# Angular Speed of a flywheel

1. Nov 30, 2005

### suspenc3

Hi,

A flywheel with a diameter of 1.2 m is rotating at an angular speed of 200 revs/min

What is the angular speed of the flywheel in radians per second?

Would I just do this?

200 revs/min = 3.33 revs/sec

2. Nov 30, 2005

### Fermat

Yes, that's right.

3. Nov 30, 2005

### suspenc3

ok..and is angular speed measure in revs/sec or rad/sec

What constant angular acceleration (in rev/min^2) will increase the wheels angular speed to 100 rev/min in 60 seconds?

I used:

$$\alpha = \frac{\Delta \omega}{ \Delta t}$$

1000 revs/min = [2000(pi)revs/min - 400(pi)revs/min] / 1(minute)

$$\alpha = 1600 \frac{revs}{min^2}$$

4. Nov 30, 2005

### suspenc3

and it also asks how many rotations did this take could I just assume since its a constant acceleration that i took 1600 revs?

5. Nov 30, 2005

### Fermat

Angular speed is usually given in rads/sec, but, depending upon the application, it can also be measured in rpm or cps, for example.

alpha = 1600 revs/min² is correct.
But you have changed the initial angular velocity from 200 rev/min to 2000 rev/min.
Is this a typo, or a different situation with a different speed.

To find the number of rotations. You are being asked to find the (angular) distance travelled.
Do you remember this,

v² - u² = 2as ??

What do you think is its equivalent in circular motion ?

Use that.

6. Nov 30, 2005

### suspenc3

its suppose to be increased to 1000 rpm

isnt it $$\omega_f - \omega_i$$

7. Nov 30, 2005

### suspenc3

What do the variables stand for...i.e - u, a, s, v?

8. Nov 30, 2005

### Fermat

u = initial (linear) speed
v = final (linear) speed
a = (linear) acceleration
s = (linear) distance travelled

Can you now transpose that (linear) eqn into its circular/rotational equivalent ?

9. Nov 30, 2005

### suspenc3

so:

$$\frac{ \Delta \omega}{2 \alpha} = d$$

10. Nov 30, 2005

### Fermat

Almost there, but not quite :(

$$\mbox{linear: } v^2 - u^2 = 2as$$
$$\mbox{circular: } \omega _f^2 - \omega _i^2 = 2\alpha\theta$$

$$\omega _f^2$$ is the final angular velocity
$$\omega _i^2$$ is the initial angular velocity
$$\alpha$$ is the angular acceleration
$$\theta$$ is the rotational displacement

Last edited: Nov 30, 2005
11. Nov 30, 2005

### suspenc3

Ok so (2000^2 - 400^2)/2(1600) = $$\theta$$

= 1200

circumferance of wheel = 2(pi)r = 3.77m

1200m/3.77m = 318revs?

12. Nov 30, 2005

### Fermat

You are working in revs/min and revs/min² for velocity and acceleration so your displacement (theta) will be in revs.

i.e. theta = 1200 revs (rotations)