Results 1 to 4 of 4

Math Help - Cycling and Trig

  1. #1
    Newbie
    Joined
    Dec 2005
    Posts
    2

    Cycling and Trig

    Can somebody help please??

    The height h in cm of a bicycle pedal above the ground at time t seconds is given by

    h = 30 + 15 cos 90 t

    How do I find the number of revolutions in one minute?

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by radius
    Can somebody help please??

    The height h in cm of a bicycle pedal above the ground at time t seconds is given by

    h = 30 + 15 cos 90 t

    How do I find the number of revolutions in one minute?

    Thanks
    The height is a periodic function of t, as it should be. The frequency
    (cycles per second) of the pedal is the frequency of the  \cos (90.t)
    part of the height formula. A cosine based signal of frequency f is:
    k. \cos (2 \pi.f.t), so in this case:

    90=2 \pi.f .

    So

    f=45/ \pi cycles per second,

    or

    \frac{45\times 60}{\pi} cycles per minute.

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2005
    Posts
    2
    Quote Originally Posted by CaptainBlack

    \frac{45\times 60}{\pi} cycles per minute.

    RonL
    Thanks.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by radius
    Can somebody help please??

    The height h in cm of a bicycle pedal above the ground at time t seconds is given by

    h = 30 + 15 cos 90 t

    How do I find the number of revolutions in one minute?

    Thanks
    Here is one way.

    h = 30 +15cos(90t)
    is a cosine curve whose:
    ---"original" or "basic" origin is shifted to 30cm above the (0,0) of the t,h-axes setup. There is no horizontal shift.
    Its amplitude is 15cm. That means at maximum height, the pedal is (30+15) = 45cm above ground, and at minimum height, the pedal is (30-15) = 15cm above the ground. Very reasonable distances from the ground.

    Its period is 360/90 = 4sec per cycle. One period = one cycle = one revolution of the pedal.
    So one revolution is 4sec. In one minute, that will be 60/4 = 15 revs per minute.
    Therefore, the pedal moves at 15 rpm. --------answer.
    Reasonable speed.

    Let us check that.
    At t = 0 sec----h = 30 +15cos(90*0) = 30 +15cos(0) = 30 +15(1) = 45 cm.
    At t = 1 sec----h = 30 +15cos(90*1) = 30 +15cos(90deg) = 30 +15(0) = 30 cm.
    At t = 2 sec----h = 30 +15cos(90*2) = 30 +15cos(180deg) = 30 +15(-1) = 15 cm.
    At t = 3 sec----h = 30 +15cos(90*3) = 30 +15cos(270deg) = 30 +15(0) = 30 cm.
    At t = 4 sec----h = 30 +15cos(90*4) = 30 +15cos(360deg) = 30 +15(1) = 45 cm.

    See, at 4sec, the pedal is back at the initial 45cm above ground. That is 4sec per revolution. In one minute, that is 15 revs. That is 15 rpm.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Compute Trig Function Values, Solve Trig Equation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: September 8th 2011, 07:00 PM
  2. Trig word problem - solving a trig equation.
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: March 14th 2011, 07:07 AM
  3. Replies: 7
    Last Post: April 15th 2010, 08:12 PM
  4. Replies: 6
    Last Post: November 20th 2009, 04:27 PM
  5. Trig Equations with Multiple Trig Functions cont.
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 7th 2008, 05:50 PM

Search Tags


/mathhelpforum @mathhelpforum