# Angular motion using Newton's Laws

1. Jun 19, 2007

### Sparky500

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

A flywheel takes 5 seconds to rotate through 5 revolutions starting from rest. If it is accelerating at a constant rate during that time, what is the angular acceleration?

2. Relevant equations

∅=ω t+at^2

3. The attempt at a solution

having done my calculations i have concluded that the angular acceleration is zero, please can someone confirm this?

Cheers

2. Jun 19, 2007

### malawi_glenn

how did you perform your calculations?

3. Jun 19, 2007

### Sparky500

i inserted the numbers into the equation above and in this resulted in:

0 = 12.5 a

0/12.5 = a

therefore a = 0

4. Jun 19, 2007

### malawi_glenn

If the weel starts from rest, the angular velocity = 0. yes, but 5 revolutions is how many radians?..

If it during constant acceleration makes 5 revolutions in 5 seconds, how can the acceleration be zero then? There should ring a bell..

5. Jun 19, 2007

### malawi_glenn

$$\theta = \theta _{0} + \omega_{0} t + \frac{1}{2} \alpha t^{2}$$

6. Jun 19, 2007

### Sparky500

thats what i did think (hence the post), however does it make difference that its asking for angular acceleration?(sorry if this is a daft question, but due to me working on these numbers all day my head is in a bad state currently)

7. Jun 19, 2007

### malawi_glenn

But after 5 relevolutions, how many radians have you then covered?..

not 0...

The units for angular acceleration is [rad/(s^2)], and angular velocity [rad/s] , and of course angle [rad]

8. Jun 19, 2007

### D H

Staff Emeritus
1. If the flywheel makes 5 revolutions in 5 seconds starting from rest, how can the acceleration be zero?
2. How many radians is 5 revolutions. Hint: It is not zero.

9. Jun 19, 2007

### Sparky500

ok after acquainting my head with the wall to gather my thoughts i now have a new answer.. brace yourself

a= -6.28

10. Jun 19, 2007

### malawi_glenn

well it should not be negataive because the weel is increasing its velocity..

and your choice for angle covered in 5 revolutions was?

Last edited: Jun 19, 2007
11. Jun 19, 2007

### Sparky500

∅=ω t+at^2

∅=not sure, i thought it was 78.55, but since this gives a negative result then i guess its wrong.

ω t = 157.10

t^2 = 25

can you point me in the right direction to working out ∅ as my chain of thought is wrong

12. Jun 19, 2007

### malawi_glenn

well do you know how many radians ONE revolution is?

Your negative result does not come from that. And I only see a "
Square = not sure"
Can you write in latex?

Just plug in the total radians covered in 5 revs, in the forumla:

theta = 0.5*alpha*t^2

13. Jun 19, 2007

### Sparky500

[ tex ] theta = ω t+at^2 [ /tex ]

testing the latex

14. Jun 19, 2007

### malawi_glenn

15. Jun 19, 2007

### malawi_glenn

But do you know how many radians ONE revolution is?

16. Jun 19, 2007

### Sparky500

1 rev = 2 pi rad

17. Jun 19, 2007

### malawi_glenn

good.

5 rev = 10pi

10 pi = (1/2) * alpha * 5^2

20 pi / 25 = a = ?

18. Jun 19, 2007

### Sparky500

$$\theta = \omega t + {1/2} \alpha t^{2}$$

19. Jun 19, 2007

### Sparky500

i cant relate this to me resolving theta?

20. Jun 19, 2007

### malawi_glenn

You were asked to calculate the angular acceleration, given that the weel started from rest and accelerated constant so it fulfilled 5 revs in 5 sec...