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I think this is correct

  1. Nov 25, 2003 #1
    Can anyone verify my results below

    To measure the acceleration due to gravity on a distant planet, an astronaut hangs a 0.085-kg ball from the end of a wire. The wire has a length of 1.5 m and a linear density of 3.1 ´ 10-4 kg/m. Us-ing electronic equipment, the astronaut measures the time for a transverse pulse to travel the length of the wire and obtains a value of 0.083 s. The mass of the wire is negligible compared to the mass of the ball. Recalling that the speed, v, of a small-amplitude wave on a string is given by
    v=squareroot(F/(m/L))

    mball= .085 kg
    L= 1.5 m
    m/L = 3.1 * 10^-4 kg/m (linear density)
    t= 0.083 s
    F=Tension=mg= (.085 kg)(9.8 m/s^2) F=.833 N

    v=squareroot(.833(3.1*10-4) = 51.8 m/s

    v= initial velocity + at

    initial velocity=0

    51.8 = 0 + a(.083)

    51.8/.083 = a

    acceleration= 624 m/s^2
     
  2. jcsd
  3. Nov 25, 2003 #2
    No.

    In this situation, g is not 9.8 m/s^2.

    9.8 m/s^2 is the acceleration due gravity near the surface of the earth. You are trying to find the equivalent quantity at the surface of that distant planet. The equation you are using, I would write as:

    v = √(T/μ)

    where v is the velocity of the wave, T is the tension in the wire, and μ is the mass per unit length of the wire. Note that T = mg in this situation is the weight of the ball only because the ball is suspended from, and therefore pulling down on, the wire. Since you are told that mass of the wire is negliglble, the weight of the ball is the only source of tension.

    The weight of the ball THERE is mg. Therefore, g is the answer you are looking for. V, you are given (since you have been told the length of the wire and the amount of time it takes for the wave to traverse it).

    Try again & see what you get.

    PS: this equation:
    v= initial velocity + at
    has no applicability here at all. It applies to a moving object undergoing a constant rate of acceleration, such as an object in free fall, for example. (You understand that the ball is just hanging from the wire, not falling, don't you?)
     
    Last edited: Nov 25, 2003
  4. Nov 25, 2003 #3
    Alright this should be it then

    thank you for the guidance


    does this look better


    v=distance/time 18.07 m/s

    18.07 m/s = ã(F/(m/L))

    18.07= ã((m*g)/(m/L))

    18.07 = ã((.083*g)/(3.1 * 10^-4))

    ((18.07^2)*(3.1 * 10^-4))/.083 = g

    g =1.22 m/s^2
     
  5. Nov 26, 2003 #4
    Close. I get 1.19 m/s2

    The difference is that you used rounded intermediate results in your calculation.

    If, instead, you rearrange the equations like this:
    v = Δs/Δt
    and
    v = √(mg/μ)
    Δs/Δt = √(mg/μ)
    (Δs/Δt)2 = mg/μ
    g = μ(Δs/Δt)2/m

    Now enter the original amounts in your calculator:
    g = 3.1x10-4(1.5/.083)2/.085
    you'll get
    g = 1.19...

    It's better to hold off rounding until the end, or at least keep one or two extra decimal places in the intermediate results & then finish rounding at the end.
     
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