# Angular Acceleration - What am I doing wrong?

1. Sep 24, 2009

### ctamasi

I am trying to answer the following question:

A cyclist starts from rest and pedals such that the wheels of the bike have a constant angular acceleration. After 12.0 s, the wheels have made 26 revolutions. What is the angular acceleration of the wheels?

My attempt at a solution:

This problem seems extremely simple, but for some reason I'm not getting the answer.

From the question I can determine that the initial angular velocity of the cyclist is equal to zero at t = 0.

When t = 12.0s:

I worked out the 26 revolutions in 12 seconds to 13.6 rad/s by

$$\frac{26}{12.0}$$ = 2.17 rev/s = $$\frac{130 rev}{1 min}$$

Therefore, $$\frac{130 rev}{1 min}$$ x $$\frac{2\pi rad}{1 rev}$$ x $$\frac{1 min}{60 s}$$ = 13.6 rad/s

$$\omega$$f= 13.6 rad/s

So, $$\alpha$$ = $$\frac{\Delta\omega}{\Delta(t)}$$

$$\alpha$$ = $$\frac{13.6 rad/s - 0 rad/s}{12.0 s}$$

$$\alpha$$ = 1.13 rad/s2

Also, I'm assumining that the wheels are moving in a clockwise direction therefore making the velocity and the acceleration negative; giving me a final answer of -1.13 rad/s2

Now I know this answer is wrong, I just don't know why.

Can someone give me a hand? Thanks in advance.

2. Sep 24, 2009

### kuruman

Hold on a minute! Angle divided by time is angular velocity only if the acceleration is zero (constant angular speed). You have to use the rotational kinematics equations.

3. Sep 24, 2009

### ctamasi

Oh wow! Thank you so much.