Angular Acceleration Of A Wheel

  • Thread starter Warmacblu
  • Start date
  • #1
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Homework Statement



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.

Homework Equations



w = omega
a = alpha

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

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
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
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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
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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
rl.bhat
Homework Helper
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wft - theta = at2 - 1/2at2

wft - theta / t2 = a - 1/2at2
This step is wrong. It should be
wft - theta = 1/2*at2
 
  • #5
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wft - theta = at2 - 1/2at2

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

How come the at2 can just be dropped?

Following your correction:

wft - theta = 1/2*at2

wft - theta / t2 = 1/2a

2wft - theta / t2 = a
 
  • #6
rl.bhat
Homework Helper
4,433
8
How come the at2 can just be dropped?

Following your correction:

wft - theta = 1/2*at2

wft - theta / t2 = 1/2a

2wft - theta / t2 = a

It should be
2[wft - theta] / t2 = a[
 
  • #7
103
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It should be
2[wft - theta] / t2 = a[

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
rl.bhat
Homework Helper
4,433
8
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
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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) = ?]

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
rl.bhat
Homework Helper
4,433
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See whether the angular velocity increasing or decreasing. If it is decreasing then the acceleration is negative.
 
  • #11
103
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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.
 

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