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Math Help - Tension & Rads

  1. #1
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    Exclamation Tension & Rads

    Hi, I've got a question that I can't see how to do it, it is:

    The tension T in a belt passing round a pulley wheel and in contact with the pulley over an angle of θ radians is given by T=Toe^μθ
    Where To and μ are constant. Experimental results obtained are:

    T Newtons: 47.9 52.8 60.3 70.1 80.9

    θ radians : 1.12 1.48 1.97 2.53 3.06

    Determine approximate values of To and μ. Hence find the tension when θ is 2.25 radians and the value of θ when the tension is 50 Newtons.

    Any help would be greatly appreciated,

    Thanks.
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  2. #2
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    Quote Originally Posted by gram1210 View Post
    Hi, I've got a question that I can't see how to do it, it is:

    The tension T in a belt passing round a pulley wheel and in contact with the pulley over an angle of θ radians is given by T=Toe^μθ
    Where To and μ are constant. Experimental results obtained are:

    T Newtons: 47.9 52.8 60.3 70.1 80.9

    θ radians : 1.12 1.48 1.97 2.53 3.06

    Determine approximate values of To and μ. Hence find the tension when θ is 2.25 radians and the value of θ when the tension is 50 Newtons.

    Any help would be greatly appreciated,

    Thanks.
    Rewrite your equation:
    T = T_0 e^{\mu \theta}

    ln(T) = \mu \theta + ln(T_0)

    If you plot your data as ln(T) vs. \theta (y vs. x) then your data should form a line with slope \mu and intercept ln(T_0). So calculate ln(T) for your data set and do a linear regression on ln(T) vs. \theta.

    -Dan
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  3. #3
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    So, as μ and To are constants, how do I calculate their values?

    That then links into the graph..where does the interception between μ and ln(To) come into existance?

    Thanks again.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by gram1210 View Post
    So, as μ and To are constants, how do I calculate their values?

    That then links into the graph..where does the interception between μ and ln(To) come into existance?

    Thanks again.
    ln(T) = \mu \theta + ln(T_0)


    Do the linear regression. That means you are modeling the data for ln(T) and \theta as a line. The computer/calculator will give you a value for the slope and the y - intercept. The slope will be \mu and the intercept will be ln(T_0). So T_0 = e^{\text{intercept}}.

    -Dan
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