# Angular deceleration

1. Mar 25, 2007

### xXmarkXx

1. The problem statement, all variables and given/known data

A circular saw blade completes 1200 revolutions in 40s while coasting to a stop after being turned off. Assuming constant deceleration, what are the angular deceleration, and the initial angular speed??

2. Relevant equations
v=wr
w^2=w0^2 + 2alpha*delta(theta)

i can't figure out how to start this problem off.

2. Mar 25, 2007

### turdferguson

You need 2 kinematics that have time as a variable

3. Mar 25, 2007

### xXmarkXx

yeah...could you possibly elaborate for me???

4. Mar 25, 2007

### turdferguson

What kinematic equations do you know?

5. Mar 25, 2007

### xXmarkXx

all of them...and they are in my book too. But i think i solved the initial angular speed, but i don't know how to find the angular deceleration...without using time.

6. Mar 25, 2007

### turdferguson

Time is given, why wouldnt you want to use it???

7. Mar 25, 2007

### Werg22

You need to know how long it takes for it to come to a stop after turning it off.

8. Mar 25, 2007

### xXmarkXx

isn't that 40s then?

9. Mar 25, 2007

### Werg22

Without the initial speed, you can't calculate the angular acceleration. Are you sure you don't have any other information?

10. Mar 25, 2007

### xXmarkXx

which is 188rad/sec i think. I took (1800revs/min)(2pie rad/rev)(60s/min) to get 188rad/sec. Isn't that the initial angular speed/velocity???

11. Mar 25, 2007

### xXmarkXx

that is the entire problem.

12. Mar 25, 2007

### Werg22

... 1800revs/min, you never mentioned that before. Is it in the problem?

13. Mar 25, 2007

### xXmarkXx

no, i figured it out...i just showed you the work.

14. Mar 25, 2007

### Werg22

Figured it out??? The disk comes to a stop in 40 sec. This means that the velocity constantly decelerates during those 40 seconds, and 1200/40 is just the average angular velocity during those 40 seconds. This has nothing to do with the actual initial angular velocity.

15. Mar 25, 2007

### xXmarkXx

ok, then do you have any ideas???

16. Mar 25, 2007

### hage567

This is how I would approach this question:

You don't need to know the initial angular speed. What you need to do is find two kinematic equations that link the things you know and the things you don't. You want angular acceleration (alpha). You have the time it takes the blade to stop, the number of revolutions (and therefore the angular displacement) it goes before it stops, and you know the final angular speed (it's zero). You can put the initial angular speed in terms of the other variables so you don't need to know it directly. Put the two equations together, and solve for alpha.

17. Mar 25, 2007

### xXmarkXx

the question asks for initial angular speed also...but i kind of know where your going with this.

18. Mar 25, 2007

### xXmarkXx

i don't quite get the angular displacement...if it spends 1200 times...its going to end up in the same place it was spinning from. So technically the dispacement is 0 right??

19. Mar 25, 2007

### hage567

OK, maybe think of it as angular distance. The point is, you can find out how many radians it goes through while it is decelerating to rest.

Once you find alpha, you can solve for the initial angular speed.

20. Mar 25, 2007

### Werg22

hage567, I believe you're wrong. Angular velocity and acceleration are the equivalent of linear acceleration and velocity along the circumference of the unit circle. This means that if we represent the angle covered by a revolving body on a straight line, we would have case of linear acceleration. For example, if we have the number of revolutions per sec before deceleration, $$k$$ we would have an angular velocity of $$v = 2{\pi}k$$. If the body came to a stop in 40 sec, the angle covered is $$1400*2{\pi}$$

The angular acceleration, $$a$$, would then depend on of this velocity since we have by linear acceleration,

$$1400*2{\pi} = \frac{1}{2}a 40^{2} + v*40$$

Obviously we see that the acceleration is dependent on the choice of the initial velocity.