Question

You are riding your bicycle down the road and decide to stop to take in the scenery.Initially, your bicycle tires are rotating at a frequency of 3 times per second. If it takesyour bicycle 60 meters to come to a full stop, what is the magnitude of the angularacceleration (assume your acceleration is constant). Give your answer in units ofradians per second squared.

Answer 1

We know that the initial frequency of the tires is 3 times per second, the frequency and angular velocity are related by:

$f=\frac{\omega}{2\pi}$

Then the initial angular velocity is:

$\begin{gathered} 3=\frac{\omega}{2\pi} \\ \omega=6\pi \end{gathered}$

Now, we know that we stop at the end, that means that the final angular velocity is:

$\omega_f=0$

Now, since the angular acceleration is constant we can use the formula:

$\omega^2_f-\omega^2_0=2\alpha(\theta-\theta_0)$

We know that the change in angular position is related to the change in linear position by:

$s=\theta r$

where r is the radius of the motion. Then:

$\theta=\frac{s}{r}$

Hence the change in angular position in this problem is:

$\theta-\theta_0=\frac{60}{r}$

Then the angular accelaration is:

$\begin{gathered} 0-(6\pi)^2=2\alpha(\frac{60}{r}) \\ -36\pi^2=\frac{120}{r}\alpha \\ -\frac{36}{120}\pi^2r=\alpha \\ \alpha=-\frac{3}{10}\pi^2r \end{gathered}$

Therefore the angular accelaration is:

$\alpha=-\frac{3}{10}\pi^2r$

radians per second squared.

(We leave the answer is terms of the radius since we can't relate the angular motion with a linear motion in other way).