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Math Help - Spin Balancing Tires

  1. #1
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    Spin Balancing Tires

    A spin balancer rotates the wheel of a car at 480 revolutions per minute. If the diamteer of the wheel is 26 inches, what road speed is being tested? Express your answer in miles per hour. At how many revolutions per minute should the balancer be set to test a road speed of 80 miles per hour?

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  2. #2
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    Quote Originally Posted by magentarita View Post
    A spin balancer rotates the wheel of a car at 480 revolutions per minute. If the diamteer of the wheel is 26 inches, what road speed is being tested? Express your answer in miles per hour.

    If an object moves along a circular path of radius r units, then its linear velocity, v, is given by:

    v=r\frac{\theta}{t}

    where \frac{\theta}{t} represents the angular velocity in radians per unit of time.

    480 revolutions = 480 \cdot 2\pi radians per minute or 960 \pi radians per minute.

    r=13in

    Here it is in one fell swoop:

    13in \times \frac{1ft}{12in} \times \frac{1mi}{5280ft} \times \frac{480rev}{1min} \times \frac{2\pi}{1rev} \times \frac{60min}{1h}\approx37.13mph

    And here it is all broken out:

    v=13in \cdot \frac{440 \cdot 2\pi \ \ rad}{1 \ \ min}=39207.0763 \ \ in/min

    This translates to 39207.0763 \times 60min = 2352424.579\ \ in/hr.

    1\ \ mi = 5280 \ \ ft = 63360 \ \ in

    2352424.579 \ \ in/hr \div 63360\ \ in/mile\approx 37.13\ \ mph
    Quote Originally Posted by magentarita View Post
    At how many revolutions per minute should the balancer be set to test a road speed of 80 miles per hour?
    Find the circumference of the wheel

    C = \pi D=26\pi in

    \frac{80mi}{1hr}\times\frac{5280ft}{1mi}\times\fra  c{12in}{1ft}\times\frac{1rev}{26\pi in}\times\frac{1hr}{60min}\approx1034.26 rev/min

    The miles, inches, feet, and hours cancel leaving you with revolutions over minutes
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  3. #3
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    ok....

    Quote Originally Posted by masters View Post
    If an object moves along a circular path of radius r units, then its linear velocity, v, is given by:

    v=r\frac{\theta}{t}

    where \frac{\theta}{t} represents the angular velocity in radians per unit of time.

    480 revolutions = 480 \cdot 2\pi radians per minute or 960 \pi radians per minute.

    r=13in

    Here it is in one fell swoop:

    13in \times \frac{1ft}{12in} \times \frac{1mi}{5280ft} \times \frac{480rev}{1min} \times \frac{2\pi}{1rev} \times \frac{60min}{1h}\approx37.13mph

    And here it is all broken out:

    v=13in \cdot \frac{440 \cdot 2\pi \ \ rad}{1 \ \ min}=39207.0763 \ \ in/min

    This translates to 39207.0763 \times 60min = 2352424.579\ \ in/hr.

    1\ \ mi = 5280 \ \ ft = 63360 \ \ in

    2352424.579 \ \ in/hr \div 63360\ \ in/mile\approx 37.13\ \ mph


    Find the circumference of the wheel

    C = \pi D=26\pi in

    \frac{80mi}{1hr}\times\frac{5280ft}{1mi}\times\fra  c{12in}{1ft}\times\frac{1rev}{26\pi in}\times\frac{1hr}{60min}\approx1034.26 rev/min

    The miles, inches, feet, and hours cancel leaving you with revolutions over minutes
    I knew this question had some work to it.
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