# What'd I do wrong? [Angular Kinematics]

1. Oct 28, 2004

### AERam

The problem is:

"Pilots can be tested for the stresses of flying high-speed jets in a whirling "human centrifuge" which takes 1.0 min to turn through 20 complete revolutions before reaching its final speed. (a) What was its angular acceleration (assume constant), and (b) what was its final speed in rpm?"

I converted 20 rpm to rps by dividing 20 by 60, and got .333 Hz. Then I plug that into:

omega=2pi f
omega=2pi (.333)

I plug that into alpha = delta omega / delta t, so that's (2.09/60), and I get roughly .035. But the book says the answer is .07! What did I do wrong?

There's also another problem I'm stumped with; I would appreciate a hint of some sort, because I don't know what to do with some of the numbers.

"The tires of a car make 65 revolutions as the car reduces its speed uniformly from 100km/h to 50 km/h. The tires have a diameter of .8m. (a) What was the angular acceleration? (b) If the car continues to decelerate at this rate, how much more time until it stops?"

Any help is appreciated!

2. Oct 29, 2004

### vsage

You found the average speed. Under constant acceleration, omegaaverage = 1/2(omega final) assuming you started from rest. For the second problem you should at least start by finding the circumference of the tire and converting k/h to m/s.

Last edited by a moderator: Oct 29, 2004
3. Oct 29, 2004

### robphy

AERam,
the simplest way to do your first problem is to write down the rotational kinematic equations for constant angular acceleration:
$$\theta=\theta_0+\omega_0 t +\frac{1}{2}\alpha t^2$$
$$\omega=\omega_0+\alpha t$$. (Hopefully, these look familiar.)
You can set your starting point $\theta_0=0$.
If you start from rest, $\omega_0=0$.

Yes, you could use the average-angular-acceleration formula
$$\alpha_\text{avg}=\frac{\Delta \omega}{\Delta t}= \frac{\omega_f-\omega_0}{t_f-t_0}$$. However, you must use the final angular-velocity $\omega_f$---not the average-angular-velocity $\omega_\text{avg}=\frac{\Delta \theta}{\Delta t}$ (as vsage pointed out).

recall the relationship between linear displacement $s$ and angular displacement $\theta$ (with a constant radius $r$):
$$s=r\theta$$.