Angular displacement and bicyclist

[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.

 

[cepheid]

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

 

[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...

 

[cepheid]

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

 

[Linus Pauling]

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

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

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

+ 1 = 0.918


?

 

[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)

??

 

[cepheid]

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​

 

[Linus Pauling]

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

Thank you.