Angular Acceleration Of A Wheel

AI Thread Summary
A wheel with a constant angular acceleration completes 22 revolutions in 9 seconds, ending with an angular velocity of 15 rad/s. The initial angular velocity is unknown, complicating the calculation of angular acceleration. The discussion involves algebraic manipulation of the equations relating angular velocity, angular displacement, and time to isolate the angular acceleration. The calculations suggest that the wheel has a negative angular acceleration, indicating a decrease in angular velocity over time. Participants clarify algebraic steps and confirm the units for angular acceleration as rad/sec².
Warmacblu
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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
 
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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.
 
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.
 
wft - theta = at2 - 1/2at2

wft - theta / t2 = a - 1/2at2
This step is wrong. It should be
wft - theta = 1/2*at2
 
rl.bhat said:
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
 
Warmacblu said:
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[
 
rl.bhat said:
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
 
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) = ?]
 
rl.bhat said:
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
See whether the angular velocity increasing or decreasing. If it is decreasing then the acceleration is negative.
 
  • #11
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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