# Homework Help: Angular displacement and bicyclist

1. Oct 24, 2009

### Linus Pauling

1.

An exhausted bicyclist pedals somewhat erratically, so that the angular velocity of his tires follows the equation

omega(t) = 0.5t - 0.25sin(2t), t greater than/equal to zero.

where t represents time (measured in seconds).

2. omega is defined as the first derivative of theta.

3. I integrated omega(t), obtaining 0.25t^2 + (1/8)cos(2t), plugged in 2 and got my incorrect answer.

2. Oct 24, 2009

### cepheid

Staff Emeritus
Hey, can you post the actual question? Is it to find the angular displacement after 2 seconds?

3. Oct 24, 2009

### Linus Pauling

What angular displacement theta has the spot of paint undergone between time 0 and 2 seconds?

That is the question. Apparently I integrated incorrectly, but I do not see where I made a mistake.......

4. Oct 24, 2009

### cepheid

Staff Emeritus
I'm really not sure what the problem is either. Is your calculator in the right mode?

5. Oct 24, 2009

### Linus Pauling

Yes, it is in radians. So, cos(4) = -.6536

(1/8)cos(4) = -.0817

+ 1 = 0.918

??????

6. Oct 24, 2009

### Linus Pauling

BTW, it says that I did my integral wrong. In other words, the angular displacement does NOT equal:

0.25t^2 + (1/8)cos(2t)

???????????

7. Oct 24, 2009

### cepheid

Staff Emeritus
You calculated the *indefinite* integral:

$$\theta(t) = \int \omega(t) \, dt$$

$$= \int \left(\frac{1}{2}t - \frac{1}{4}\sin(2t) \right) \, dt$$

$$= \frac{1}{4}t^2 + \frac{1}{8}\cos(2t) + C$$ ​

where C = $\theta(0)$

The question is asking you for the angular *displacement* between t = 0 and t = 2, which is given by $\theta(2) - \theta(0)$, which is given by the *definite* integral:

$$\theta(2) - \theta(0) = \int_0^2 \omega(t) \, dt$$

$$= \int_0^2 \left(\frac{1}{2}t - \frac{1}{4}\sin(2t) \right) \, dt$$

$$= \left[\frac{1}{4}t^2 + \frac{1}{8}\cos(2t)\right]_0^2$$ ​

8. Oct 24, 2009

### Linus Pauling

Damn it, I forgot the blood 1/8 from taking cos(0).

Thank you.