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Homework Help: Centripetal Force Lab

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data

    In this experiment, we used an apparatus that had a shaft that could be rotated with a bob attached. The bob was connected by a string and a spring to the shaft. The shaft was rotated so that the bob would be moving at a fixed distance away from the shaft. This fixed distance was equivalent to the radius of the circular path of motion. The time that it took for the bob to pass the needle (in other words, one full revolution) 25 times was recorded. This time, after being adjusted to account for only one revolution, along with the fixed distance (radius), was used to calculate the centripetal force on the bob. In the next portion of the experiment, a string was connected to the bob and passed over a pulley. Mass was hung from the string until the spring connecting the bob to the shaft stretched enough so that the bob was situated directly over the needle. The force on the bob in this instance was also calculated and then a percent difference calculation between the forces was made. In subsequent trials, mass is added to the bob and the needle is moved further away from the shaft which requires the shaft to be spun at a faster rate to compensate.

    2. Relevant equations

    m1 = 0.1454kg = mass of the bob
    r = 17cm = 0.17m = radius of circular motion
    t1 = 17.96s = first trial of 25 revolutions
    t2 = 17.67s = second trial of 25 revolutions
    t3 = 17.55s = third trial of 25 revolutions
    t_average=(17.96s+17.67s+17.55s)/3=17.73s = average of 25 revolutions
    m2 = mhanger + mweights = 0.0694kg + 0.560kg = 0.6294kg = mass of hanger + weights to exert force to pull bob to r

    3. The attempt at a solution

    t= t_average/25=17.73s/25=0.7092s = average time per revolution
    v=2πr/t=(2π∙0.17m)/0.7092s=1.50(m )⁄s = velocity of bob
    F_1=(m_1 v^2)/r=(0.1454kg ∙〖〖 1.50 m⁄s〗^2〗^2)/0.17m=1.92N = centripetal force of bob motion
    F_2=m_2 g=0.6294kg ∙ 9.8m/s^2 = 6.17N = force (of tension?) exerted by mass of hanger + weights
    % Difference=200|F_1- F_2 |/(F_1+ F_2 )= 200|1.92N- 6.17N|/(1.92N+ 6.17N)=105.07%

    We, I'm assuming, are supposed to have a low percent difference here. In all three of our trials, our percent errors are enormous. I have no idea why... anyone? :(
  2. jcsd
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