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

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- Dec 29th 2005, 08:04 AMradiusCycling 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 - Dec 29th 2005, 08:22 AMCaptainBlackQuote:

Originally Posted by**radius**

(cycles per second) of the pedal is the frequency of the

part of the height formula. A cosine based signal of frequency is:

, so in this case:

.

So

cycles per second,

or

cycles per minute.

RonL - Dec 29th 2005, 10:10 AMradiusQuote:

Originally Posted by**CaptainBlack**

- Jan 1st 2006, 01:19 AMticbolQuote:

Originally Posted by**radius**

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.