# Angular Displacement of bike pedals

1. Oct 12, 2006

### snowmx0090

An exhausted bicyclist pedals somewhat erratically, so that the angular velocity of his tires follows the equation,
ω(t)=(1/2)t - (1/4)sin(2t)
where represents time (measured in seconds).
There is a spot of paint on the front tire of the bicycle. Take the position of the spot at time to be at angle radians with respect to an axis parallel to the ground (and perpendicular to the axis of rotation of the tire) and measure positive angles in the direction of the tire's rotation. What angular displacement has the spot of paint undergone between time 0 and 2 seconds?

I thought you would take the integral of the equation from t=0 to t=2 in which the equation would then be (1/4)t^2 + (1/2)cos(2t) but this was wrong. Where should I start with this equation?

2. Oct 12, 2006

### Staff: Mentor

Check your integration--the second term has an error.

3. Oct 12, 2006

### snowmx0090

Would the equation after integrating be (1/4)t^2 + (1/4)cos(2t)? Was this my mistake?

4. Oct 13, 2006

There is still a mistake. What does $$\frac{d}{dt}(\frac{1}{4}\cos (2t))$$ equal?

5. Mar 11, 2008

### faller217

you would have to use a u substitute. Set u=2t, and du=2. So instead of it being 1/4cos(2t), it would be 1/8cos(2t).

final : ((1/4)(t^2))+((1/8)(cos2t)) from 0 to 2 seconds

.918295 - .125 = .793 radians

Last edited: Mar 11, 2008