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Math Help - Lawn Mower Cord Problem (Radians and Angular Velocity)

  1. #1
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    Unhappy Lawn Mower Cord Problem (Radians and Angular Velocity)

    Hi, this is my first time on this website and I hope to help other people while getting help myself. First off, I was absent the day this homework was assigned; therefore I have no idea how to do it. Anyway, here is the problem, THANKS.

    Yank Hardy pulls the cord on his power mower. in order for the engine to start, the pulley must turn at 180 revolutions per minute. The pulley has a radius of 0.2 ft.

    1. How many radians per second must the pulley turn?

    2. How fast must Yank pull the cord to start the mower?

    3. When Yank pulls this hard, what is the angular velocity of the center of the pulley?

    Thanks for looking I appreciate all help!!!!!!!!
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  2. #2
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    Perhaps this is enough to get you started:

    There are pi radians in 180 degrees.
    1 revolution = 360 degrees.

    The linear speed at the edge of a pulley is
    (angular speed in radians) * (radius of pulley)

    As long as the pulley is rigid the angular speed from the center is the same throughout the pulley.
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  3. #3
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    Quote Originally Posted by snowtea View Post
    Perhaps this is enough to get you started:

    There are pi radians in 180 degrees.
    1 revolution = 360 degrees.

    The linear speed at the edge of a pulley is
    (angular speed in radians) * (radius of pulley)

    As long as the pulley is rigid the angular speed from the center is the same throughout the pulley.
    This is useful information. I can answer number 1 from what you told me, and is my math correct?

    180 revs/min = 3 revs/sec

    pi radians = 180degrees
    2pi = 360 degrees
    2pi = 1 rev
    6pi radians/sec = 3 revs/sec
    So number one is 6pi radians/sec?

    Also, thank you for the linear speed equation, but I cannot complete it because I do not know how to get angular speed. Can you please provide me with some additional information on angular speed so I can calculate the linear speed? Thank you I greatly appreciate your help.
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  4. #4
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    I think 6pi rads/sec is correct.
    And this *is* your angular speed
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  5. #5
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    Quote Originally Posted by snowtea View Post
    I think 6pi rads/sec is correct.
    And this *is* your angular speed
    Thank you again, but another question . Since I have to multiply my angular speed by radius to get linear speed, does it matter what unit the radius is in? for example, in the problem it is .2feet, but I could also use 2.4 inches or something else and depending on which unit I use, the answer will differ.
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  6. #6
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    Yup. So the final speed will have different units.
    E.g. feet / sec or inches / sec
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  7. #7
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    Quote Originally Posted by snowtea View Post
    Yup. So the final speed will have different units.
    E.g. feet / sec or inches / sec
    OK so I'm still a little confused, but would this be correct:

    (6pi rads/sec) * (0.2 ft) = 1.2pi feet/ sec
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  8. #8
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    Yes, 1.2pi ft/sec looks correct.

    The idea in using radians is to simplify calculations of arc length.
    For example you want to find the distance around a circle: this is 2*pi*r
    so they defined 360 degrees to be 2*pi radians
    and you can directly get the distance by multiplying.

    If you want to find the distance of a half circle pi*r
    Radians for 180 degrees is pi

    If you want to find the distance of a quarter circle pi/2 * r
    Radians for 90 degrees is pi/2

    Also, speed = distance / time, so radians work the same way
    speed = distance / time = radians * r / time = (radians / time) * r
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