Angular Acceleration of a Merry Go Round: Solving a Physics Problem

[ixerr]

Homework Statement

A merry go round with a radius of 3 meters is at rest. The operator starts the merry go round up and it accelerates at a constant rate for 20 seconds. At the end of that time, the merry go round is moving at 10 revolutions per minute.

1) What was the angular acceleration of the merry go round?

2) What is the speed of one of the horses at the edge of the merry go round when the merry go round reaches its maximum rotation rate?

Homework Equations

let a=angular acceleration
let w=angular velocity
let r= radius

a=rw^2

The Attempt at a Solution

So, I solved #1 out, but I'm not sure it's right at all. Here is my work:

a=rw^2
=(3)((10*2*(22/7))/60)^2 = 3.31 rad/s^2

And I don't know how to do part 2. Please help.

 

[SammyS]

You found the centripetal acceleration of the edge of the merry go round. (BTW: 22/7 is a poor approximation for pi. 3.1416 is better. 355/113 is even better.)

The problem asks you to find angular acceleration.

 

[ixerr]

I know, I used pi in my actual calculation. But thanks!
And how the heck do I find angular acceleration then?! Ahh this is all too confusing :-(

 

[SammyS]

The equations for rotational kinematics are much like those for one dimensional linear kinematics.

If α is constant (uniform), then:

$\theta=\theta_0+\omega_0 t+(1/2)\alpha t^2$

$\omega=\omega_0+\alpha t$

$\omega^2={\omega_0}^2+2\alpha\theta$

$\displaystyle \omega_\text{Average}=\frac{\omega+\omega_0}{2}$​