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Helium balloon and simple harmonic motion

  1. Mar 10, 2008 #1

    ~christina~

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    Gold Member

    1. The problem statement, all variables and given/known data
    A light baloon filled wiht helium and with a density of 0.180kg/m^3 is tied to a light string of lenght L= 3.00m. The string is tied to the ground, forming an "inverted" simple pendulum. The balloon is displaced slightly from it's equillibrium position, as shown in picture.
    a) show that the balloon executes simple harmonic motion when it is released, i.e., show that the restoring force is a Hooke's law force.

    b) What is the period of the motion?

    Ignore energy lost due to air friction and take the density of air to be 1.29kg/m^3
    [​IMG]

    2. Relevant equations????
    3. The attempt at a solution

    a) show that the balloon executes simple harmonic motion when it is released, i.e., show that the restoring force is a Hooke's law force.

    not sure how to prove this and how it should look if I do prove it when I do.
    in the begining position a):

    balloon has:
    buoyant force: up
    gravity: down
    tension: down

    position b:
    balloon
    buoyant force: up
    gravity: down?
    tension: at an angle

    forces on balloon
    [tex]\Sum F= B-mg -T= 0 [/tex]

    [tex]\Sum F'= B-mg- Tsin \theta [/tex]

    I realy don't know... do I combine them??

    b) period of the motion

    don't know how to find this as well...

    please help me on this.

    Thank you
     
  2. jcsd
  3. Mar 10, 2008 #2
    (B-mg) is a constant force directed upwards. This must be balanced by the vertical component of the tension in the rope: T cos(theta). The horizontal component of the
    tension in the rope is your restoring force.

    For small values of theta you can replace cos(theta) by 1 and sin(theta) by theta
    (if theta is in radians).
     
  4. Mar 10, 2008 #3

    ~christina~

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    Gold Member

    Um isn't it supposed to be Tsin(theta) for the restoring force?

    so it would be

    [tex]\Sum F= B-mg -T= 0 [/tex]

    [tex]\Sum F'= B-mg- Tsin \theta [/tex]

    subtract them

    [tex]-B+mg+T= 0[/tex]
    [tex]B-mg-Tsin \theta[/tex]
    _______________________
    [tex]F_{net}= T-Tsin \theta [/tex]

    I'm not sure how do I know it's simple harmonic motion.

    and how do I find part b)
    The period of the motion? using the density they gave?
     
    Last edited: Mar 10, 2008
  5. Mar 10, 2008 #4

    rock.freak667

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    The restoring force is [itex]Tsin\theta[/itex] and if [itex]theta[/itex] is small and in radians then [itex]sin\theta \approx \theta[/itex]

    So that in the horizontal direction.

    [tex]ma= -T\theta[/tex]


    and the distance it is displaced is given by the length of the arc (x) multiplied by the length,L, i.e. [itex]\theta=xL[/itex]
     
  6. Mar 10, 2008 #5

    ~christina~

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    but how would that help in my proving that the balloon undergoes simple harmonic motion?

    was what I did above incorrect?
     
  7. Mar 10, 2008 #6

    rock.freak667

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    Well to show that something undergoes SHM, you need to show that [itex]a=-\omega^2x[/itex] where x=displacement,a=acceleration.
     
  8. Mar 10, 2008 #7

    ~christina~

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    Gold Member

    ok..I'm assuming that you were going off what I did before so:

    [tex]\Sum F= B-mg -T= 0 [/tex]

    [tex]\Sum F'= B-mg- Tsin \theta [/tex]

    subtract them

    [tex]-B+mg+T= 0[/tex]
    [tex]B-mg-Tsin \theta[/tex]
    _______________________
    [tex]F_{net}= T-Tsin \theta [/tex]

    and including what you said where [tex] sin\theta\aprox \theta[/tex] then

    [tex]F_{net}= T-Tsin \theta [/tex]

    I know that F= ma however you showed that
    why is it not [tex]ma= T-T\theta [/tex] ?
     
  9. Mar 10, 2008 #8

    rock.freak667

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    Because the restoring force is given by Tsin. When you split T into its components you must use the components and not the Resultant vector T
     
  10. Mar 10, 2008 #9

    ~christina~

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    ok.. so it cancels out...

    I'm assuming that you were going off what I did before so:

    [tex]\Sum F= B-mg -T= 0 [/tex]

    [tex]\Sum F'= B-mg- Tsin \theta [/tex]

    subtract them

    [tex]-B+mg+T= 0[/tex]
    [tex]B-mg-Tsin \theta[/tex]
    _______________________
    [tex]F_{net}= T-Tsin \theta [/tex]

    and including what you said where [tex] sin\theta\aprox \theta[/tex] then

    [tex]F_{net}= T-Tsin \theta [/tex]

    I know that F= ma however you showed that
    [tex]ma= -T\theta [/tex]

    They should really tell us this but they do not and always ask to prove but never tell us what we are looking for....

    Oh well I know that [tex]\omega = \sqrt{g} {L} [/tex]

    so... [tex]\theta= x L[/tex]

    then plugging in..

    [tex]ma= -T\theta = ma[/tex]

    [tex]ma= -TxL [/tex]

    but do I just divide over the m? I'm not sure what that would accomplish for finding
    [tex]a= \omega^2x [/tex]

    I know that [tex]\omega= \sqrt{g} {L} [/tex] and I'm not sure what relation I can find from this unfortunately.
    Is there another one I need to think of to make it equal to having simple harmonic motion?

    Thank you
     
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