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Centripetal acceleration and forces - Lab help wanted?

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data
    This will be kind of long because it's not just a simple problem, it's a lab. Sorry about that.

    So the lab is testing the concepts of forces and centripetal force, for starters. We are supposed to be "investigating the relationship between period and acceleration."

    In class we performed an experiment which went sort of like this: A drain stopper was attached to a string, which had a makeshift handle (a glass tube) and a paperclip on the end. So:

    stopper--------------tube handle-----------------paperclip
    where ------ is string.

    We were supposed to add washers (little metal things) to the paperclip to increase the downward force imposed, via gravity, on the stopper. There was a mark on the string so that when we swung it, the radius of the resulting circle would be constant. But acceleration of the stopper was not constant, because with more washers (more force downwards), it had to be swung faster to keep from falling through the glass tube. This would be bad because it would make radius of the swinging circular motion non-constant.

    The teacher gave us the equation F(net)=mass*acceleration. We haven't done the forces unit yet, so this wasn't explained too much. But apparently there's a relationship between net force (force downwards imposed by the washers), mass of the stopper, and the speed at which the stopper must be swung to maintain movement in a circular path.

    The data being collected was # washers and time it took to go around once. The washers were representative of the net force downward. The time to go around once was interpreted as one period.

    We were measuring period (time taken to go 2pi or around a circle?) in seconds.

    I was given the equation
    acceleration = ( 4pi^2 *r ) / (period^2)

    Oh, also the teacher wrote "we would not expect a graph of t vs. a to be linear. A derived quantity of T would produce a linear graph. What is this quantity?" T is period, by the way. Through class discussion that I totally didn't understand, we determined that this quantity was 1/(T^2). I really have no idea why - if the idea is to take the period down a power, wouldn't 1/T be sufficient?

    About my specific problems with the lab:
    From a graph of 1/(T^2) versus force in N, I have to determine the mass of the stopper. (The slope of the graph is supposed to have something to do with it?) I have no idea how to do this. I get the feeling I am supposed to have been making conceptual leaps this whole time, yet I have not been.

    I guess this has to do with the [Net Force = mass * acceleration] equation? Equation translates to Net force=m * 4(pi^2)r(1/period^2) - this was suggested by the teacher. I don't know how this is helpful but for posterity I'll copy it here.
    2. Relevant equations
    Copied from above:
    acceleration = ( 4pi^2 *r ) / (period^2)

    3. The attempt at a solution
    I don't suppose it would be helpful to copy my data here?
    Newtons ------ Period (T) ------ 1/(T^2)
    I converted washers into Newtons and the slope I ended up obtaining from the graph was 3.75, where the y axis is force in N and the x axis is the transformed period, (1/T^2). The equation of the line was y = 3.75x - 5.205. Imperfect data, because it was supposed to be a direct proportional relationship and thus come from (0,0).
    The radius of the swing was about .76 meters.
    The mass of the stopper was 14.8 g.

    We can start with
    Net force = mass*acceleration
    or washers = mass of stopper*acceleration

    so that goes to:
    mass of stopper = washers/acceleration

    Then the question becomes, how do you find acceleration?
    Acceleration is equal to ( 4pi^2 *r ) / (period^2).
    So mass of the stopper = washers * (( 4pi^2 *r ) / (period^2))

    I can see solving this for all different forces and periods on my data table. Would I then have to average all the values? And this is where I get stuck: how does this relate to the slope of my graph of acceleration versus 1/T^2 ? I get the feeling the answer to the question 'why did we transform the period like that in the first place anyway?' is the key to fixing this stuckness?

    I'm not sure if this is really supposed to go into introductory physics. I'm in AP physics, though it is the first physics class I've taken. Apologies if it's the wrong section.

    Thanks so much for any help given in advance. I feel like I'm asking the world of some poor anonymous person...
    Last edited: Oct 26, 2009
  2. jcsd
  3. Oct 26, 2009 #2

    Andrew Mason

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    The purpose of the lab is to show the relationship between the centripetal force and the angular speed. The paperclip ensures that the mass that is swinging around is at the same distance from the centre. The weights hanging down through the glass tube are proportional to the tension in the string - ie. the centripetal force that keeps the mass moving in a circle.

    The hypothesis is that the centripetal force is: [itex]F = ma = m\frac{\Delta \vec v}{\Delta t} = mv^2/r[/itex]. The purpose of the experiment is to see if that hypothesis is correct.

    The mass is moving at constant speed. It travels a distance of [itex]2\pi r[/itex] in one period, T. So [itex]v = 2\pi r/T[/itex]. Putting this into the equation for centripetal force:

    [tex]F = mv^2/r = m(2\pi r/T)^2/r = m\frac{4\pi^2 r}{T^2}[/tex]

    In the experiment, r and m are kept the same. So if the hypothesis is correct, you should have a linear relationship between F (represented by the number of washers hanging on the string) and 1/T^2 (the reciprocal of the period squared).

    Is that what your data shows?

  4. Oct 26, 2009 #3
    Thanks, Andrew!

    The relationship is vaguely linear and can be somewhat fit by a line.

    I'm still kind of confused: if the relationship is linear, what does the slope of the line have to do with the mass of the stopper?

    If F=ma,
    F=m(v^2/r)= (m(2pi*r/T)^2 )/r = (m(4pi^2) *r) / (T^2) , as you said.

    If slope = delta y / delta x, and 1/T^2 is the x-variable, does it follow that delta y = m(4pi^2) *r ? But on my graph, force (in N) is on the y-axis...

    Is m*(4pi^2)r = delta y = F (on y-axis)? Doesn't this contradict F=ma?

    Or... Am I misinterpreting your suggestion completely?
  5. Oct 27, 2009 #4
    Never mind, I did end up figuring the direct relationship - if y = mx (+ b, but not in this case, since it's a direct proportion), and y is F and x is 1/T^2, the relationship is easy to find. You just compare y = mx to F = ma = m(4pi^2) *r (1/T^2). When you take away the bits you know to be represented by the variables, what's left as the slope is m(4pi^2) *r.

    Thanks again for your help.
  6. Oct 27, 2009 #5

    Andrew Mason

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    You are welcome.

    This is how one verifies complicated relationships - you try to find the relationship that makes the graph a straight line. There is a linear relationship between F and 1/T^2 here.

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