    the next part of the question is: the moment of the inertia of the winding drum.
    do you know which formula is the correct one for me to use, i have these 3:
    T = Iα
    I = mk^2 (where k = radius of gyration)
    I = mr^2/2 (friction gravitational?)

    edit: found a formula under for simple discs:
    for angular motion newtons 2nd law (f = ma) becomes:
    T =Ia
    where I = mr^2/2

    to find applied torque i need alpha. using the formula applied torque = I alpha / inertia (+friction torque)
    how do i find alpha

    w = v/r?
    so w1 = 1.1/1.5? and w2 = 4.6/1.5?
     
    Last edited: Jun 27, 2014
  13. Jun 27, 2014 #12

    CAF123

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    It would be in radians. A quick check on this is to note that the formula for the circumference of the disc is a special case of s=rθ. The arc length s is equal to the circumference of the circle when θ=2π.
    Use the last one for the moment of inertia of the disc. The first equation may come in handy later on. The moment of inertia is a geometrical property of the disc. You can attribute many different formulae to its MoI depending on the axis of rotation. The system acts like a pulley, so the disc is rotating about its centre and the corresponding MoI is the one provided in the question. See http://en.wikipedia.org/wiki/List_of_moments_of_inertia for more info.

    You can find alpha by knowing the net torque on the disc. One such torque driving the rotation of the disc is induced by the tension in the rope. The other is the frictional torque which opposes the rotation of the disc.
     
  14. Jun 27, 2014 #13
    the question a asks for the angle turned.
    i have
    s = 0.5 (u + v) t
    s = 0.5 (1.1 + 4.6) x 20 = 57 (m?)

    θ = s/r
    = 57(m?)/0.75m = 76 (radians?)
    since the question asks for an angle will i need to turn the radians into degrees, if so how?

    b.) I = mr^2/2 = 1200 x 0.75^2 / 2 = 337.5

    can you confirm this is right or highlight/correct where im wrong please?
    and how do i find the net torque sorry?
     
  15. Jun 27, 2014 #14

    CAF123

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    Yes, are you unsure of the units? To check that it is metres, carry the units along in your calculation. e.g $$s =0.5 \left(1.1 \frac{m}{s} + 4.6 \frac{m}{s}\right) \times 20 s = 57m.$$ The units behave like algebraic variables and so the s terms cancel.
    Use the fact that 2π radians correspond to 360 degrees. The calculation assumes there is enough string in the first place to bring the cage up by 57m (I.e without the cage coming into contact with the disc). The circumference of the disc is only about 4.5m, so it does about 12 revolutions. So it is a little unclear what is meant by 'angle' in this sense.
    Using what I said above about units, try to find the units here.

    A force acting at a distance r from some point creates a torque about that point. Consider the centre of the disc. There is a tension force where the string meets the disc at a distance r from the centre of the disc. What formula do you have for torque?
     
  16. Jun 27, 2014 #15
    don't understand what you means regarding working out radians? is θ not in radians unit? meaning the 76 will be in radians already so that will be the final answer?

    b. will just be metres?

    for torque i have: Iα/inertia. + friction torque (which will be 48Nm) onto final answer
     
  17. Jun 27, 2014 #16

    CAF123

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    Yes, I believe that would do fine as an answer. You can write down the equivalent number of degrees also using the conversion ##2\pi\,\, \text{rad} \rightarrow 360 \,\,\text{degrees}##.

    No, can you show your working? You can also work out the units just by using your formula I=mr2/2. Ignore the factor of 1/2 for determining units. Notice that in every formula in the wiki link from my last post has the same form: mx2, where x is a length. The differing axes of rotation are manifest in the different numerical constants.

    Ok, but how about torque = r F where r is the perpendicular distance from a point to the point of application of force F.
     
  18. Jun 27, 2014 #17
    ah so will the units be kg/m^2?

    do you know which formula I can use to find F for that formula? I don't have any examples.
    the r will be 0.75m right?
     
  19. Jun 27, 2014 #18

    CAF123

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    Close! Why did you divide by m^2? In the formula, I=mr2, the m and r2 are multiplying and so to do the units.
    See end of post #4 for this. To guide you:
    1) What is the upwards acceleration of the cage? (Find using info in the first sentence of the question)
    2) What is Newton's second law for the cage?
    With 1) and 2) answered, you should be able to find the tension force.

    Yes, r corresponds to the radius of the drum.
     
  20. Jun 27, 2014 #19
    kgm^2?
    the formula is I = mr^2/2, right?

    i need to find w1 and w2 don't I, also?

    F = ma
    a = v x t? = 20 x 4.6 = 92 m/s^2
    F = 1500 x 92 = 138000 Nm
    Torque = r x F
    T = 0.75m x 138000 Nm = 103500 Nm
    Applied torque = 103500 Nm + 48 Nm?

    d.) total work done
    W = Tθ =
    = 1500kg x 76rads = 114000 J = 114 kJ

    e.) the maximum power developed by the driving motor.
    P = T x ω
    (how do i find ω?)

    edit: think ive missed gravity out somewhere (9.81)
     
  21. Jun 27, 2014 #20

    CAF123

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    Correct. Yes.

    It depends, what would you need them for?

    Quick unit check: Dimension of v: m/s , dimension of t: s. So their product vxt is (m/s)(s) = m, i.e a distance so your formula cannot be right. Consider using one of the kinematic SUVAT equations here.

    Have you applied Newton's second law before?
     
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