Angular Acceleration Of A Wheel

1. Oct 16, 2009

Warmacblu

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

A wheel rotating with a constant angular acceleration turns through 22 revolutions during a 9 s time interval. Its angular velocity at the end of this interval is 15 rad/s.

What is the angular acceleration of the wheel? Note that the initial angular velocity is not zero.

2. Relevant equations

w = omega
a = alpha

wf = wi + at
theta = wit + 1/2at2

3. The attempt at a solution

wf = 15
wi = ?
a = ?
theta = 2pi * 22 = 128.23
t = 9

So I have two unknowns and two equations but I can't seem to solve for the two equations. Both equations have two unknowns in them which is throwing me off. Any help is appreciated.

Thanks

2. Oct 16, 2009

rl.bhat

The first equation can be written as
wf*t = wi*t + a*t^2and subtract it from the second equation.wi*t get canceled out.Then solve for a.

3. Oct 16, 2009

Warmacblu

How does this look?

wft = wit + at2
theta = wit + 1/2at2

wft - theta = at2 - 1/2at2

wft - theta / t2 = a - 1/2at2

wft - theta / t4 = a - 1/2a

2wft - theta / t4 = a

Thanks for the help.

4. Oct 16, 2009

rl.bhat

wft - theta = at2 - 1/2at2

wft - theta / t2 = a - 1/2at2
This step is wrong. It should be
wft - theta = 1/2*at2

5. Oct 16, 2009

Warmacblu

How come the at2 can just be dropped?

wft - theta = 1/2*at2

wft - theta / t2 = 1/2a

2wft - theta / t2 = a

6. Oct 16, 2009

rl.bhat

It should be
2[wft - theta] / t2 = a[

7. Oct 18, 2009

Warmacblu

Here is what I got:

2((15*9)-138.23) / 92 = -.0798

Does that mean that the disc has a negative acceleration?

Also, I still do not understand how you just dropped the at2:

wft - theta = at2 - 1/2at2

wft - theta / t2 = a - 1/2at2
This step is wrong. It should be
wft - theta = 1/2*at2

8. Oct 18, 2009

rl.bhat

Also, I still do not understand how you just dropped the at^2
I am not dropping at^2. It is a simple algebra.
wft-θ = at^2 - 1/2*at^2
= 2*1/2*at^2 - 1/2at^2....= ? [Just like (2x - x) = ?]

9. Oct 18, 2009

Warmacblu

Ah, okay. I see it now.

I am unsure if my answer is in rad/sec2 which the question specifies for. I multiplied 15 rad/s times 9 sec - 138.23 rad all divided by 9 sec. The terms seem correct but the negative is still throwing me off.

10. Oct 18, 2009

rl.bhat

See whether the angular velocity increasing or decreasing. If it is decreasing then the acceleration is negative.

11. Oct 18, 2009

Warmacblu

22 rev / 9 sec = 2.44 rev/sec = 15.36 rad / sec. At the end of the time interval it is at 15 rad/s, so I guess the acceleration is negative.