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Mass Swinging on a Rope

  1. Jun 21, 2015 #1

    Drakkith

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    1. The problem statement, all variables and given/known data
    A mass hangs on the end of a massless rope. The pendulum is held horizontal and released from rest. When the mass reaches the bottom of its path it is moving at a speed v = 2.7 m/s and the tension in the rope is T = 19.8 N.

    1. How long is the rope?
    2. What is the mass of the object?

    pendulum1.png
    2. Relevant equations


    3. The attempt at a solution

    I found Q1 by the following method:

    ΔE = ΔU + ΔK
    ΔU = U2 - U1
    ΔK = K2 - K1

    At the initial postion, U1 = mgh and K1 = 0
    At the 2nd position U2 = 0 and K2 = 1/2 MV2

    ΔU = 0 - mgh
    ΔK = 1/2mv2 - 0
    So ΔE = 1/2mv2 - mgh

    Since only conservative forces are doing work here, E1 = E2 and ΔE = 0
    That means 0 = 1/2mv2 - mgh
    mgh = 1/2mv2
    Canceling m: gh = 1/2v2
    h = v2/2g
    h = 2.72 / (2 x 9.81) = 0.372 meters, which is correct.

    For Q2, I tried to find the mass the following way:
    T(tension) = Fc - mg
    Fc = mac
    ac = v2 / r
    So T = mv2/r - mg
    Then T = m(v2/r - g)
    Finally: m = T/[(v2/r) - g]
    m = 19.8/[(2.72/0.372) - 9.81]
    m = 2.02 kg

    But 2.02 kg is wrong and I don't know why. Using 2.02 kg in my equation T = Fc - mg gives me 19.769 N, which rounds to 19.8 N, which is what was given for tension.
     
  2. jcsd
  3. Jun 21, 2015 #2

    Drakkith

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    Should this be Fc + mg?
     
  4. Jun 21, 2015 #3

    haruspex

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    Check your signs.
    Remember, centripetal force is (the radial component of) the resultant of the applied forces. It's generally less prone to error if you write it that way: centripetal = ##\Sigma F## (since there is no other acceleration here).
     
  5. Jun 21, 2015 #4

    haruspex

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    yes.
     
  6. Jun 21, 2015 #5

    Drakkith

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    You're saying Fc = mg + mac?
     
  7. Jun 21, 2015 #6

    haruspex

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  8. Jun 21, 2015 #7

    Drakkith

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  9. Jun 21, 2015 #8

    Nathanael

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    Did you write this equation off the top of your head? Fc is not a force which acts in the system, rather it is what the other forces must combine to in order for circular motion to occur.

    I find it analogous to how if the velocity is constant then you know that no matter what that sum of the forces is zero: ΣF=0

    Similarly, if an object travels at speed v in a circular arc of radius R, then no matter what the sum of the forces in the centripetal direction must be: ΣFcentripetal.component=mv2/R

    This is why I find it strange that you wrote an equation for the tension straight away. If you did the algebra in your head that's fine, but the first equation that should come to mind is an equation for Fc, not an equation directly for T.

    I don't know if I explained very well what I meant but maybe it helps.
     
  10. Jun 21, 2015 #9

    Drakkith

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    Probably.
     
  11. Jun 21, 2015 #10

    Drakkith

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    I don't understand. Why would I consider Fc prior to T?
     
  12. Jun 21, 2015 #11

    SammyS

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    When the mass reaches the bottom of its path, what is its acceleration? Don't forget the direction.

    What forces are available to produce that acceleration ?
     
  13. Jun 21, 2015 #12

    haruspex

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    There are two actual forces acting on the mass, gravity and tension. The net force is the sum of these.
    ##\vec F_{net}=\vec T + m\vec g##.
    In order to follow a path which is an arc of a circle radius r, the component of acceleration orthogonal to the velocity must have magnitude ##\|\vec a_c\|=\frac{\vec v^2}r##. There may also be tangential acceleration, in general.
    ##\vec F_{net}. \vec r = -m\vec v^2##
    (The minus sign is because the net force is towards the centre of the arc, whereas the vector r points away from it.)
    The 'centripetal force' is this radial component: ##\|\vec F_c\|r = -\vec F_{net}. \vec r##, and ##\vec F_c = m\vec a_c##.
    (As I wrote at the page I linked to, it may be better never to think in terms of centripetal force, only centripetal acceleration.)

    In the problem in this thread, at the point of interest, there are no tangential (=horizontal) forces, so no tangential acceleration.
    So the equation reduces to ##m\frac{\vec v^2}r=\|\vec T + m\vec g\|##.
    Note that T and mg act in opposite directions, and T wins, so ##m\frac{\vec v^2}r=\|\vec T\| - m\|\vec g\|##.
     
  14. Jun 21, 2015 #13

    Nathanael

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    Sorry I wasn't clear. You're still considering them at the same time (the equation for Fc will involve T).
    My point was this: instead of starting with the equation T=mg+mac you should start with the equation mac=T-mg and then get to the other equation by algebra.
    The latter equation for mac makes sense (it says the forces must produce a centripetal acceleration) but the former equation for T makes less sense.

    Sorry if I'm being confusing. :blushing:

    Basically I'm just trying to emphasize that mac is not a real force, it is the result of the other forces.
     
  15. Jun 21, 2015 #14

    Drakkith

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    Of course. As soon as I realized I made a sign error, I managed to get the right answer.

    Well, in order to solve for m we have to get mg and mac on the same side. Plus, on my force diagram I have T = Fc + mg acting upwards on the mass. So my equation for T comes right from my force diagram. (Which I put there from memory, which is why I answered your question about if the equation came from the top of my head. Did I confuse you?)

    Where did this equation come from?
     
  16. Jun 21, 2015 #15

    haruspex

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    ##\vec F_{net}. \vec r = -m\vec v^2## is just the ##F_c=m\frac{v^2}r## equation you are familiar with, but written in a vectorial form which allows for some unknown tangential acceleration to be present. If we divide it through by r, the magnitude of the radius vector, we get ##\vec F_{net}. \frac {\vec r}r = -m\frac{\vec v^2}r##.
    ##\vec F_{net}. \frac {\vec r}r ## is the component of ##\vec F_{net}## in the radially outward direction.
     
  17. Jun 21, 2015 #16

    Drakkith

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    Ah, okay.
     
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