Suppose you are driving a car around in a circle of radius 212 ft, at a velocity which has the constant magnitude of 43 ft/s. A string hangs

Suppose you are driving a car around in a circle of radius 212 ft, at a velocity which has the constant magnitude of 43 ft/s. A string hangs from the ceiling of the car with a mass of 1.8 kg suspended from it. What angle (in degrees) will the string make with the vertical

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  1. To solve this problem we will apply the concepts related to dynamic equilibrium. The vertical component of the tension is equivalent to the centripetal Force while the horizontal component of the tension is equivalent to the weight, therefore we will have the two equations,

    [tex]Tsin\alpha = \frac{mv^2}{r}[/tex]

    [tex]Tcos\alpha = mg[/tex]

    Our values are,

    [tex]v = 43ft/s = 13.11m/s[/tex]

    [tex]r = 212ft = 64.62m[/tex]

    Therefore if we divide both equation we have,

    [tex]tan \alpha = \frac{v^2}{rg}[/tex]

    Replacing,

    [tex]tan\alpha = \frac{13.11^2}{(64.62)(9.8)}[/tex]

    [tex]tan\alpha = 0.2714[/tex]

    [tex]\alpha = tan^{-1} (0.2714)[/tex]

    [tex]\alpha = 15.18\°[/tex]

    Therefore the angle required is 15.18°

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