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=(\omega)/(2\pi)`

Then the initial angular velocity is:

`\begin{gathered} 3=(\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=(s)/(r)`

Hence the change in angular position in this problem is:

`\theta-\theta_0=(60)/(r)`

Then the angular accelaration is:

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

Therefore the angular accelaration is:

`\alpha=-(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).