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Centripetal Force Lab Percent Error and Accepted Values Help PLEASE

  1. Jan 16, 2012 #1
    Hi! I need help and please do not delete this! What I need help with i cannot put into the template provided. Im sorry but I need help with a lab report. PLEASE I AM BEGGING YOU TO NOT DELETE THIS BECAUSE IT DOES NOT FOLLOW THE TEMPLATE!!!!!! I REALLY REALLY NEED HELP!!!!!!!!!!!!

    I need to find the accepted values and percent errors for the velocity and I don't know what else. I have the theoretical values. My teacher gave us a bunch of equations in class but I don't know what they go with. I attempted to attach the lab and the equations from class because I thought it would be easier. I REALLY need help!!!!!!!!!!!!!!!!! Let me know if the attachments don't work!

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Jan 16, 2012 #2


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    Welcome to PF!
    This is a very interesting experiment; glad to see you doing it.

    Check your derivation of the "accepted value" formula. You have T*sin(θ) = mg and T*cos(θ) = mv²/r. When I solve the first for T and sub into the second, I get v = square root of rg/tan(θ) rather than your cos(θ).
    It is odd that you don't have the values for d and θ on your data sheet. I guess you have another sheet for raw data. I am especially wondering how you varied d and θ.
    So v = sqrt[rg/tan(θ)] if I haven't bungled the derivation. It should be easy to work out those if you have the r and θ values. What do you mean by "percent errors"? Do you mean the difference between the v = sqrt[rg/tan(θ)] and the v = 2πr/T values calculated from your time measurements? What trouble are you having in working these out? As for whether the errors are "reasonable", that would depend on how accurately the measurements were made. Did you happen to repeat one of the trials and get slightly different results? I think the angle measurement is the cause of most of the error because it is so difficult to measure - you might estimate how accurately you measured the angle and calculate how much that changes the v = sqrt[rg/tan(θ)].

    I would sure like to see your raw data for one trial so I can work out the two v values for myself. That would include r and θ as well as the values on your data sheet.

    I am struggling to understand why you are asked to graph mg vs v².
    It seems to me it would be much more interesting to "check" the theory that Fc = mv²/r using
    Fc = Tcos(θ) = mg*cos(θ)/sin(θ) = mg*tan(θ)
    and graphing Fc vs v²/r.
    Last edited: Jan 16, 2012
  4. Jan 16, 2012 #3
    1st of all thank you soooooooooooo much for responding to me!

    I was told d = 0.80m and we have to find [itex]\theta[/itex].

    All of the equations were ones my teacher gave in class, so I do not know where he got cos[itex]\theta[/itex]. For percent error, I mean [itex]\left\|[/itex] (theoritcal velocity - actual velocity) / actual velocity [itex]\right\|[/itex] X 100. I do not know if if use 2[itex]\pi[/itex]dcos.[itex]\theta[/itex]/T or [itex]\sqrt{(RMgcos\theta)/m}[/itex]. I am also unsure if I should use the equation [itex]\theta [/itex] = sin[itex]^{-1}[/itex] (m/M) to find [itex]\theta[/itex].

    The first document I attached had the diagram my teacher gave to me on it. He said r=dcos[itex]\theta[/itex]. d = 0.08m, and I do not know how to find [itex]\theta[/itex].

    The raw data I have for trial one is: Mass of Washers (m): 0.2256kg
    Mass of rubber stopper (M): 0.117kg ​
    Time for 20 Revolutions: 21s 27ms​

    these were the answers I got for the theoretical tangential velocity and square of tangential velocity for trial one: tangential velocity: 4.78m/s
    square of tangential velocity: 22.8(m/s)[itex]^{2}[/itex]​

    I do not know why it asks us to graph the square of the tangential velocity either

    Thank you again sooooo much!!!
  5. Jan 16, 2012 #4


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    I don't like the words "theoretical" and "actual" but it is clear you are to find the per cent difference between the v calculated using the measured period and the v calculated from the mass, angle and radius.

    The big problem is you didn't measure the angle, which is needed for those calculations. No fair calculating the angle from an assumption that the two v's should be identical. A real scientist would run the experiment again, devising a clever way to measure the angle as accurately as possible. The usual method is to run the experiment in front of a white board or wall with a bright light in front to cast a shadow of the apparatus on the board. You can mark an angle scale on the board ahead of time or mark the spots where the shadows of the bob and glass tube are, then measure the angle after the run.

    Think about it - any tube or pipe will do, so you could easily do the experiment again at home. It will go very easily and quickly the second time and you will have complete and better data.
  6. Jan 16, 2012 #5
    what would I use v = [itex]\sqrt{(RMgcos\theta)/m}[/itex] for?
  7. Jan 16, 2012 #6


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    That would be your "theoretical" velocity calculation. You can't do it because you didn't measure θ in the experiment. Note: I think it should be v = sqrt[rg/tan(θ)]. I don't understand why you have two different m's in your expression.
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