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Math Help - help me on solving this problem please

  1. #1
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    help me on solving this problem please

    hi guys ! i don't wanna disturbe you but i truly need to solve this problem:
    An object is travelling around a circle with a radius of 10 cm. If in 20 seconds a central angle of 1/3 radian is swept out, what is the linear speed of the object?
    thanks in advance
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  2. #2
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    Quote Originally Posted by alessandromangione View Post
    hi guys ! i don't wanna disturbe you but i truly need to solve this problem:
    An object is travelling around a circle with a radius of 10 cm. If in 20 seconds a central angle of 1/3 radian is swept out, what is the linear speed of the object?
    thanks in advance
    If the circle has radius of 10cm, then

    C = 2\pi \cdot 10\,\textrm{cm}

    C = 20 \pi \,\textrm{cm}.


    It has travelled \frac{1}{3} of a radian.

    Since there are 2\pi radians on the circumference, that means 1 radian  = \frac{C}{2\pi}.

    Therefore \frac{1}{3}\,\textrm{radian} = \frac{C}{6\pi}.


    Since the circumference was 20 \pi \,\textrm{cm}

    \frac{1}{3}\,\textrm{radian} = \frac{20\pi\,\textrm{cm}}{6\pi}

    \frac{1}{3}\,\textrm{radian} = \frac{10}{3}\,\textrm{cm}.


    Since it travels this far in 20 seconds, or \frac{1}{3}\,\textrm{min}, this means that the speed is

    \frac{10}{3}\,\textrm{cm} per \frac{1}{3}\,\textrm{min}

    or

    10\,\textrm{cm}\, / \, \textrm{min}.
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  3. #3
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    first of all thank you very much for answwering.
    But i tried to do it in another way and the result is different.
    since the angular speed is obtained by 1/3 : 20 seconds ( angle divided per time) i got 1 /60 . Then i applied the formula v= radius angular speed , so v= 1/60 multiplied per 10 i got 1/6...the teacer told me to use the formulas above written but the result is different from yours...where did i mistake?
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  4. #4
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    Quote Originally Posted by alessandromangione View Post
    first of all thank you very much for answwering.
    But i tried to do it in another way and the result is different.
    since the angular speed is obtained by 1/3 : 20 seconds ( angle divided per time) i got 1 /60 . Then i applied the formula v= radius angular speed , so v= 1/60 multiplied per 10 i got 1/6...the teacer told me to use the formulas above written but the result is different from yours...where did i mistake?
    Your result is the speed in cm/s. Multiply by 60 s/min to get the speed in cm/min.

    You'll get the same result as ProveIt.
    Last edited by earboth; December 12th 2009 at 11:04 PM.
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  5. #5
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    Quote Originally Posted by alessandromangione View Post
    first of all thank you very much for answwering.
    But i tried to do it in another way and the result is different.
    since the angular speed is obtained by 1/3 : 20 seconds ( angle divided per time) i got 1 /60 . Then i applied the formula v= radius angular speed , so v= 1/60 multiplied per 10 i got 1/6...the teacer told me to use the formulas above written but the result is different from yours...where did i mistake?
    Speed is not the ratio of angle to time. It's the ratio of distance to time.

    So you need to work out the distance travelled.
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  6. #6
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    Quote Originally Posted by earboth View Post
    ...
    so..1/3 x 2 greek pi ( i don't know how to put the symbol sorry ) divided by 20 and i get 2 greek pi/ 60 then i multiplied this per 10 and i get 2 greek pi / 6...the result isn't 10 cm/min i'm confused.......can you explain me the procedure using this way?
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  7. #7
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    Did you read my post?

    It has not travelled \frac{1}{3} of 2\pi.

    It has travelled \frac{1}{3} of a RADIAN.


    Like I said, 2\pi^r = C, so 1^r = \frac{C}{2\pi}.

    So it has travelled \frac{1}{3}^r = \frac{C}{6\pi}.
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  8. #8
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    Quote Originally Posted by alessandromangione View Post
    so..1/3 x 2 greek pi ( i don't know how to put the symbol sorry ) divided by 20 and i get 2 greek pi/ 60 then i multiplied this per 10 and i get 2 greek pi / 6...the result isn't 10 cm/min i'm confused.......can you explain me the procedure using this way?
    I've made a very silly mistake in my previous post. It's already "repaired"!
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  9. #9
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    Quote Originally Posted by earboth View Post
    I've made a very silly mistake in my previous post. It's already "repaired"!
    ok thanks anyway!
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